[PHP-DB] parsing PHP from blob varaibles
When a php statement is part of a markup snippet in a mysql database table field, what must be done to have the php parse in the html document. A simple demo demonstrating my problem is at http://www.boclair.com/test/blobvars.php Louise
SV: [PHP-DB] PHP and MySql question
Hi,
Your problem is that your variables $Lat1,$Lat2,$Lon1,$Lon2 all reffer
to resultsets and not the Latitudes or Longitudes.
Try something like this:
$res = mysql_query(
SELECT ZG_LATITUDE, ZG-LONGITUDE FROM zip_code where
zg_zipcode = '$zip');
List($Lat,$Lon) = mysql_fetch_row($res);
$Lat1=Lat-2; $Lat2=Lat+2; $Lon1=Lon-2; $Lon2=Lat+2;
Now you can use the values in your final query.
Hth Henrik Hornemann
-Oprindelig meddelelse-
Fra: ReClMaples [mailto:[EMAIL PROTECTED]
Sendt: 16. april 2005 22:38
Til: php-db@lists.php.net
Emne: [PHP-DB] PHP and MySql question
Hello, I'm kinda new at PHP programming with MySQL. I am having an
issue and am not sure if this is the corret way to do this: Here is my
code, I'll start there:
?php
$Lat1 = mysql_query(
SELECT (ZG_LATITUDE-2) FROM zip_code where zg_zipcode =
'$zip');
if (!$Lat1) {
echo(PError performing query: .
mysql_error() . /P);
exit();
}
$Lat2 = mysql_query(
SELECT (ZG_LATITUDE+2) FROM zip_code where zg_zipcode =
'$zip');
if (!$Lat2) {
echo(PError performing query: .
mysql_error() . /P);
exit();
}
$Lon1 = mysql_query(
SELECT (ZG_LONGITUDE-2) FROM zip_code where zg_zipcode =
'$zip');
if (!$Lon1) {
echo(PError performing query: .
mysql_error() . /P);
exit();
}
$Lon2 = mysql_query(
SELECT (ZG_LONGITUDE+2) FROM zip_code where zg_zipcode =
'$zip');
if (!$Lon2) {
echo(PError performing query: .
mysql_error() . /P);
exit();
}
$ZipCode = mysql_query(
SELECT * FROM zip_code where ZG_LATITUDE = '$Lat1'
and ZG_LATITUDE = '$Lat2' and ZG_LONGITUDE = '$Lon1' and ZG_LONGITUDE
= '$Lon2');
if (!$Zipcodesearch) {
echo(PError performing query: .
mysql_error() . /P);
exit();
}
while ( $row = mysql_fetch_array($ZipCode) ) {
echo(bCity: /b . $row[ZG_CITY]. br);
echo(bState: /b . $row[ZG_STATE]. br); echo(bZip Code:
/b . $row[ZG_ZIPCODE]. br); echo(bArea Code: /b .
$row[ZG_AREACODE]. br); echo(bCounty FIPS: /b .
$row[ZG_COUNTY_FIPS]. br); echo(bCounty Name: /b .
$row[ZG_COUNTY_NAME]. br); echo(bTime Zone: /b .
$row[ZG_TIME_ZONE]. br); echo(bDay Light Savings?: /b .
$row[ZG_DST]. br);
echo(bLatitude: /b . $row[ZG_LATITUDE]. br);
echo(bLongitude: /b . $row[ZG_LONGITUDE]. P); } ?
Basically I'm trying to have a user input a zip code and then have a php
script pull all zip codes that are in a region of that submitted zip
code.
Can you have a look at my code and see what I'm doing wrong? When using
this it returns no results but makes the connection to the database so I
have to believe that it's within here that I have my issue.
I have Apache/2.0.47 (Win32) PHP/4.3.9 Server and MySql 4.0.14 (I know I
can upgrade and be able to do subselects but I would like to know what
I'm doing wrong here.
Thanks in advance for any help.
Thanks
-rich
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[PHP-DB] Parsing PHP sourced from blob
When a php statement is part of a markup snippet in a mysql database table field, what must be done to have the php parse in the html A simple demo demonstrating my problem is at http://www.boclair.com/test/blobvars.php Louise -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Parsing PHP sourced from blob
try eval (http://ca3.php.net/manual/en/function.eval.php) Bastien From: boclair [EMAIL PROTECTED] To: php-db@lists.php.net php-db@lists.php.net Subject: [PHP-DB] Parsing PHP sourced from blob Date: Mon, 18 Apr 2005 17:52:52 +1000 When a php statement is part of a markup snippet in a mysql database table field, what must be done to have the php parse in the html A simple demo demonstrating my problem is at http://www.boclair.com/test/blobvars.php Louise -- 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
[PHP-DB] Re: php and tables
There are two places you can figure out the average: in MySQL or in PHP. Then you want to sort the results by this Average: in MySql use ORDER BY and in php use array_multisort in MySQL your SQL might look like: SELECT TEAM, PLAYED, WIN, LOSS, DRAW, POINTS, POINTS / PLAYED as AVERAGE FROM your table ORDER BY AVERAGE desc in PHP you'd have to loop through your results loading them into a local array with an additional column for average and then use array_multisort which has many options (and I don't exactly remember how to make it work exactly but look at: http://www.phpbuilder.com/manual/function.array-multisort.php to see). I'd do it in MySQL but that's me... Good luck, Frank On Apr 17, 2005, at 2:13 PM, [EMAIL PROTECTED] wrote: From: Vikas Nanda [EMAIL PROTECTED] Date: April 17, 2005 7:39:57 AM PDT To: php-db@lists.php.net Subject: php and tables Hi I am new to php and mysql and am having a few problems that I hope someone can help with. What I am trying to do is to display ratings for particular teams using some basic code. I needed to print the information out on a table with the highest average at the top and then ascending. I have all this info stored in a mysql db. eg TEAMPLAYED WIN LOSS DRAW POINTS AVERAGE ENGLAND 16 6 6 4 16 1.00 I don't know how to print variables out in a table and I don't know how to make it ascending. Any help would be much appreciated. Thanks Vikas
RE: [PHP-DB] PHP and MySql question
Ok, I tried what you saidHere's the code:
$res = mysql_query(
SELECT ZG_LATITUDE, ZG_LONGITUDE FROM zip_code where
zg_zipcode = '$zip');
List($Lat,$Lon) = mysql_fetch_row($res);
$Lat1 = ($Lat-2);
$Lat2= ($Lat+2);
$Lon1= ($Lon-2);
$Lon2= ($Lon+2);
//echo ($Lat1);
//echo ($Lat2);
//echo ($Lon1);
//echo ($Lon2);
$zipcode = mysql_query(
SELECT * FROM zip_code where ZG_LATITUDE = $Lat1
and ZG_LATITUDE = $Lat2 and ZG_LONGITUDE = $Lon1 and ZG_LONGITUDE =
$Lon2);
if (!$zipcode) {
echo(PError performing query: .
mysql_error() . /P);
exit();
}
while ( $row = mysql_fetch_array($zipcode) )
{
echo(bCity: /b . $row[ZG_CITY]. br);
echo(bState: /b . $row[ZG_STATE]. br);
echo(bZip Code: /b . $row[ZG_ZIPCODE]. br);
echo(bArea Code: /b . $row[ZG_AREACODE]. br);
echo(bCounty FIPS: /b . $row[ZG_COUNTY_FIPS]. br);
echo(bCounty Name: /b . $row[ZG_COUNTY_NAME]. br);
echo(bTime Zone: /b . $row[ZG_TIME_ZONE]. br);
echo(bDay Light Savings?: /b . $row[ZG_DST]. br);
echo(bLatitude: /b . $row[ZG_LATITUDE]. br);
echo(bLongitude: /b . $row[ZG_LONGITUDE]. P);
}
I get no results with this still, it acutally doesn't even go to the this
page. When I uncomment out:
//echo ($Lat1);
//echo ($Lat2);
//echo ($Lon1);
//echo ($Lon2);
and comment out everything below this, I get the expected results for
$Lat1,$Lat2,$Lon1, and $Lon2. So I have to believe my issue is below those
lines, can anyone see what I have that is incorrect here?
-Original Message-
From: Henrik Hornemann [mailto:[EMAIL PROTECTED]
Sent: Monday, April 18, 2005 2:44 AM
To: php-db@lists.php.net
Subject: SV: [PHP-DB] PHP and MySql question
Hi,
Your problem is that your variables $Lat1,$Lat2,$Lon1,$Lon2 all reffer
to resultsets and not the Latitudes or Longitudes.
Try something like this:
$res = mysql_query(
SELECT ZG_LATITUDE, ZG-LONGITUDE FROM zip_code where
zg_zipcode = '$zip');
List($Lat,$Lon) = mysql_fetch_row($res);
$Lat1=Lat-2; $Lat2=Lat+2; $Lon1=Lon-2; $Lon2=Lat+2;
Now you can use the values in your final query.
Hth Henrik Hornemann
-Oprindelig meddelelse-
Fra: ReClMaples [mailto:[EMAIL PROTECTED]
Sendt: 16. april 2005 22:38
Til: php-db@lists.php.net
Emne: [PHP-DB] PHP and MySql question
Hello, I'm kinda new at PHP programming with MySQL. I am having an
issue and am not sure if this is the corret way to do this: Here is my
code, I'll start there:
?php
$Lat1 = mysql_query(
SELECT (ZG_LATITUDE-2) FROM zip_code where zg_zipcode =
'$zip');
if (!$Lat1) {
echo(PError performing query: .
mysql_error() . /P);
exit();
}
$Lat2 = mysql_query(
SELECT (ZG_LATITUDE+2) FROM zip_code where zg_zipcode =
'$zip');
if (!$Lat2) {
echo(PError performing query: .
mysql_error() . /P);
exit();
}
$Lon1 = mysql_query(
SELECT (ZG_LONGITUDE-2) FROM zip_code where zg_zipcode =
'$zip');
if (!$Lon1) {
echo(PError performing query: .
mysql_error() . /P);
exit();
}
$Lon2 = mysql_query(
SELECT (ZG_LONGITUDE+2) FROM zip_code where zg_zipcode =
'$zip');
if (!$Lon2) {
echo(PError performing query: .
mysql_error() . /P);
exit();
}
$ZipCode = mysql_query(
SELECT * FROM zip_code where ZG_LATITUDE = '$Lat1'
and ZG_LATITUDE = '$Lat2' and ZG_LONGITUDE = '$Lon1' and ZG_LONGITUDE
= '$Lon2');
if (!$Zipcodesearch) {
echo(PError performing query: .
mysql_error() . /P);
exit();
}
while ( $row = mysql_fetch_array($ZipCode) ) {
echo(bCity: /b . $row[ZG_CITY]. br);
echo(bState: /b . $row[ZG_STATE]. br); echo(bZip Code:
/b . $row[ZG_ZIPCODE]. br); echo(bArea Code: /b .
$row[ZG_AREACODE]. br); echo(bCounty FIPS: /b .
$row[ZG_COUNTY_FIPS]. br); echo(bCounty Name: /b .
$row[ZG_COUNTY_NAME]. br); echo(bTime Zone: /b .
$row[ZG_TIME_ZONE]. br); echo(bDay Light Savings?: /b .
$row[ZG_DST]. br);
echo(bLatitude: /b . $row[ZG_LATITUDE]. br);
echo(bLongitude: /b . $row[ZG_LONGITUDE]. P); } ?
Basically I'm trying to have a user input a zip code and then have a php
script pull all zip codes that are in a region of that submitted zip
code.
Can you have a look at my code and see what I'm doing wrong? When using
this it returns no results but makes the connection to the database so I
have to believe that it's within here that I have my issue.
I have Apache/2.0.47 (Win32) PHP/4.3.9 Server and MySql 4.0.14 (I know I
can upgrade and be able to do subselects but I would like to know what
I'm doing wrong here.
Thanks in advance for any help.
Thanks
-rich
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http://www.php.net/unsub.php
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To unsubscribe, visit: http://www.php.net/unsub.php
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[PHP-DB] PHPMySQL left/substring etc problem
Hello,
I need your help about PHP (ver 4.3.1) and MySQL (ver 3.23) database.
I am struggling to customise a database (using left command etc), and am
stuck now.
Please have a brief look at my scripts below.
What I do like to do is to get first 3 letters from "loc1" column from
openart_table (MySQL).
And show the data in the follwoing part...
td? echo $row["loc1"]; ?br/td
but also I would like to make hyperlinks to jpg files to images/"the first 3
letters from loc1".jpg
So it will be something like this...
td
a href="/image/? echo $row["loc1"]; ?.jpg"? echo $row["loc1"];
?/abr
/td
But when I tried, "select left (loc1,3) from openart_table where.."
doesnt work properly.
I do not want to destroy anything for the rest of the table.
I need all data from "tit1", "pub1", "id1", and "ref1" in each column.
Could you help me, please?
-PHP script from here
html
body
table border="1" bordercolor="black" cellspacing="0" align="center"
BGCOLOR="#CC" width="900"
tr bgcolor="#ffcc66"
td align="center"bID No/bbr/td
td align="center" BGCOLOR="#CC3366"bBook Title/bbr/td
td align="center"bPublisher/bbr/td
td align="center"bID/bbr/td
td align="center"bReference/bbr/td
td align="center" width="10%"bLocation/bbr/td
/tr
?
mysql_connect(localhost,root,cheers);
mysql_select_db(openart);
if($tit == ""
$res == ""
$pub == ""
$date == ""
$id == ""
$ref == ""
$loc == ""
$type == ""
$vol == ""
$fre == ""
$note == "")
{
echo 'Type something';
}
elseif($tit == "%" || $res == "%" || $pub == "%" || $date == "%" || $id ==
"%" ||
$ref == "%" || $loc == "%" || $type == "%" || $vol == "%" || $fre ==
"%" || $note == "%"){
echo 'Not Valid';
}
else{
if($tit == ""){
$tit = '%';
}
if($res == ""){
$res = '%';
}
if($pub == ""){
$pub = '%';
}
if($date == ""){
$date = '%';
}
if($id == ""){
$id = '%';
}
if($ref == ""){
$ref = '%';
}
if($loc == ""){
$loc = '%';
}
if($type == ""){
$type = '%';
}
if($vol == ""){
$vol = '%';
}
if($fre == ""){
$fre = '%';
}
if($note == ""){
$note = '%';
}
$result = mysql_query("select * from openart_table
where tit1 like '%$tit%'
and res1 like '%$res%'
and pub1 like '%$pub%'
and date1 like '%$date%'
and id1 like '%$id%'
and ref1 like '%$ref%'
and loc1 like '%$loc%'
and type1 like '%$type%'
and vol1 like '%$vol%'
and fre1 like '%$fre%'
and note1 like '%$note%' order by tit1");
$rows = mysql_num_rows($result);
echo $rows,"Records Found p";
while($row = mysql_fetch_array($result)){
?
tr
tda href = "openart_detail.php ?iden=? echo $row["openart_id"] ?"?
echo $row["openart_id"]; ?/abr/td
td? echo $row["tit1"]; ?br/td
td? echo $row["pub1"]; ?br/td
td? echo $row["id1"]; ?br/td
td? echo $row["ref1"]; ?br/td
td? echo $row["loc1"]; ?br/td
/tr
?
}
}
?
/table
/body
/html
end of script
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