[PHP-DB] Retrieving Image Location in MySQL
Dear All, I am trying to retrieve the location of a image (not the image, just the location of the image) stored on MySQL database, using PHP to display it in my browser. One of the users has been kind enough to provide me with an example code as to how to do it. He asked me to use /src="picture.php?&qry=123"> /in the html code. I understood the part of /picture.php? /which tells us that the php code used to retrieve the image is located in the file picture.php (If what I have understood is correct). But I did not quite understand the second part /&qry=123 /. What does this mean? Of what use is this second part? Does the variable &qry has something assigned to it in the picture.php code? I would greatly appreciate it if any one of you can answer my questions. Thanks, Sashi -- ~ ~ Sashikanth Gurram Graduate Research Assistant Department of Civil and Environmental Engineering Virginia Tech Blacksburg, VA 24060, USA -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Retrieving Image Location in MySQL
Sashi, This (likely) means you have a some generic page (i.e., picture.php) that displays some picture. The picture it displays depends on the parameter passed when the page is called (i.e., 123). Sashi's Test Page ", $row['firstname']); } //Clean up mysql_free_result($output); ?> Please note, I haven't tried this. It just seems plausible. Good luck. A- -Original Message- From: Sashikanth Gurram [mailto:sashi...@vt.edu] Sent: Sunday, March 01, 2009 10:27 PM To: php-db@lists.php.net Subject: [PHP-DB] Retrieving Image Location in MySQL Dear All, I am trying to retrieve the location of a image (not the image, just the location of the image) stored on MySQL database, using PHP to display it in my browser. One of the users has been kind enough to provide me with an example code as to how to do it. He asked me to use / /in the html code. I understood the part of /picture.php? /which tells us that the php code used to retrieve the image is located in the file picture.php (If what I have understood is correct). But I did not quite understand the second part /&qry=123 /. What does this mean? Of what use is this second part? Does the variable &qry has something assigned to it in the picture.php code? I would greatly appreciate it if any one of you can answer my questions. Thanks, Sashi -- ~ ~ Sashikanth Gurram Graduate Research Assistant Department of Civil and Environmental Engineering Virginia Tech Blacksburg, VA 24060, USA -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Retrieving Image Location in MySQL
Fortuno, Adam írta: //Write a query to pull out the picture's path $sql = "SELECT path FROM Image WHERE ID = %s"; mysql_real_escape_string($value); Sorry, but this won't work, since you don't map the value of the escaped $value to the %s, lets say with sprintf()... Regards, Matya -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Retrieving Image Location in MySQL
Matya, Ha, ha, ha! Thank you good friend. I did say I didn't try the code :-) AF> Please note, I haven't tried this. It just seems plausible. I apologize if I confused anyone. I just meant to show how the parameter could help retrieve a picture. I wasn't too concerned with the particulars. Hopefully the concept is sound. If not, please do flame my note so no one attempts it. Be Well, A- -Original Message- From: Mattyasovszky Janos [mailto:m...@matya.hu] Sent: Monday, March 02, 2009 9:29 AM To: php-db@lists.php.net Subject: Re: [PHP-DB] Retrieving Image Location in MySQL Fortuno, Adam írta: > //Write a query to pull out the picture's path > $sql = "SELECT path FROM Image WHERE ID = %s"; > mysql_real_escape_string($value); Sorry, but this won't work, since you don't map the value of the escaped $value to the %s, lets say with sprintf()... Regards, Matya -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Retrieving Image Location in MySQL
Hello guys, Thanks to you all for your kind replies. I will try the steps and get back to you if I encounter more problems. Thanks, Sashi Fortuno, Adam wrote: Matya, Ha, ha, ha! Thank you good friend. I did say I didn't try the code :-) AF> Please note, I haven't tried this. It just seems plausible. I apologize if I confused anyone. I just meant to show how the parameter could help retrieve a picture. I wasn't too concerned with the particulars. Hopefully the concept is sound. If not, please do flame my note so no one attempts it. Be Well, A- -Original Message- From: Mattyasovszky Janos [mailto:m...@matya.hu] Sent: Monday, March 02, 2009 9:29 AM To: php-db@lists.php.net Subject: Re: [PHP-DB] Retrieving Image Location in MySQL Fortuno, Adam írta: //Write a query to pull out the picture's path $sql = "SELECT path FROM Image WHERE ID = %s"; mysql_real_escape_string($value); Sorry, but this won't work, since you don't map the value of the escaped $value to the %s, lets say with sprintf()... Regards, Matya -- ~ ~ Sashikanth Gurram Graduate Research Assistant Department of Civil and Environmental Engineering Virginia Tech Blacksburg, VA 24060, USA -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Retrieving Image Location in MySQL
Dear all,
I have been trying to retrieve the location of a image from database and
display the image in the browser using PHP. I have written a sort of
code for the purpose. All I am getting in my browser after using the
code is a long set of ASCII characters with the following warning
*Warning*: Cannot modify header information - headers already sent by
(output started at C:\wamp\www\mysqli.php:65)
in *C:\wamp\www\mysqli.php* on line *237
*
I am herewith attaching my code. If anyone can point out the error or
give some kind of advice that would be really great. The location of the
image is stored in a table by the name IMAGE under the column name
Location and here in the code, the location is retrieved under the
varibale name $Location.
thanks,
Sashi
/
Building Name:
Select a Building
Agnew Hall
Rector Field House
Richard B. Talbot Educational Resources Center
Robeson Hall
Sandy Hall
Saunders hall
Seitz Hall
Shanks Hall
Shultz Hall
Skelton Conference
Center
Williams Hall
Women's Softball Field
Wright House
=00 && $minute<=14)
{
$minute=00;
}
elseif ($minute>=15 && $minute<=29)
{
$minute=15;
}
elseif ($minute>=30 && $minute<=44)
{
$minute=30;
}
else
{
$minute=45;
}
$times="$hour:$minute";
$sql="SELECT buildingname, parking_lot_name, empty_spaces, distance,
round(distance/($low_speed*60),1) AS low_time,
round(distance/($high_speed*60),1) AS high_time, Location FROM
buildings, buildings_lots, parkinglots, occupancy2, Image where
(buildings.buildingcode=occupancy2.building AND
buildings.buildingcode=buildings_lots.building_code AND
parkinglots.parking_lot_code=buildings_lots.parking_lot_code AND
parkinglots.parking_lot_code=occupancy2.parking_lot AND
Buildings.BuildingCode=Image.BuildingCode) AND buildingname='$build' AND
month='$Month' AND day='$Today' AND Time='$times'";
$data = mysqli_query($cxn,$sql);
if (!$data=mysqli_query($cxn,$sql))
{
$error=mysqli_error($cxn);
echo "$error";
die();
}
else
{
echo "";
echo "Query sent successfully";
}
echo "";
echo " PARKING LOT INFORMATION ";
echo "";
echo "\n
Building\n
Parking Lot\n
Estimated Number of Empty Spaces\n
Distance (Feet)\n
Estimated walking time to the building\n
\n";
while ($row=mysqli_fetch_array($data))
{
extract($row);
$building = $row[0];
$parking_lot = $row[1];
$Number_of_Empty_Spaces = $row[2];
$Distance = $row[3];
$time_l = $row[4];
$time_h=$row[5];
$location=$row[6];
echo "\n
$building\n
$parking_lot\n
$Number_of_Empty_Spaces\n
$Distance\n
$time_h - $time_l mins\n
$location\n
\n";
}
echo "\n";
$err=1;
if ($img = file_get_contents($location, FILE_BINARY))
{
if ($img = imagecreatefromstring($img)) $err = 0;
}
if ($err)
{
header('Content-Type: text/html');
echo 'Error getting
image...';
}
else
{
header('Content-Type: image/jpeg');
imagejpeg($img);
imagedestroy($img);
}
?>
/
Sashikanth Gurram wrote:
Hello guys,
Thanks to you all for your kind replies. I will try the steps and get
back to you if I encounter more problems.
Thanks,
Sashi
Fortuno, Adam wrote:
Matya,
Ha, ha, ha! Thank you good friend. I did say I didn't try the code :-)
AF> Please note, I haven't tried this. It just seems plausible.
I apologize if I confused anyone. I just meant to show how the
parameter could help retrieve a picture. I wasn't too concerned with
the particulars. Hopefully the concept is sound. If not, please do
flame my note so no one attempts it.
Be Well,
A-
-Original Message-
From: Mattyasovszky Janos [mailto:m...@matya.hu] Sent: Monday, March
02, 2009 9:29 AM
To: php-db@lists.php.net
Subject: Re: [PHP-DB] Retrieving Image Location in MySQL
Fortuno, Adam írta:
//Write a query to pull out the picture's path
$sql = "SELECT path FROM Image WHERE ID = %s";
mysql_real_escape_string($value);
Sorry, but this won't work, since you don't map the value of the
escaped $value to the %s, lets say with sprintf()...
Regards,
Matya
--
~
~
Sashikanth Gurram
Graduate Research Assistant
Department of Civil and Environmental Engineering
Virginia Tech
Blacksburg, VA 24060, USA
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Retrieving Image Location in MySQL
Hi Sashi:
You're incurring in this error because the headers for the page have
already been issued in the head section of your page.
When you issue the php header() function, it will generate an error.
Anyway, you are storing a path to a file in the database. My question
is: do you want to put your images on a subfolder of your site, where
they will become acessible through regular URLS, or you want to keep
them private and only accessible by scripting?
Cheers.
JP
-Original Message-
From: Sashikanth Gurram
To: php-db@lists.php.net
Subject: Re: [PHP-DB] Retrieving Image Location in MySQL
Date: Fri, 06 Mar 2009 12:23:31 -0500
Dear all,
I have been trying to retrieve the location of a image from database and
display the image in the browser using PHP. I have written a sort of
code for the purpose. All I am getting in my browser after using the
code is a long set of ASCII characters with the following warning
*Warning*: Cannot modify header information - headers already sent by
(output started at C:\wamp\www\mysqli.php:65)
in *C:\wamp\www\mysqli.php* on line *237
*
I am herewith attaching my code. If anyone can point out the error or
give some kind of advice that would be really great. The location of the
image is stored in a table by the name IMAGE under the column name
Location and here in the code, the location is retrieved under the
varibale name $Location.
thanks,
Sashi
/
Building Name:
Select a Building
Agnew Hall
Rector Field House
Richard B. Talbot Educational Resources Center
Robeson Hall
Sandy Hall
Saunders hall
Seitz Hall
Shanks Hall
Shultz Hall
Skelton Conference
Center
Williams Hall
Women's Softball Field
Wright House
=00 && $minute<=14)
{
$minute=00;
}
elseif ($minute>=15 && $minute<=29)
{
$minute=15;
}
elseif ($minute>=30 && $minute<=44)
{
$minute=30;
}
else
{
$minute=45;
}
$times="$hour:$minute";
$sql="SELECT buildingname, parking_lot_name, empty_spaces, distance,
round(distance/($low_speed*60),1) AS low_time,
round(distance/($high_speed*60),1) AS high_time, Location FROM
buildings, buildings_lots, parkinglots, occupancy2, Image where
(buildings.buildingcode=occupancy2.building AND
buildings.buildingcode=buildings_lots.building_code AND
parkinglots.parking_lot_code=buildings_lots.parking_lot_code AND
parkinglots.parking_lot_code=occupancy2.parking_lot AND
Buildings.BuildingCode=Image.BuildingCode) AND buildingname='$build' AND
month='$Month' AND day='$Today' AND Time='$times'";
$data = mysqli_query($cxn,$sql);
if (!$data=mysqli_query($cxn,$sql))
{
$error=mysqli_error($cxn);
echo "$error";
die();
}
else
{
echo "";
echo "Query sent successfully";
}
echo "";
echo " PARKING LOT INFORMATION ";
echo "";
echo "\n
Building\n
Parking Lot\n
Estimated Number of Empty Spaces\n
Distance (Feet)\n
Estimated walking time to the building\n
\n";
while ($row=mysqli_fetch_array($data))
{
extract($row);
$building = $row[0];
$parking_lot = $row[1];
$Number_of_Empty_Spaces = $row[2];
$Distance = $row[3];
$time_l = $row[4];
$time_h=$row[5];
$location=$row[6];
echo "\n
$building\n
$parking_lot\n
$Number_of_Empty_Spaces\n
$Distance\n
$time_h - $time_l mins\n
$location\n
\n";
}
echo "\n";
$err=1;
if ($img = file_get_contents($location, FILE_BINARY))
{
if ($img = imagecreatefromstring($img)) $err = 0;
}
if ($err)
{
header('Content-Type: text/html');
echo 'Error getting
image...';
}
else
{
header('Content-Type: image/jpeg');
imagejpeg($img);
imagedestroy($img);
}
?>
/
Sashikanth Gurram wrote:
> Hello guys,
>
> Thanks to you all for your kind replies. I will try the steps and get
> back to you if I encounter more problems.
>
> Thanks,
> Sashi
>
> Fortuno, Adam wrote:
>> Matya,
>>
>> Ha, ha, ha! Thank you good friend. I did say I didn't try the code :-)
>>
>> AF> Please note, I haven't tried this. It just seems plausible.
>>
>> I apologize if I confused anyone. I just meant to show how the
>> parameter could help retrieve a picture. I wasn't too concerned with
>> the particulars. Hopefully the concept is sound. If not, please do
>> flame my note so no one attempts it.
>>
>> Be Well,
>> A-
>>
>> -Original Message-
>> From: Mattyasovszky Janos [mailto:m...@matya.hu] Sent: Monday, March
>&
Re: [PHP-DB] Retrieving Image Location in MySQL
On Fri, Mar 6, 2009 at 1:06 PM, Joao Gomes Madeira wrote:
> Hi Sashi:
>
> You're incurring in this error because the headers for the page have
> already been issued in the head section of your page.
> When you issue the php header() function, it will generate an error.
>
> Anyway, you are storing a path to a file in the database. My question
> is: do you want to put your images on a subfolder of your site, where
> they will become acessible through regular URLS, or you want to keep
> them private and only accessible by scripting?
>
> Cheers.
> JP
>
> -Original Message-
> From: Sashikanth Gurram
> To: php-db@lists.php.net
> Subject: Re: [PHP-DB] Retrieving Image Location in MySQL
> Date: Fri, 06 Mar 2009 12:23:31 -0500
>
> Dear all,
>
> I have been trying to retrieve the location of a image from database and
> display the image in the browser using PHP. I have written a sort of
> code for the purpose. All I am getting in my browser after using the
> code is a long set of ASCII characters with the following warning
> *Warning*: Cannot modify header information - headers already sent by
> (output started at C:\wamp\www\mysqli.php:65)
> in *C:\wamp\www\mysqli.php* on line *237
> *
> I am herewith attaching my code. If anyone can point out the error or
> give some kind of advice that would be really great. The location of the
> image is stored in a table by the name IMAGE under the column name
> Location and here in the code, the location is retrieved under the
> varibale name $Location.
>
> thanks,
> Sashi
>
> /
>
>
>
>
> Building Name:
> Select a Building
> Agnew Hall
> Rector Field House
> Richard B. Talbot Educational Resources Center
> Robeson Hall
> Sandy Hall
> Saunders hall
> Seitz Hall
> Shanks Hall
> Shultz Hall
> Skelton Conference
> Center
> Williams Hall
> Women's Softball Field
> Wright House
>
>
>
>
>
>
> // Connects to your Database
> $host="***";
> $user="*";
> $password="**";
> $dbname="***";
> $cxn=mysqli_connect($host, $user, $password, $dbname) ;
> if (!$cxn=mysqli_connect($host, $user, $password, $dbname))
>{
>$error=mysqli_error($cxn);
>echo "$error";
>die();
>}
> else
>{
>echo "Connection established successfully";
>}
> //Define the variables for Day and Month
> $Today=date("l");
> $Month=date("F");
> $build=$_POST["name"];
> $low_speed=2.5;
> $high_speed=4;
> $hour=date("G");
> $minute=date("i");
> if ($minute>=00 && $minute<=14)
>{
>$minute=00;
>}
> elseif ($minute>=15 && $minute<=29)
>{
>$minute=15;
>}
> elseif ($minute>=30 && $minute<=44)
>{
>$minute=30;
>}
> else
>{
>$minute=45;
>}
> $times="$hour:$minute";
> $sql="SELECT buildingname, parking_lot_name, empty_spaces, distance,
> round(distance/($low_speed*60),1) AS low_time,
> round(distance/($high_speed*60),1) AS high_time, Location FROM
> buildings, buildings_lots, parkinglots, occupancy2, Image where
> (buildings.buildingcode=occupancy2.building AND
> buildings.buildingcode=buildings_lots.building_code AND
> parkinglots.parking_lot_code=buildings_lots.parking_lot_code AND
> parkinglots.parking_lot_code=occupancy2.parking_lot AND
> Buildings.BuildingCode=Image.BuildingCode) AND buildingname='$build' AND
> month='$Month' AND day='$Today' AND Time='$times'";
> $data = mysqli_query($cxn,$sql);
> if (!$data=mysqli_query($cxn,$sql))
>{
>$error=mysqli_error($cxn);
>echo "$error";
>die();
>}
> else
>{
>echo "";
>echo "Query sent successfully";
>}
> echo "";
> echo " PARKING LOT INFORMATION ";
> echo "";
> echo "\n
>Building\n
>Parking Lot\n
>Estimated Number of Empty Spaces\n
>Distance (Feet)\n
>Estimated walking time to the building\n
>\n";
> while ($row=mysqli_fetch_array($data))
>{
>extract($row);
> $building = $row[0];
> $parking_lot = $row[1];
> $Number_of_Empty_Spaces = $row[2];
> $Distance = $row[3];
> $time_l = $row[4];
> $time_h=$row[5];
> $location=$row[6];
> echo "\n
> $building\n
> $parking_lot\
Re: [PHP-DB] Retrieving Image Location in MySQL
Hi JP,
I am just storing the location of the image on the database. I want to
keep them private and retrieve them through scripting. The images are
stored in a folder on my PC. Just to give you an example, this is the
location of a particular image as stored in the database.
C:\Users\Sashikanth\Desktop\Bldgs_lots\Burruss.jpg
Thanks,
Sashi
Joao Gomes Madeira wrote:
Hi Sashi:
You're incurring in this error because the headers for the page have
already been issued in the head section of your page.
When you issue the php header() function, it will generate an error.
Anyway, you are storing a path to a file in the database. My question
is: do you want to put your images on a subfolder of your site, where
they will become acessible through regular URLS, or you want to keep
them private and only accessible by scripting?
Cheers.
JP
-Original Message-
From: Sashikanth Gurram
To: php-db@lists.php.net
Subject: Re: [PHP-DB] Retrieving Image Location in MySQL
Date: Fri, 06 Mar 2009 12:23:31 -0500
Dear all,
I have been trying to retrieve the location of a image from database and
display the image in the browser using PHP. I have written a sort of
code for the purpose. All I am getting in my browser after using the
code is a long set of ASCII characters with the following warning
*Warning*: Cannot modify header information - headers already sent by
(output started at C:\wamp\www\mysqli.php:65)
in *C:\wamp\www\mysqli.php* on line *237
*
I am herewith attaching my code. If anyone can point out the error or
give some kind of advice that would be really great. The location of the
image is stored in a table by the name IMAGE under the column name
Location and here in the code, the location is retrieved under the
varibale name $Location.
thanks,
Sashi
/
Building Name:
Select a Building
Agnew Hall
Rector Field House
Richard B. Talbot Educational Resources Center
Robeson Hall
Sandy Hall
Saunders hall
Seitz Hall
Shanks Hall
Shultz Hall
Skelton Conference
Center
Williams Hall
Women's Softball Field
Wright House
=00 && $minute<=14)
{
$minute=00;
}
elseif ($minute>=15 && $minute<=29)
{
$minute=15;
}
elseif ($minute>=30 && $minute<=44)
{
$minute=30;
}
else
{
$minute=45;
}
$times="$hour:$minute";
$sql="SELECT buildingname, parking_lot_name, empty_spaces, distance,
round(distance/($low_speed*60),1) AS low_time,
round(distance/($high_speed*60),1) AS high_time, Location FROM
buildings, buildings_lots, parkinglots, occupancy2, Image where
(buildings.buildingcode=occupancy2.building AND
buildings.buildingcode=buildings_lots.building_code AND
parkinglots.parking_lot_code=buildings_lots.parking_lot_code AND
parkinglots.parking_lot_code=occupancy2.parking_lot AND
Buildings.BuildingCode=Image.BuildingCode) AND buildingname='$build' AND
month='$Month' AND day='$Today' AND Time='$times'";
$data = mysqli_query($cxn,$sql);
if (!$data=mysqli_query($cxn,$sql))
{
$error=mysqli_error($cxn);
echo "$error";
die();
}
else
{
echo "";
echo "Query sent successfully";
}
echo "";
echo " PARKING LOT INFORMATION ";
echo "";
echo "\n
Building\n
Parking Lot\n
Estimated Number of Empty Spaces\n
Distance (Feet)\n
Estimated walking time to the building\n
\n";
while ($row=mysqli_fetch_array($data))
{
extract($row);
$building = $row[0];
$parking_lot = $row[1];
$Number_of_Empty_Spaces = $row[2];
$Distance = $row[3];
$time_l = $row[4];
$time_h=$row[5];
$location=$row[6];
echo "\n
$building\n
$parking_lot\n
$Number_of_Empty_Spaces\n
$Distance\n
$time_h - $time_l mins\n
$location\n
\n";
}
echo "\n";
$err=1;
if ($img = file_get_contents($location, FILE_BINARY))
{
if ($img = imagecreatefromstring($img)) $err = 0;
}
if ($err)
{
header('Content-Type: text/html');
echo 'Error getting
image...';
}
else
{
header('Content-Type: image/jpeg');
imagejpeg($img);
imagedestroy($img);
}
?>
/
Sashikanth Gurram wrote:
Hello guys,
Thanks to you all for your kind replies. I will try the steps and get
back to you if I encounter more problems.
Thanks,
Sashi
Fortuno, Adam wrote:
Matya,
Ha, ha, ha! Thank you good friend. I did say I didn't try the code :-)
AF> Please note, I haven't tried this. It just seems plausible.
I apologize if I confused anyone. I just meant to show how the
parameter could help retrieve a picture. I wasn't too concerned w
Re: [PHP-DB] Retrieving Image Location in MySQL
echo 'Error getting
> image...';
> }
> else
> {
> header('Content-Type: image/jpeg');
> imagejpeg($img);
> imagedestroy($img);
> }
> ?>
>
> /
>
> Sashikanth Gurram wrote:
>> Hello guys,
>>
>> Thanks to you all for your kind replies. I will try the steps and get
>> back to you if I encounter more problems.
>>
>> Thanks,
>> Sashi
>>
>> Fortuno, Adam wrote:
>>> Matya,
>>>
>>> Ha, ha, ha! Thank you good friend. I did say I didn't try the code :-)
>>>
>>> AF> Please note, I haven't tried this. It just seems plausible.
>>>
>>> I apologize if I confused anyone. I just meant to show how the
>>> parameter could help retrieve a picture. I wasn't too concerned with
>>> the particulars. Hopefully the concept is sound. If not, please do
>>> flame my note so no one attempts it.
>>>
>>> Be Well,
>>> A-
>>>
>>> -Original Message-
>>> From: Mattyasovszky Janos [mailto:m...@matya.hu] Sent: Monday, March
>>> 02, 2009 9:29 AM
>>> To: php-db@lists.php.net
>>> Subject: Re: [PHP-DB] Retrieving Image Location in MySQL
>>>
>>> Fortuno, Adam írta:
>>>
>>>
>>>> //Write a query to pull out the picture's path
>>>> $sql = "SELECT path FROM Image WHERE ID = %s";
>>>> mysql_real_escape_string($value);
>>>>
>>>
>>> Sorry, but this won't work, since you don't map the value of the
>>> escaped $value to the %s, lets say with sprintf()...
>>>
>>> Regards,
>>> Matya
>>>
>>>
>>
>>
>
>
--
Extra details:
OSS:Gentoo Linux-2.6.25-r8
profile:x86
Hardware:msi geforce 8600GT asus p5k-se
location:/home/muhsin
language(s):C/C++,VB,VHDL,bash
Typo:40WPM
url:http://mambo-tech.net
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Re: [PHP-DB] Retrieving Image Location in MySQL
Hi,
Thanks a lot for the suggestion. It is working fine but there is some
problem. When I introduced the piece of code you have suggested into my
code, I am getting a space for the picture but nothing is there in that
space. It is just blank. I do not know the reason for this. Further,
this is the piece of code I have written to display the pic.
echo "alt='$build' >";
Any help would be greatly appreciated.
Thanks,
Sashi
mrfroasty wrote:
Hello,
Storing only location of the image in dB wouldnt be any different than
other string/data retrieved from the dB.That means if you have something
like this in html documents whereby $image_location comes from the dB
It should work without header stuffs
Example
P:S
Replace $image_location with image location from your dB.
GR
Muhsin
Sashikanth Gurram wrote:
Dear all,
I have been trying to retrieve the location of a image from database
and display the image in the browser using PHP. I have written a sort
of code for the purpose. All I am getting in my browser after using
the code is a long set of ASCII characters with the following warning
*Warning*: Cannot modify header information - headers already sent by
(output started at C:\wamp\www\mysqli.php:65) in
*C:\wamp\www\mysqli.php* on line *237
*
I am herewith attaching my code. If anyone can point out the error or
give some kind of advice that would be really great. The location of
the image is stored in a table by the name IMAGE under the column name
Location and here in the code, the location is retrieved under the
varibale name $Location.
thanks,
Sashi
/
Building Name:
Select a Building
Agnew Hall
Rector Field House
Richard B. Talbot Educational Resources Center
Robeson Hall
Sandy Hall
Saunders hall
Seitz Hall
Shanks Hall
Shultz Hall
Skelton Conference
Center
Williams Hall
Women's Softball
Field
Wright House
=00 && $minute<=14)
{
$minute=00;
}
elseif ($minute>=15 && $minute<=29)
{
$minute=15;
}
elseif ($minute>=30 && $minute<=44)
{
$minute=30;
}
else
{
$minute=45;
}
$times="$hour:$minute";
$sql="SELECT buildingname, parking_lot_name, empty_spaces, distance,
round(distance/($low_speed*60),1) AS low_time,
round(distance/($high_speed*60),1) AS high_time, Location FROM
buildings, buildings_lots, parkinglots, occupancy2, Image where
(buildings.buildingcode=occupancy2.building AND
buildings.buildingcode=buildings_lots.building_code AND
parkinglots.parking_lot_code=buildings_lots.parking_lot_code AND
parkinglots.parking_lot_code=occupancy2.parking_lot AND
Buildings.BuildingCode=Image.BuildingCode) AND buildingname='$build'
AND month='$Month' AND day='$Today' AND Time='$times'";
$data = mysqli_query($cxn,$sql);
if (!$data=mysqli_query($cxn,$sql))
{
$error=mysqli_error($cxn);
echo "$error";
die();
}
else
{
echo "";
echo "Query sent successfully";
}
echo "";
echo " PARKING LOT INFORMATION ";
echo "";
echo "\n
Building\n
Parking Lot\n
Estimated Number of Empty Spaces\n
Distance (Feet)\n
Estimated walking time to the building\n
\n";
while ($row=mysqli_fetch_array($data))
{
extract($row);
$building = $row[0];
$parking_lot = $row[1];
$Number_of_Empty_Spaces = $row[2];
$Distance = $row[3];
$time_l = $row[4];
$time_h=$row[5];
$location=$row[6]; echo "\n
$building\n
$parking_lot\n
$Number_of_Empty_Spaces\n
$Distance\n
$time_h - $time_l mins\n
$location\n
\n";
}
echo "\n";
$err=1;
if ($img = file_get_contents($location, FILE_BINARY))
{
if ($img = imagecreatefromstring($img)) $err = 0;
}
if ($err)
{
header('Content-Type: text/html');
echo 'Error getting
image...';
}
else
{
header('Content-Type: image/jpeg');
imagejpeg($img);
imagedestroy($img);
}
?>
/
Sashikanth Gurram wrote:
Hello guys,
Thanks to you all for your kind replies. I will try the steps and get
back to you if I encounter more problems.
Thanks,
Sashi
Fortuno, Adam wrote:
Matya,
Ha, ha, ha! Thank you good friend. I did say I didn't try the code :-)
AF> Please note, I haven't tried this. It just seems plausible.
I apologize if I confused anyone. I just meant to show how the
parameter could help retrieve a picture. I wasn't too concerned with
the particulars. Hopefully the concept is sound. If not, please do
flame my note so no one attempts it.
Be Well,
A-
-Original Message-
From: Mattyasovszky Janos [mailto:m...@matya.hu] Sent: Monday, March
02, 2009 9:29 AM
To: php-db@lists.php.net
Subject: Re:
Re: [PHP-DB] Retrieving Image Location in MySQL
ks,
Sashi
/
Building Name:
Select a Building
Agnew Hall
Rector Field House
Richard B. Talbot Educational Resources Center
Robeson Hall
Sandy Hall
Saunders hall
Seitz Hall
Shanks Hall
Shultz Hall
Skelton Conference
Center
Williams Hall
Women's Softball
Field
Wright House
=00 && $minute<=14)
{
$minute=00;
}
elseif ($minute>=15 && $minute<=29)
{
$minute=15;
}
elseif ($minute>=30 && $minute<=44)
{
$minute=30;
}
else
{
$minute=45;
}
$times="$hour:$minute";
$sql="SELECT buildingname, parking_lot_name, empty_spaces, distance,
round(distance/($low_speed*60),1) AS low_time,
round(distance/($high_speed*60),1) AS high_time, Location FROM
buildings, buildings_lots, parkinglots, occupancy2, Image where
(buildings.buildingcode=occupancy2.building AND
buildings.buildingcode=buildings_lots.building_code AND
parkinglots.parking_lot_code=buildings_lots.parking_lot_code AND
parkinglots.parking_lot_code=occupancy2.parking_lot AND
Buildings.BuildingCode=Image.BuildingCode) AND buildingname='$build'
AND month='$Month' AND day='$Today' AND Time='$times'";
$data = mysqli_query($cxn,$sql);
if (!$data=mysqli_query($cxn,$sql))
{
$error=mysqli_error($cxn);
echo "$error";
die();
}
else
{
echo "";
echo "Query sent successfully";
}
echo "";
echo " PARKING LOT INFORMATION ";
echo "";
echo "\n
Building\n
Parking Lot\n
Estimated Number of Empty Spaces\n
Distance (Feet)\n
Estimated walking time to the building\n
\n";
while ($row=mysqli_fetch_array($data))
{
extract($row);
$building = $row[0];
$parking_lot = $row[1];
$Number_of_Empty_Spaces = $row[2];
$Distance = $row[3];
$time_l = $row[4];
$time_h=$row[5];
$location=$row[6]; echo "\n
$building\n
$parking_lot\n
$Number_of_Empty_Spaces\n
$Distance\n
$time_h - $time_l mins\n
$location\n
\n";
}
echo "\n";
$err=1;
if ($img = file_get_contents($location, FILE_BINARY))
{
if ($img = imagecreatefromstring($img)) $err = 0;
}
if ($err)
{
header('Content-Type: text/html');
echo 'Error getting
image...';
}
else
{
header('Content-Type: image/jpeg');
imagejpeg($img);
imagedestroy($img);
}
?>
/
Sashikanth Gurram wrote:
Hello guys,
Thanks to you all for your kind replies. I will try the steps and get
back to you if I encounter more problems.
Thanks,
Sashi
Fortuno, Adam wrote:
Matya,
Ha, ha, ha! Thank you good friend. I did say I didn't try the code
:-)
AF> Please note, I haven't tried this. It just seems plausible.
I apologize if I confused anyone. I just meant to show how the
parameter could help retrieve a picture. I wasn't too concerned with
the particulars. Hopefully the concept is sound. If not, please do
flame my note so no one attempts it.
Be Well,
A-
-Original Message-
From: Mattyasovszky Janos [mailto:m...@matya.hu] Sent: Monday, March
02, 2009 9:29 AM
To: php-db@lists.php.net
Subject: Re: [PHP-DB] Retrieving Image Location in MySQL
Fortuno, Adam írta:
//Write a query to pull out the picture's path
$sql = "SELECT path FROM Image WHERE ID = %s";
mysql_real_escape_string($value);
Sorry, but this won't work, since you don't map the value of the
escaped $value to the %s, lets say with sprintf()...
Regards,
Matya
--
~
~
Sashikanth Gurram
Graduate Research Assistant
Department of Civil and Environmental Engineering
Virginia Tech
Blacksburg, VA 24060, USA
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Retrieving Image Location in MySQL
gt;>>>>{
>>>>>$minute=00;
>>>>>}
>>>>> elseif ($minute>=15 && $minute<=29)
>>>>>{
>>>>>$minute=15;
>>>>>}
>>>>> elseif ($minute>=30 && $minute<=44)
>>>>>{
>>>>>$minute=30;
>>>>>}
>>>>> else
>>>>>{
>>>>>$minute=45;
>>>>>}
>>>>> $times="$hour:$minute";
>>>>> $sql="SELECT buildingname, parking_lot_name, empty_spaces, distance,
>>>>> round(distance/($low_speed*60),1) AS low_time,
>>>>> round(distance/($high_speed*60),1) AS high_time, Location FROM
>>>>> buildings, buildings_lots, parkinglots, occupancy2, Image where
>>>>> (buildings.buildingcode=occupancy2.building AND
>>>>> buildings.buildingcode=buildings_lots.building_code AND
>>>>> parkinglots.parking_lot_code=buildings_lots.parking_lot_code AND
>>>>> parkinglots.parking_lot_code=occupancy2.parking_lot AND
>>>>> Buildings.BuildingCode=Image.BuildingCode) AND buildingname='$build'
>>>>> AND month='$Month' AND day='$Today' AND Time='$times'";
>>>>> $data = mysqli_query($cxn,$sql);
>>>>> if (!$data=mysqli_query($cxn,$sql))
>>>>>{
>>>>>$error=mysqli_error($cxn);
>>>>>echo "$error";
>>>>>die();
>>>>>}
>>>>> else
>>>>>{
>>>>>echo "";
>>>>>echo "Query sent successfully";
>>>>>}
>>>>> echo "";
>>>>> echo " PARKING LOT INFORMATION ";
>>>>> echo "";
>>>>> echo "\n
>>>>>Building\n
>>>>>Parking Lot\n
>>>>>Estimated Number of Empty Spaces\n
>>>>>Distance (Feet)\n
>>>>>Estimated walking time to the building\n
>>>>>\n";
>>>>> while ($row=mysqli_fetch_array($data))
>>>>>{
>>>>>extract($row);
>>>>> $building = $row[0];
>>>>> $parking_lot = $row[1];
>>>>> $Number_of_Empty_Spaces = $row[2];
>>>>> $Distance = $row[3];
>>>>> $time_l = $row[4];
>>>>> $time_h=$row[5];
>>>>> $location=$row[6]; echo "\n
>>>>> $building\n
>>>>> $parking_lot\n
>>>>> $Number_of_Empty_Spaces\n
>>>>> $Distance\n
>>>>> $time_h - $time_l mins\n
>>>>> $location\n
>>>>> \n";
>>>>>}
>>>>>echo "\n";
>>>>> $err=1;
>>>>> if ($img = file_get_contents($location, FILE_BINARY))
>>>>> {
>>>>> if ($img = imagecreatefromstring($img)) $err = 0;
>>>>> }
>>>>> if ($err)
>>>>> {
>>>>> header('Content-Type: text/html');
>>>>> echo 'Error getting
>>>>> image...';
>>>>> }
>>>>> else
>>>>> {
>>>>> header('Content-Type: image/jpeg');
>>>>> imagejpeg($img);
>>>>> imagedestroy($img);
>>>>> }
>>>>> ?>
>>>>>
>>>>> /
>>>>>
>>>>> Sashikanth Gurram wrote:
>>>>>
>>>>>> Hello guys,
>>>>>>
>>>>>> Thanks to you all for your kind replies. I will try the steps and
>>>>>> get
>>>>>> back to you if I encounter more problems.
>>>>>>
>>>>>> Thanks,
>>>>>> Sashi
>>>>>>
>>>>>> Fortuno, Adam wrote:
>>>>>>
>>>>>>> Matya,
>>>>>>>
>>>>>>> Ha, ha, ha! Thank you good friend. I did say I didn't try the code
>>>>>>> :-)
>>>>>>>
>>>>>>> AF> Please note, I haven't tried this. It just seems plausible.
>>>>>>>
>>>>>>> I apologize if I confused anyone. I just meant to show how the
>>>>>>> parameter could help retrieve a picture. I wasn't too concerned
>>>>>>> with
>>>>>>> the particulars. Hopefully the concept is sound. If not, please do
>>>>>>> flame my note so no one attempts it.
>>>>>>>
>>>>>>> Be Well,
>>>>>>> A-
>>>>>>>
>>>>>>> -Original Message-
>>>>>>> From: Mattyasovszky Janos [mailto:m...@matya.hu] Sent: Monday,
>>>>>>> March
>>>>>>> 02, 2009 9:29 AM
>>>>>>> To: php-db@lists.php.net
>>>>>>> Subject: Re: [PHP-DB] Retrieving Image Location in MySQL
>>>>>>>
>>>>>>> Fortuno, Adam írta:
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>> //Write a query to pull out the picture's path
>>>>>>>> $sql = "SELECT path FROM Image WHERE ID = %s";
>>>>>>>> mysql_real_escape_string($value);
>>>>>>>>
>>>>>>> Sorry, but this won't work, since you don't map the value of the
>>>>>>> escaped $value to the %s, lets say with sprintf()...
>>>>>>>
>>>>>>> Regards,
>>>>>>> Matya
>>>>>>>
>>>>>>>
>>>>>>
>>>>>
>>>>
>>>
>>>
>>>
>>>
>>>
>>
>>
>>
>
>
--
Extra details:
OSS:Gentoo Linux-2.6.25-r8
profile:x86
Hardware:msi geforce 8600GT asus p5k-se
location:/home/muhsin
language(s):C/C++,VB,VHDL,bash
Typo:40WPM
url:http://mambo-tech.net
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Retrieving Image Location in MySQL
ocation from your dB.
GR
Muhsin
Sashikanth Gurram wrote:
Dear all,
I have been trying to retrieve the location of a image from database
and display the image in the browser using PHP. I have written a sort
of code for the purpose. All I am getting in my browser after using
the code is a long set of ASCII characters with the following warning
*Warning*: Cannot modify header information - headers already sent by
(output started at C:\wamp\www\mysqli.php:65) in
*C:\wamp\www\mysqli.php* on line *237
*
I am herewith attaching my code. If anyone can point out the
error or
give some kind of advice that would be really great. The location of
the image is stored in a table by the name IMAGE under the column
name
Location and here in the code, the location is retrieved under the
varibale name $Location.
thanks,
Sashi
/
Building Name:
Select a Building
Agnew Hall
Rector Field House
Richard B. Talbot Educational Resources Center
Robeson Hall
Sandy Hall
Saunders hall
Seitz Hall
Shanks Hall
Shultz Hall
Skelton Conference
Center
Williams Hall
Women's Softball
Field
Wright House
=00 && $minute<=14)
{
$minute=00;
}
elseif ($minute>=15 && $minute<=29)
{
$minute=15;
}
elseif ($minute>=30 && $minute<=44)
{
$minute=30;
}
else
{
$minute=45;
}
$times="$hour:$minute";
$sql="SELECT buildingname, parking_lot_name, empty_spaces, distance,
round(distance/($low_speed*60),1) AS low_time,
round(distance/($high_speed*60),1) AS high_time, Location FROM
buildings, buildings_lots, parkinglots, occupancy2, Image where
(buildings.buildingcode=occupancy2.building AND
buildings.buildingcode=buildings_lots.building_code AND
parkinglots.parking_lot_code=buildings_lots.parking_lot_code AND
parkinglots.parking_lot_code=occupancy2.parking_lot AND
Buildings.BuildingCode=Image.BuildingCode) AND buildingname='$build'
AND month='$Month' AND day='$Today' AND Time='$times'";
$data = mysqli_query($cxn,$sql);
if (!$data=mysqli_query($cxn,$sql))
{
$error=mysqli_error($cxn);
echo "$error";
die();
}
else
{
echo "";
echo "Query sent successfully";
}
echo "";
echo " PARKING LOT INFORMATION ";
echo "";
echo "\n
Building\n
Parking Lot\n
Estimated Number of Empty Spaces\n
Distance (Feet)\n
Estimated walking time to the building\n
\n";
while ($row=mysqli_fetch_array($data))
{
extract($row);
$building = $row[0];
$parking_lot = $row[1];
$Number_of_Empty_Spaces = $row[2];
$Distance = $row[3];
$time_l = $row[4];
$time_h=$row[5];
$location=$row[6]; echo "\n
$building\n
$parking_lot\n
$Number_of_Empty_Spaces\n
$Distance\n
$time_h - $time_l mins\n
$location\n
\n";
}
echo "\n";
$err=1;
if ($img = file_get_contents($location, FILE_BINARY))
{
if ($img = imagecreatefromstring($img)) $err = 0;
}
if ($err)
{
header('Content-Type: text/html');
echo 'Error getting
image...';
}
else
{
header('Content-Type: image/jpeg');
imagejpeg($img);
imagedestroy($img);
}
?>
/
Sashikanth Gurram wrote:
Hello guys,
Thanks to you all for your kind replies. I will try the steps and
get
back to you if I encounter more problems.
Thanks,
Sashi
Fortuno, Adam wrote:
Matya,
Ha, ha, ha! Thank you good friend. I did say I didn't try the code
:-)
AF> Please note, I haven't tried this. It just seems plausible.
I apologize if I confused anyone. I just meant to show how the
parameter could help retrieve a picture. I wasn't too concerned
with
the particulars. Hopefully the concept is sound. If not, please do
flame my note so no one attempts it.
Be Well,
A-
-Original Message-
From: Mattyasovszky Janos [mailto:m...@matya.hu] Sent: Monday,
March
02, 2009 9:29 AM
To: php-db@lists.php.net
Subject: Re: [PHP-DB] Retrieving Image Location in MySQL
Fortuno, Adam írta:
//Write a query to pull out the picture's path
$sql = "SELECT path FROM Image WHERE ID = %s";
mysql_real_escape_string($value);
Sorry, but this won't work, since you don't map the value of the
escaped $value to the %s, lets say with sprintf()...
Regards,
Matya
--
~
~
Sashikanth Gurram
Graduate Research Assistant
Department of Civil and Environmental Engineering
Virginia Tech
Blacksburg, VA 24060, USA
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Retrieving Image Location in MySQL
Hello again Sashi The answer provided by Mushin won't work because images are not acessible from the outside. You can't have because that's not a valid URL... You can have something like which when interpreted by the browser will become something like http://yoursite/Bldgs_lots/Burruss.jpg Now, for this to happen you need to have Bldgs_lots as a subfolder or alias for your site. Otherwise, there is the way I mentioned in the first place, involving a php script just to get the image, which has to be called from within your script, passing it the info from the database echo ''; Now, I don't recommend passing local paths like this to scripts ... Hope this helps C ya JP -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Retrieving Image Location in MySQL
Hi JP,
Thanks for chipping in again. I have been using the following code
(after retrieving the local location of my image to the variable
$location. I am not really looking to host the pictures onto a site as
of now), which u have earlier provided me with
/
$err=1;
if ($img = file_get_contents("$location", FILE_BINARY))
{
if ($img = imagecreatefromstring($img)) $err = 0;
}
if ($err)
{
header('Content-Type: text/html');
echo 'Error getting
image...';
}
else
{
header('Content-Type: image/jpeg');
imagejpeg($img);
imagedestroy($img);
}
/
I am just using this and nothing more, and I have been getting a warning
of sort "CANNOT MODIFY HEADER INFORMATION-HEADERS ALREADY SENT ..ON
LINE 234". I have looked about this warning online and i found that this
occurs when the output is sent to the browser before the headers. From
the look of the above code, I dont think I am sending the info to the
browser before the headers. Also, when I use the above code, I see a lot
of ASCII characters in my browser. May be, the image is encrypted into
some form and the encrypted form is being displayed instead of the
picture. Further, you have always mentioned about using
img src="picture.php?&img=1234. I did not clearly understand where to put it in
the code. Is it a part of HTML code or is it a part of PHP code?
Thanks,
Sashi
Joao Gomes Madeira wrote:
Hello again Sashi
The answer provided by Mushin won't work because images are not
acessible from the outside. You can't have
because that's not a valid URL...
You can have something like
which when interpreted by the browser will become something like
http://yoursite/Bldgs_lots/Burruss.jpg
Now, for this to happen you need to have Bldgs_lots as a subfolder or
alias for your site.
Otherwise, there is the way I mentioned in the first place, involving a
php script just to get the image, which has to be called from within
your script, passing it the info from the database
echo 'alt="' . $build . '">';
Now, I don't recommend passing local paths like this to scripts ...
Hope this helps
C ya
JP
--
~
~
Sashikanth Gurram
Graduate Research Assistant
Department of Civil and Environmental Engineering
Virginia Tech
Blacksburg, VA 24060, USA
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Retrieving Image Location in MySQL
comes from
the dB
It should work without header stuffs
Example
P:S
Replace $image_location with image location from your dB.
GR
Muhsin
Sashikanth Gurram wrote:
Dear all,
I have been trying to retrieve the location of a image from
database
and display the image in the browser using PHP. I have written
a sort
of code for the purpose. All I am getting in my browser after
using
the code is a long set of ASCII characters with the following
warning
*Warning*: Cannot modify header information - headers already
sent by
(output started at C:\wamp\www\mysqli.php:65) in
*C:\wamp\www\mysqli.php* on line *237
*
I am herewith attaching my code. If anyone can point out the
error or
give some kind of advice that would be really great. The
location of
the image is stored in a table by the name IMAGE under the
column
name
Location and here in the code, the location is retrieved under
the
varibale name $Location.
thanks,
Sashi
/
Building Name:
Select a Building
Agnew Hall
Rector Field Houseoption>
Richard B. Talbot Educational Resources Center
Robeson Hall
Sandy Hall
Saunders hall
Seitz Hall
Shanks Hall
Shultz Hall
Skelton Conference
Center
Williams Hall
Women's Softball
Field
Wright House
=00 && $minute<=14)
{
$minute=00;
}
elseif ($minute>=15 && $minute<=29)
{
$minute=15;
}
elseif ($minute>=30 && $minute<=44)
{
$minute=30;
}
else
{
$minute=45;
}
$times="$hour:$minute";
$sql="SELECT buildingname, parking_lot_name, empty_spaces,
distance,
round(distance/($low_speed*60),1) AS low_time,
round(distance/($high_speed*60),1) AS high_time, Location FROM
buildings, buildings_lots, parkinglots, occupancy2, Image where
(buildings.buildingcode=occupancy2.building AND
buildings.buildingcode=buildings_lots.building_code AND
parkinglots.parking_lot_code=buildings_lots.parking_lot_code AND
parkinglots.parking_lot_code=occupancy2.parking_lot AND
Buildings.BuildingCode=Image.BuildingCode) AND
buildingname='$build'
AND month='$Month' AND day='$Today' AND Time='$times'";
$data = mysqli_query($cxn,$sql);
if (!$data=mysqli_query($cxn,$sql))
{
$error=mysqli_error($cxn);
echo "$error";
die();
}
else
{
echo "";
echo "Query sent successfully";
}
echo "";
echo " PARKING LOT INFORMATION ";
echo "";
echo "\n
Building\n
Parking Lot\n
Estimated Number of Empty Spaces\n
Distance (Feet)\n
Estimated walking time to the building\n
\n";
while ($row=mysqli_fetch_array($data))
{
extract($row);
$building = $row[0];
$parking_lot = $row[1];
$Number_of_Empty_Spaces = $row[2];
$Distance = $row[3];
$time_l = $row[4];
$time_h=$row[5];
$location=$row[6]; echo "\n
$building\n
$parking_lot\n
$Number_of_Empty_Spaces\n
$Distance\n
$time_h - $time_l mins\n
$location\n
\n";
}
echo "\n";
$err=1;
if ($img = file_get_contents($location, FILE_BINARY))
{
if ($img = imagecreatefromstring($img)) $err = 0;
}
if ($err)
{
header('Content-Type: text/html');
echo 'Error getting
image...';
}
else
{
header('Content-Type: image/jpeg');
imagejpeg($img);
imagedestroy($img);
}
?>
/
Sashikanth Gurram wrote:
Hello guys,
Thanks to you all for your kind replies. I will try the steps
and
get
back to you if I encounter more problems.
Thanks,
Sashi
Fortuno, Adam wrote:
Matya,
Ha, ha, ha! Thank you good friend. I did say I didn't try
the code
:-)
AF> Please note, I haven't tried this. It just seems
plausible.
I apologize if I confused anyone. I just meant to show how the
parameter could help retrieve a picture. I wasn't too
concerned
with
the particulars. Hopefully the concept is sound. If not,
please do
flame my note so no one attempts it.
Be Well,
A-
-Original Message-
From: Mattyasovszky Janos [mailto:m...@matya.hu] Sent: Monday,
March
02, 2009 9:29 AM
To: php-db@lists.php.net
Subject: Re: [PHP-DB] Retrieving Image Location in MySQL
Fortuno, Adam írta:
//Write a query to pull out the picture's path
$sql = "SELECT path FROM Image WHERE ID = %s";
mysql_real_escape_string($value);
Sorry, but this won't work, since you don't map the value of
the
escaped $value to the %s, lets say with sprintf()...
Regards,
Matya
---
---
--
--
~
~
Sashikanth Gurram
Graduate Research Assistant
Department of Civil and Environmental Engineering
Virginia Tech
Blacksburg, VA 24060, USA
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
--
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To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Retrieving Image Location in MySQL
ieved from the dB.That means if you have
something
like this in html documents whereby $image_location comes from
the dB
It should work without header stuffs
Example
P:S
Replace $image_location with image location from your dB.
GR
Muhsin
Sashikanth Gurram wrote:
Dear all,
I have been trying to retrieve the location of a image from
database
and display the image in the browser using PHP. I have written
a sort
of code for the purpose. All I am getting in my browser after
using
the code is a long set of ASCII characters with the following
warning
*Warning*: Cannot modify header information - headers already
sent by
(output started at C:\wamp\www\mysqli.php:65) in
*C:\wamp\www\mysqli.php* on line *237
*
I am herewith attaching my code. If anyone can point out the
error or
give some kind of advice that would be really great. The
location of
the image is stored in a table by the name IMAGE under the column
name
Location and here in the code, the location is retrieved under the
varibale name $Location.
thanks,
Sashi
/
Building Name:
Select a Building
Agnew Hall
Rector Field House
Richard B. Talbot Educational Resources Center
Robeson Hall
Sandy Hall
Saunders hall
Seitz Hall
Shanks Hall
Shultz Hall
Skelton Conference
Center
Williams Hall
Women's Softball
Field
Wright House
=00 && $minute<=14)
{
$minute=00;
}
elseif ($minute>=15 && $minute<=29)
{
$minute=15;
}
elseif ($minute>=30 && $minute<=44)
{
$minute=30;
}
else
{
$minute=45;
}
$times="$hour:$minute";
$sql="SELECT buildingname, parking_lot_name, empty_spaces,
distance,
round(distance/($low_speed*60),1) AS low_time,
round(distance/($high_speed*60),1) AS high_time, Location FROM
buildings, buildings_lots, parkinglots, occupancy2, Image where
(buildings.buildingcode=occupancy2.building AND
buildings.buildingcode=buildings_lots.building_code AND
parkinglots.parking_lot_code=buildings_lots.parking_lot_code AND
parkinglots.parking_lot_code=occupancy2.parking_lot AND
Buildings.BuildingCode=Image.BuildingCode) AND
buildingname='$build'
AND month='$Month' AND day='$Today' AND Time='$times'";
$data = mysqli_query($cxn,$sql);
if (!$data=mysqli_query($cxn,$sql))
{
$error=mysqli_error($cxn);
echo "$error";
die();
}
else
{
echo "";
echo "Query sent successfully";
}
echo "";
echo " PARKING LOT INFORMATION ";
echo "";
echo "\n
Building\n
Parking Lot\n
Estimated Number of Empty Spaces\n
Distance (Feet)\n
Estimated walking time to the building\n
\n";
while ($row=mysqli_fetch_array($data))
{
extract($row);
$building = $row[0];
$parking_lot = $row[1];
$Number_of_Empty_Spaces = $row[2];
$Distance = $row[3];
$time_l = $row[4];
$time_h=$row[5];
$location=$row[6]; echo "\n
$building\n
$parking_lot\n
$Number_of_Empty_Spaces\n
$Distance\n
$time_h - $time_l mins\n
$location\n
\n";
}
echo "\n";
$err=1;
if ($img = file_get_contents($location, FILE_BINARY))
{
if ($img = imagecreatefromstring($img)) $err = 0;
}
if ($err)
{
header('Content-Type: text/html');
echo 'Error getting
image...';
}
else
{
header('Content-Type: image/jpeg');
imagejpeg($img);
imagedestroy($img);
}
?>
/
Sashikanth Gurram wrote:
Hello guys,
Thanks to you all for your kind replies. I will try the steps and
get
back to you if I encounter more problems.
Thanks,
Sashi
Fortuno, Adam wrote:
Matya,
Ha, ha, ha! Thank you good friend. I did say I didn't try the
code
:-)
AF> Please note, I haven't tried this. It just seems plausible.
I apologize if I confused anyone. I just meant to show how the
parameter could help retrieve a picture. I wasn't too concerned
with
the particulars. Hopefully the concept is sound. If not,
please do
flame my note so no one attempts it.
Be Well,
A-
-Original Message-
From: Mattyasovszky Janos [mailto:m...@matya.hu] Sent: Monday,
March
02, 2009 9:29 AM
To: php-db@lists.php.net
Subject: Re: [PHP-DB] Retrieving Image Location in MySQL
Fortuno, Adam írta:
//Write a query to pull out the picture's path
$sql = "SELECT path FROM Image WHERE ID = %s";
mysql_real_escape_string($value);
Sorry, but this won't work, since you don't map the value of the
escaped $value to the %s, lets say with sprintf()...
Regards,
Matya
--
~
~
Sashikanth Gurram
Graduate Research Assistant
Dep
Re: [PHP-DB] Retrieving Image Location in MySQL
t;>>>>> ../ One step back from the script location >>>>>>> >>>>>>> >>>>>>> GR >>>>>>> Muhsin >>>>>>> >>>>>>> >>>>>>> Sashikanth Gurram wrote: >>>>>>> >>>>>>>> Hi, >>>>>>>> Thanks a lot for the suggestion. It is working fine but there is some >>>>>>>> problem. When I introduced the piece of code you have suggested into >>>>>>>> my code, I am getting a space for the picture but nothing is there in >>>>>>>> that space. It is just blank. I do not know the reason for this. I am >>>>>>>> just attaching a picture just in case it might give you a better idea. >>>>>>>> Further, this is the piece of code I have written to display the pic. >>>>>>>> >>>>>>>> /echo ">>>>>>> alt='$build' >";/ >>>>>>>> >>>>>>>> Any help would be greatly appreciated. >>>>>>>> Thanks, >>>>>>>> Sashi >>>>>>>> >>>>>>>> mrfroasty wrote: >>>>>>>> >>>>>>>>> Hello, >>>>>>>>> >>>>>>>>> Storing only location of the image in dB wouldnt be any different than >>>>>>>>> other string/data retrieved from the dB.That means if you have >>>>>>>>> something >>>>>>>>> like this in html documents whereby $image_location comes from the dB >>>>>>>>> It should work without header stuffs >>>>>>>>> >>>>>>>>> Example >>>>>>>>> >>>>>>>> alt="Image" ISMAP> >>>>>>>>> P:S >>>>>>>>> Replace $image_location with image location from your dB. >>>>>>>>> >>>>>>>>> >>>>>>>>> GR >>>>>>>>> Muhsin >>>>>>>>> >>>>>>>>> >>>>>>>>> >>>>>>>>> Sashikanth Gurram wrote: >>>>>>>>> >>>>>>>>>> Dear all, >>>>>>>>>> >>>>>>>>>> I have been trying to retrieve the location of a image from database >>>>>>>>>> and display the image in the browser using PHP. I have written a sort >>>>>>>>>> of code for the purpose. All I am getting in my browser after using >>>>>>>>>> the code is a long set of ASCII characters with the following warning >>>>>>>>>> *Warning*: Cannot modify header information - headers already sent by >>>>>>>>>> (output started at C:\wamp\www\mysqli.php:65) in >>>>>>>>>> *C:\wamp\www\mysqli.php* on line *237 >>>>>>>>>> * >>>>>>>>>> I am herewith attaching my code. If anyone can point out the >>>>>>>>>> error or >>>>>>>>>> give some kind of advice that would be really great. The location of >>>>>>>>>> the image is stored in a table by the name IMAGE under the column >>>>>>>>>> name >>>>>>>>>> Location and here in the code, the location is retrieved under the >>>>>>>>>> varibale name $Location. >>>>>>>>>> >>>>>>>>>> thanks, >>>>>>>>>> Sashi >>>>>>>>>> >>>>>>>>>> / >>>>>>>>>> >>>>>>>>>> >>>>>>>>>> >>>>>>>>>> >>>>>>>>>> Building Name: >>>>>>>>>> Select a Building >>>>>>>>>> Agnew Hall >>>>>>>>>>Rector Field House >>>>>>>>>>Richard B. Talbot Educational Resources Center >>>>>>>>>>Robeson Hall >>>>>>>>>>Sandy Hall >>>>>>>>>>Saunders hall >>>>>>>&g
