Re: [PHP-DB] Auto increament
Yes there is an auto increment field in mySQL and most other databases. Can any of you guide me how to do auto increament in a field? I think there is one datatype for auto increament. But I dont know how to do it ? Rinku -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] [PHP]: session problem
Are you getting any errors displayed? The php.ini file has to have a directory specified for the session temp variables. I want to use sessions to check wether a user is logged in or not. Therefore,I use session_start and session_register. When I registere the values on the same page, they all seem to work. But when I ask the value of a registered session variable on another page, it is empty?!?! Does anyone know how this comes? i am using session_start() at every page.. i believe it must be in my php.ini setting because on my hosting provider it does work all correctly.. can anyone helpme? -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] [PHP]: session problem
What are you setting $_SESSION['sid'] to? i do ($_SESSION['sid']), but the variables remain empty... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Date help needed
A drop down with 365 days !?!? Isn't that a little big? I have a problem, I currently have some code which populates a dropdown box - this code gives me every day for the next x amount of days (EG: a years worth of days), however what I really need to be able to do, is to find a way to display this data in the dropdown box but ONLY show 3 days a week, IE: Mondays, Fridays and Sundays, so it would show the dates for each Monday, Friday and Sunday for X amount of days (IE: 365 days in the dropdown). Does anyone have any idea how to do this? I would really appreciate any help, I'd send my sample code only I'm not at my home/work computer ATM. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Multiple Inserts
Good point. Since it's form data, what about $_POST['TimesheetID'] ?
Don't see anything obviously wrong with your query string. Are the inserts
happening in the same block of code, i.e., are you sure that
_$TimesheetID_ has a value in it when you're performing the second insert?
-dave
I am using a form to Insert data into 2 tables in the same database.
$TimesheetID needs to be in each table. However, it is not being
inserted
into the second table, tblTimesheetDetails . Any advise?
$result_timesheet=mysql_query(INSERT INTO tblTimesheet (TimesheetID,
WorkerID, ClientID, TimesheetDate, ProspectiveOrRetrospective) VALUES
('$TimeSheetID','$WorkerID','$ClientID','$TimesheetDate','$ProspectiveOrRetr
ospective'))or die(Insert Error: .mysql_error());
$result_timesheetdetails=mysql_query(INSERT INTO tblTimesheetDetails
(TimesheetID, ActivityTypeID, TimeSpentHours, TimeSpentMinutes) VALUES
('$TimeSheetID','$ActivityTypeID','$TimeSpentHours','$TimeSpentMinutes'))or
die(Insert Error: .mysql_error());
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Re: [PHP-DB] problem....
What error are you getting?
why doesn't this work:
$pic=mysql_query('SELECT Rune, username FROM RuneRunner
RuneRunner_1 WHERE (User_ID = 3)',$connection);
$pic=mysql_fetch_array($pic);
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Re: [PHP-DB] anonymous select error
I'm not sure you need a semi-colon after the table name.
$result = mysql_query( 'Select * From newsletter_subscribers', $db );
To: [EMAIL PROTECTED]
MIME-Version: 1.0
Content-Type: text/plain; charset=us-ascii
I'm trying this query:
$link = mysql_connect( $site, $id, $pass );
if (!$link) {
die('Could not connect: ' . mysql_error());
}
echo 'Connected successfullybr';
$db = mysql_select_db( $dbname, $link);
if( !$db ) {
echo DB falsebr;
exit;
}
$result = mysql_query( 'Select * From
newsletter_subscribers;', $db );
if (!$result) {
echo DB Error, could not list tablesbr;
echo 'MySQL Error: ' . mysql_error();
exit;
}
with this output:
Connected successfully
DB Error, could not list tables
MySQL Error:
I dont understand why there is an error, but it prints
nothing for the error. Any ideas? Thanks, Josh
Re: [PHP-DB] About retrieving auto increment value
I think you want mysql_insert_id() http://us3.php.net/manual/en/function.mysql-insert-id.php I have the typical problem with retrieving auto increment value in php-mysql environment. I have search info in mysql page and I have found LAST_INSERT_ID() funtion but I don't know how to use correctly. I have one script that insert a row. And another script that must retrieve last auto_increment value for primary key on a table. I want something like this: SELECT LAST_INSERT_ID() FROM foo_table; (like curval(foo_sequence) in postgres) This doesn't work. It's possible? Where I am wrong? Thanks in advance -- Marc Soler -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] strange results in mysql with readdir.php
I'd think there where multiple files with jpg and gif extensions.
Like:
image1.jpg
image1.jpg.jpg
Anybody has an idea about what happened to me ?
The following script should just charge in a mysql db the images located in
a directory. Unfortunately everytime I call readdir.php (never mind if
there are or not new images) mysql is charging 2 or 6 copies of the same
last image uploaded into the directory. Anybody may explain to me where I'm
failing?
Thanks for all the phpers!
Alessandro
?
$dbcnx = mysql_connect(localhost, root, password);
mysql_select_db(news);
if (!$dbcnx)
{
echo( pconnection to database server failed!/p);
exit();
}
if (! @mysql_select_db(news) )
{
echo( pImage Database Not Available!/p );
exit();
}
$path = ./;
$dir_handle = @opendir($path) or die(Unable to open directory $path);
while ($file = readdir($dir_handle)) {
$filetyp = substr($file, -3);
if ($filetyp == 'gif' OR $filetyp == 'jpg') {
$handle = fopen($path . / . $file,'r');
$file_content = fread($handle,filesize($path . / . $file));
fclose($handle);
$encoded = chunk_split(base64_encode($file_content));
$sql = INSERT INTO images SET sixfourdata='$encoded';
@mysql_query($sql);
}
}
closedir($dir_handle);
echo(complete);
?
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Re: [PHP-DB] Error loading module: Unable to load dynamic library 'c:\tools\php\extensions\php_oci8.dll' - The specified procedure could not be found
Is the file in c:\tools\php\extensions\php_oci8.dll ? Hi, I am experiencing the message in the title of this message when starting or restarting Apache. I have the following configuration: - Windows XP - Apache 2.0.49 - PHP 4.3.6 - Oracle Client 8i If anyone has any information as to why this is happening and how I can solve it, I would be very grateful. Thanks. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Re: big/small letters with oracle No2
I've used Oracle for years, and am not aware of Oracle having case sensitive column
names.
Torsten Lange [EMAIL PROTECTED] wrote in message news:[EMAIL PROTECTED]
But when using queries on the USER_... data dictionary, Oracle
delivers always big letters, which is for chemical elements (NA
vs. Na) or location names (ALICE SPRINGS vs. Alice Springs) and
location codes often uncomfortable to read.
Then I see only one way: create a mapping array to map your field names to
what you want them to be *really* called.
$mapping = array('FIELD1' = 'My real field name', 'FIELD2' = 'My second
field name');
Then you get the value this way:
$realName = $mapping[$fieldNameFromDB];
A mapping is the best way. It separates the internal schema structure
(i.e column names as created by Oracle) from the display values (i.e.
the column names you want to display).
But it is possible to get PHP to return case sensitive column names from
Oracle, see below.
Chris
-
?php
// Example using case sensitive column names in Oracle.
//
// Table P1 was created in SQL*Plus using:
//
//create table p1 (MyCol number);
//insert into p1 values (1234);
//commit;
//
// The output of this PHP script is:
//
// array(1) {
// [MyCol]=
// string(4) 1234
// }
$conn = OCILogon(scott, tiger, MYDB);
$query = 'select * from p1';
$stid = OCIParse($conn, $query);
OCIExecute($stid);
OCIFetchInto($stid, $row, OCI_ASSOC);
echo pre; var_dump($row); echo /pre;
OCILogoff($conn);
?
--
Christopher Jones, Oracle Corporation, Australia.
Re: [PHP-DB] Multiple SQL queries...?
Is this what you want ??
$sql = SELECT DISTINCT(email) FROM $table;
$result = mysql_query($sql,$connection) or die(Couldn't execute query 0.);
$row1 = 0;
while ($row = mysql_fetch_array($result)) {
echo $row[email] ;
$sql1 = SELECT * FROM $table WHERE viewed = '1' AND email = '$row[email]';
$result1 = mysql_query($sql1,$connection) or die(Couldn't execute query
1.);
if (mysql_num_rows($result1) = 1) {
echo 'Viewed' ;
} else {
echo '' ;
}
}
echo $row1 ;
Right, todays fun dilema... ;-)
I've a user capture system set up for downloads on our site.
Each time a user downloads a file, their info is captured (so we'll have
multiple entries for each email address).
Also in the table, is a field to state if the result has been viewed by my
boss. (Just a 1/0 value)
So, what my boss wants me to do now, is to show her, the total No of
people captured, and how many she's viewed.
Using the 'email' field, from the table, I've done a distinct() sort in
MySQL to ensure that I have a list of emails(users) and no duplicates.
My prob is this:
I want to take each email, and see if at any point, there's a '1' in the
viewed field.
if there is, I want to add one to the total of $row1
So I can output the num_rows from the distinct request, and then show the
total of $row1...
So I'll have total No of users, and total No of viewed... as I've
mentioned above.
What am I doing wrong?
?
$sql = SELECT DISTINCT(email) FROM $table;
$result = mysql_query($sql,$connection)
or die(Couldn't execute query 0.);
$row1 = 0;
while ($row = mysql_fetch_array($result)) {
$sql1 = SELECT * FROM $table WHERE viewed = '1' AND email =
'$row[email]';
$result1 = mysql_query($sql1,$connection)
or die(Couldn't execute query 1.);
if (mysql_num_rows($result1) = 1) {
$row1++;
}
}
?
?=$row1 ?
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Re: [PHP-DB] Multiple SQL queries...?
OK, how about this?
$sql = SELECT DISTINCT(email) FROM $table;
$result = mysql_query($sql,$connection) or die(Couldn't execute query
0.);
while ($row = mysql_fetch_array($result)) {
echo $row[email] ;
$sql1 = SELECT * FROM $table WHERE viewed = '1' AND email =
'$row[email]';
$result1 = mysql_query($sql1,$connection) or die(Couldn't execute
query 1.);
echo ' ' . mysql_num_rows($result1) ;
}
echo $row1 ;
Kinda, but I just wanna count how many views in total...
Hence my $row1++; bit...
I'll have a play with your code though...!
Cheers,
Tris...
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Re: [PHP-DB] Multiple SQL queries...?
Not tested... how about something like this? SELECT t1.email, t2.count(*) FROM $table t1 LEFT JOIN $table t2 ON (t1.email = t2.email and t2.viewed = '1') GROUP BY EMAIL Nope... HHmmm, this is really getting to me... I can do distinct, I can count, but I can't combine the two? Can't be that hard ;-) -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Question about undices for inversed LIKE statesments.
The index should still work normally. Hello. I have a question about behaviour of indices in queries with inversed LIKE in MySQL. I mean something like this: select foo from bar where '$some_string' LIKE some_field where in the `some_field` I have a strings like 'symbols%' :) Do you have some ideas? WBR, Wicked -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Where Error
Can you post the code?
Something like:
WHERE status IN ('i', 'c', 'o') ;
This should be ultimately simple; it just doesn't work. I want to limit
display to rows from a single table in which the status column contains
i. Only options are i, c, o. After the select line I have where status
= i (no quote marks). I get an error message to the while line that
calls the table.
When I limit by another column where netreceipts 500 I get the desired
result. When I change it to where netreceipts = 500 (and there is a
500) I get the error message above.
What's the solution?
Ken
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Re: [PHP-DB] Where Error
SELECT ..
WHERE status IN ('i', 'c', 'o') ;
ORDER BY $sort_field $sort_order;
$get_data_query = select state, cd, representative, status, party,
netreceipts,
individualcontributions, paccommcontributions, candidatesupport,
netdisbursements
from fec33104
where status = i
order by $sort_field $sort_order;
**
On Sat, 29 May 2004 07:50:52 -0700, Daniel Clark [EMAIL PROTECTED] wrote:
Can you post the code?
Something like:
WHERE status IN ('i', 'c', 'o') ;
This should be ultimately simple; it just doesn't work. I want to
limit
display to rows from a single table in which the status column contains
i. Only options are i, c, o. After the select line I have where
status
= i (no quote marks). I get an error message to the while line
that
calls the table.
When I limit by another column where netreceipts 500 I get the
desired
result. When I change it to where netreceipts = 500 (and there is a
500) I get the error message above.
What's the solution?
Ken
Re: [PHP-DB] Undefined variable
Are you using a FORM Post or Get ? Perhaps $_POST['Name'] or $_GET['Name'] Hi!! I have a very simple and smail script with just one input text box and a submit button. After that, I have an echo function in the same code page that just displays the data entered in the text box. But I'm receiving an undefinied variable error when I execute it!! Echo ($Name); I'm using Windows 2000 Server and PHP for Windows. Regards, Miguel Guirao Servicios Datacard www.SIASA.com.mx
Re: [PHP-DB] Select news based on dates
ORDER BY date, text ?
Hi!
You can use order by date and such method:
code
$result = mysql_query(blablabla);
$date = foo;
while($news = mysql_fetch_array($result)){
if ($date==$news[date]){
echo new date;
}
echo $news[text];
$date = $news[date];
}
Monday, May 17, 2004, 3:50:04 PM, T. wrote:
THG Hello,
THG I would like to display my news like this:
THG *10.04.2004.*
THG - news 1
THG - news 2
THG - news 3
THG *14.04.2004.*
THG - news 4
THG *15.04.2004.*
THG - news 5
THG ...
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Re: [PHP-DB] Newbie inserting into a database
Remove the single quotes or change to double quotes around the variables.
Here is what I have so far?
?
$DBname = testdatabase;
$host=localhost;
$user=root;
$password = password;
$dbconnect = @mysql_connect('$host','$user','$password','$DBname');
I know this is wrong since I am not connecting.
What I want to do is INSERT several values into a database. What am I
doing
wrong and what do I need to do after I connect (The Code)? 7 day old
problem here and pulling at my hair! Appreciate it!
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Re: [PHP-DB] Newbie inserting into a database
Remove the single quotes or change to double quotes around the variables.
(in the mysql_connect() statement)
Here is what I have so far?
?
$DBname = testdatabase;
$host=localhost;
$user=root;
$password = password;
$dbconnect = @mysql_connect('$host','$user','$password','$DBname');
I know this is wrong since I am not connecting.
What I want to do is INSERT several values into a database. What am I
doing
wrong and what do I need to do after I connect (The Code)? 7 day old
problem here and pulling at my hair! Appreciate it!
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Re: [PHP-DB] inserting data into database!
$tmp isn't defined yet, so can't append to it's self.
(no .= )
$tmp = ID : $row-ID br\n;
Two questions?
First why am I getting undefined variable tmp , yet it works
Second how would I insert data into this table?
?
$db_table = 'log';
$conn = mysql_connect($db_host,$db_user,$db_pass);
if ($conn == true) {
mysql_select_db($db_name,$conn);
$result = mysql_query(SELECT * from $db_table,$conn);
while($row = mysql_fetch_object($result)) {
$tmp .= ID : $row-ID br\n;
$tmp .= Base : $row-Base br\n;
$tmp .= Date_and_Time : $row-Date_and_Time br\n;
$tmp .= Event_Type : $row-Event_Type br\n;
$tmp .= Description : $row-Description br\n;
$tmp .= Initials : $row-Initials brhr\n;
}
} else {
echo 'could not connect to database : '. mysql_error();
}
print $tmp;
?
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Re: [PHP-DB] Need some HELP
unless you have Globals turned on, I think you and $_POST['variable_name_here'] in the second page. print INPUT TYPE='hidden' NAME='UserName' VALUE='$_POST['username']\n; Hi, I am a new to php. I am using php ver 4.0.3 and MySQL. I have wrote two files and I'd like to pass some varaibles from first file to use them in the second file. but the values did not displayed in the second file(nothing print out). and also I want to pass them to anothr file ... here is what I've wrote : file1.php** ?php . $username=A; $user_num=123; $user_addres=User Address; print FORM action='file2.php' method=post; print input type=submit value='send'; print INPUT TYPE='hidden' NAME='UserName' VALUE='$username'\n; print input type=hidden name=UserNum' value='$user_num'; print input type=hidden name='Addrress' value='$user_addres'; print /td; print /FORM/tr; ? ** second file file2.php ?php ? FORM action='anothrfile.php' method=post table ? print INPUT TYPE='hidden' NAME='User_name' VALUE='$UserName'; print input type=hidden name=User_Num' value='$UserNum'; print input type=hidden name='User_Addres' value='$Address'; ? trth align=left Name:/thtd ? print $User_name; ?/td/tr trth align=left User Number:/thtd ? print $User_Num; ?/td/tr trth align=left Address:/th td ? print$User_Addres; ?/td/tr ... input type=submit value='submit' /table /FORM .. Kind Regrads Adam I hope someone can help me. Thanks _ Stay in touch with absent friends - get MSN Messenger http://www.msn.co.uk/messenger -- 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] Need some HELP (not works)
Is it User_name or UserName? Also try this string with 'UserName' in single quotes. print INPUT TYPE=\hidden\ NAME=\User_name\ VALUE=$_POST['UserName']\; I've tried but still does not work. when I put the varaibles name between ' ' I found this error: Parse error: parse error, expecting `T_STRING' or `T_VARIABLE' or `T_NUM_STRING'. I typed in this way print INPUT TYPE='hidden' NAME='User_name' VALUE='$_POST[UserName]'; but nothing a new. I have checked register_globals is on. in my first page I used session_start(); and global $HTTP_SESSION_VARS; I dont know if this cause the problem that I had. Thanks again and more help please. Adam -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Serious Problem: Eval'ing conditioned blocks!
I had something like this too.
I used something other for the start and ending blocks.
# #
Then had a replace() section.
Hi there,
I (still :-) ) have the code of my php pages stored in a database. A main
page parses this code using the eval_html function postet at php.net.
This
used to work fine, but then I tried to employ conditioned blocks, e.g.
? while (foo) { ?
pHello World/p
? }?
The function I use seperates this code in three parts (1st: ? while (foo)
{ ?) and tries to eval every single one. This of course causes an error
because neither while (foo) { nor } is valid statement.
Anybody knows a workaround (that doesn't mean putting all HTML output in
PHP
echo commands) ??
Thanks
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Re: [PHP-DB] Re: how to parse php code stored in databases?
Torsten is right. eval() works great ... and so does having your php code in the database :-) Florian Wagner [EMAIL PROTECTED] wrote in message news:[EMAIL PROTECTED] Hi there, I thought it would be a great idea to store the code of nearly all my web pages in text fields of a mysql database. Now I've got the problem that I want to use php in those pages. Let's say $row[content] contains the string ? echo 'Hello World!' ?. Then echo $row[content] only produces the unparsed ? echo 'Hello World!' ? which effectivly displays nothing, because the browser doesn't know the ? tag. How can I tell PHP to parse this string before displaying it??? Thanks for any suggestions... This can be done with eval(): http://de.php.net/manual/en/function.eval.php Take a look at the user comments there, especially regarding the inclusion of ? and ?. Regards, Torsten
Re: [PHP-DB] mysql - SELECT DISTINCT problem
Very odd. I would think it would return every record seeing as the auto_increment ID field is different for every record. I use SELECT DISTINCT tip, marca, model FROM modele to select records without duplicates on the field tip. This works ok, BUT if I use SELECT DISTINCT tip, marca, model, id FROM... (id is auto_increment and is the table's primary key) my query won't produce any result. ANyone has a clue? Thank you! -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Image / file uploader
I think you need the quotes: $_SERVER['PHP_SELF'] Try using the following: $_SERVER[PHP_SELF] Mainly, get rid of the quotes inside the brackets. Usually this error 'T_String' generally occurs when you have quotes ( or ') where they shouldn`t be. Give it a try if that thing still troubles you... Good luck -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] headers and if statements
Hum. Is $logIn null at times?!?!
?if($logIn != 1) {header(Location: loginError.php);}?
Why does this direct me to loginError.php even when $logIn = 1?
- Matt
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RE: [PHP-DB] headers and if statements
I think that would try to set $login=1.
Need two ==
Have you tried:
?php if (!$login=1) {header(Location: loginerror.php);)?
-Original Message-
From: Daniel Clark [mailto:[EMAIL PROTECTED]
Sent: Friday, April 30, 2004 4:19 PM
To: matthew perry
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] headers and if statements
Hum. Is $logIn null at times?!?!
?if($logIn != 1) {header(Location: loginError.php);}?
Why does this direct me to loginError.php even when $logIn = 1?
- Matt
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RE: [PHP-DB] headers and if statements
We've ALL done it before :-)
I am testing now testing it with the ==, silly mistake
-Original Message-
From: John W. Holmes [mailto:[EMAIL PROTECTED]
Sent: Friday, April 30, 2004 4:38 PM
To: Erik Meyer; PHP-DB
Subject: Re: [PHP-DB] headers and if statements
From: Erik Meyer [EMAIL PROTECTED]
Have you tried:
?php if (!$login=1) {header(Location: loginerror.php);)?
Uhmm... Have _YOU_ tried that???
= vs == ?
;)
---John Holmes...
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Re: [PHP-DB] Image / file uploader
I think you want single quotes around PHP_SELF.
$_SERVER['PHP_SELF']
I have a script where it uploads a image to directory. I have used
the script several times and it works when I send the form to another
page. The problem is I have changed the form to submit to itself and I
can't seem to get it to work.
echo ( form method='post' action='$_SERVER[PHP_SELF]'
encType='multipart/form-data');
The thing is all my other fields update except for the image. Further,
the image does not even show up if I echo the query out.
UPDATE users SET bio = 'Bio - Test', goals = 'Goals - Test', fav_race =
'Fav Races - test', fav_train = 'Traing -test', photo_name = '' WHERE
user_id = '1'
Here is the code of the image uploader and my query, perhaps someone
could take a look let me know what I'm missing.
Thanks - CH
input type='file' name='photo'
if($_POST[postback_bio])
{
include include/dbadmin.php;
//image uploader
$uploadpath = '/images/clients/';
$source = $HTTP_POST_FILES['photo']['tmp_name'];
$photo_name = $HTTP_POST_FILES['photo']['name'];
$dest = '';
if (($source != '') ($source != '')) {
$dest = $uploadpath.$photo_name;
if ($dest = '') {
if (move_uploaded_file($source, $dest)) {
echo (p id='message'Image and Bio has been
successfully
stored./p );
} else {
echo (p id='message'Image could not be
stored./p);
}
}
} else {
echo (p id='message'No new image supplied./p);
$photo_name = $oldimage;
}
//declare varibles
$user_id = $_POST[user_id];
$name = $_POST[name];
$bio = $_POST[bio];
$goals = $_POST[goals];
$fav_race = $_POST[fav_race];
$fav_train = $_POST[fav_train];
$photo = $_POST[photo_name];
$query = UPDATE users SET bio = '$bio', goals = '$goals', fav_race =
'$fav_race', fav_train = '$fav_train', photo_name = '$photo_name'
WHERE user_id = '$user_id';
echo $query;
$msg = span style='color:red' id='message'Could not update
record/span;
$result = mysql_query($query, $db);
}
__
Craig Hoffman - eClimb Media
v: (847) 644 - 8914
f: (847) 866 - 1946
e: [EMAIL PROTECTED]
w: www.eclimbmedia.com
_
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Re: [PHP-DB] Image / file uploader
Change to double quotes for the HTML portion.
action=\$_SERVER['PHP_SELF']\
When I put single quotes in the PHP_SELF and I get this error:
Parse error: parse error, unexpected T_ENCAPSED_AND_WHITESPACE,
expecting T_STRING or T_VARIABLE or T_NUM_STRING
and I can't use double quotes because its in an echo statement. What
am I missing?
echo (form method='post' action='$_SERVER['PHP_SELF']'
encType='multipart/form-data'
Any more thoughts?
CH
__
Craig Hoffman - eClimb Media
v: (847) 644 - 8914
f: (847) 866 - 1946
e: [EMAIL PROTECTED]
w: www.eclimbmedia.com
_
On Apr 29, 2004, at 6:49 PM, Daniel Clark wrote:
I think you want single quotes around PHP_SELF.
$_SERVER['PHP_SELF']
I have a script where it uploads a image to directory. I have used
the script several times and it works when I send the form to another
page. The problem is I have changed the form to submit to itself and
I
can't seem to get it to work.
echo ( form method='post' action='$_SERVER[PHP_SELF]'
encType='multipart/form-data');
The thing is all my other fields update except for the image.
Further,
the image does not even show up if I echo the query out.
UPDATE users SET bio = 'Bio - Test', goals = 'Goals - Test', fav_race
=
'Fav Races - test', fav_train = 'Traing -test', photo_name = '' WHERE
user_id = '1'
Here is the code of the image uploader and my query, perhaps someone
could take a look let me know what I'm missing.
Thanks - CH
input type='file' name='photo'
if($_POST[postback_bio])
{
include include/dbadmin.php;
//image uploader
$uploadpath = '/images/clients/';
$source = $HTTP_POST_FILES['photo']['tmp_name'];
$photo_name = $HTTP_POST_FILES['photo']['name'];
$dest = '';
if (($source != '') ($source != '')) {
$dest = $uploadpath.$photo_name;
if ($dest = '') {
if (move_uploaded_file($source, $dest)) {
echo (p id='message'Image and Bio has been
successfully
stored./p );
} else {
echo (p id='message'Image could not be
stored./p);
}
}
} else {
echo (p id='message'No new image supplied./p);
$photo_name = $oldimage;
}
//declare varibles
$user_id = $_POST[user_id];
$name = $_POST[name];
$bio = $_POST[bio];
$goals = $_POST[goals];
$fav_race = $_POST[fav_race];
$fav_train = $_POST[fav_train];
$photo = $_POST[photo_name];
$query = UPDATE users SET bio = '$bio', goals = '$goals', fav_race =
'$fav_race', fav_train = '$fav_train', photo_name = '$photo_name'
WHERE user_id = '$user_id';
echo $query;
$msg = span style='color:red' id='message'Could not update
record/span;
$result = mysql_query($query, $db);
}
__
Craig Hoffman - eClimb Media
v: (847) 644 - 8914
f: (847) 866 - 1946
e: [EMAIL PROTECTED]
w: www.eclimbmedia.com
_
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Re: [PHP-DB] First letter
$rest = substr(abcdef, 1, 3); // returns bcd http://www.phpbuilder.com/manual/function.substr.php What PHP function returns just the first letter of a string? - Matt -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Session
Depends on the browser. IE and Netscsape handle opening another window differently. Hi, I have a problem in handling session... so I just reply in this topic. I have created a site... when user login from one browser ..i set a session['user'] the problem is that when user open another browser ...he is not login ...he have to login again... So that 1 pc may login as 2 or more different user from different browser... Nicholas Sk2 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: [PHP] What's wrong with this IF statement?
What about removing the quotes around the numbers.
if ($cat_id != 53 || $cat_id != 54 || $cat_id != 55 etc...
My IF statement should be picking up on the numbers, and if the number
matches not be displaying out the information, however I look at the
outputted page and the information is still there, what have I got wrong
on the code?
CODE SNIPPET
//Show categories first
$get_cats = select id_num, id_name, description, cat_code from
categories order by id_name;
$get_cats_res = mysql_query($get_cats) or die(mysql_error());
if (mysql_num_rows($get_cats_res) 1)
{
$display_block = PemSorry, no categories to browse./em/P;
}
else
{
while ($cats = mysql_fetch_array($get_cats_res))
{
$cat_id = $cats[id_num];
if ($cat_id != 53 || $cat_id != 54 || $cat_id != 55 || $cat_id
!= 117 || $cat_id != 118 || $cat_id != 74)
{
$cat_title = strtoupper(stripslashes($cats[id_name]));
$cat_desc = stripslashes($cats[description]);
$display_block .= stronga
href=$_SERVER[PHP_SELF]?cat_id=$cat_id$cat_title
$cat_desc/a/strongbr\n;
while ($items = mysql_fetch_array($get_items_res))
{
$item_id = $items[id];
$item_num = $items[item_num];
$item_desc = stripslashes($items[description]);
if ($item_num != ABC-R37 || $item_num != ABC-R42 || $item_num
!= HB-99100 || $item_num != RO-PUMPS || $item_num != ML-HDGALJUG
|| $item_num != PFS-CAC21 || $item_num != PFS-CO2)
{
$item_num = ltrim($item_num);
$item_num = rtrim($item_num);
$display_block .= emstronga
href=\catalog/$item_id.html\$item_num/a/strong -
$item_desc/embr\n;
}
}
END SNIPPETS
My assumption is that BOTH IF statements are not working correctly since
the logic is that if they are built the same they would react the same.
HELP!
TIA!
Robert
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Re: [PHP-DB] logic problem
How about: SELECT convert( varchar,eventime,110) as date from events, badge, count(convert( varchar,eventime, 110)) as count WHERE events.cardnum = badge.id and devid = '1' and convert( varchar, events.eventime, 110) BETWEEN '$startdate' and 'enddate' AND type = '1' GROUP BY convert( varchar,events.eventime, 110), badge I'm using some proprietary software/hardware where when a visitor swipes their entry card, it is recorded in a mssql 2000 server. My boss wants a count of unique vistors for a range of dates. So, I need to have it give a count of unique vistors, meaning that I need to count all vists for a day as one visit (because if they go outside to smoke and come back and swipe their card again to get in, each one is a separate visit, but i need to count all visits by each person as one visit since i just want to know if they came at all each day, not how many times they came in). This is my SQL statement: select distinct count(convert( varchar,eventime, 110)) as count, convert( varchar,eventime,110) as date from events, badge wher events.cardnum = badge.id and devid = '1' and convert( varchar, events.eventime, 110) BETWEEN '$startdate' and 'enddate' and type = '1' group by convert( varchar,events.eventime, 110) for reference, devid = '1' is the hardware device, where everytime it triggers, it means someone swiped their card to get in, and type = '1' means patron (because we have a type = 2 that is for staff and we jsut want to know how many patrons visited) When I execute this statement, its returning the result for each date of the total number of card swipes (so if a person comes in twice on a date, its recording it as 2 swipes, but I just need to know that they came to the building at all on this date, so I just need it to register that there was a count of atleast one for this card that was swiped) any suggestions? thanks -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] logic problem
AND: any count =1 shows they came in that day. How about: SELECT convert( varchar,eventime,110) as date from events, badge, count(convert( varchar,eventime, 110)) as count WHERE events.cardnum = badge.id and devid = '1' and convert( varchar, events.eventime, 110) BETWEEN '$startdate' and 'enddate' AND type = '1' GROUP BY convert( varchar,events.eventime, 110), badge I'm using some proprietary software/hardware where when a visitor swipes their entry card, it is recorded in a mssql 2000 server. My boss wants a count of unique vistors for a range of dates. So, I need to have it give a count of unique vistors, meaning that I need to count all vists for a day as one visit (because if they go outside to smoke and come back and swipe their card again to get in, each one is a separate visit, but i need to count all visits by each person as one visit since i just want to know if they came at all each day, not how many times they came in). This is my SQL statement: select distinct count(convert( varchar,eventime, 110)) as count, convert( varchar,eventime,110) as date from events, badge wher events.cardnum = badge.id and devid = '1' and convert( varchar, events.eventime, 110) BETWEEN '$startdate' and 'enddate' and type = '1' group by convert( varchar,events.eventime, 110) for reference, devid = '1' is the hardware device, where everytime it triggers, it means someone swiped their card to get in, and type = '1' means patron (because we have a type = 2 that is for staff and we jsut want to know how many patrons visited) When I execute this statement, its returning the result for each date of the total number of card swipes (so if a person comes in twice on a date, its recording it as 2 swipes, but I just need to know that they came to the building at all on this date, so I just need it to register that there was a count of atleast one for this card that was swiped) any suggestions? thanks -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] logic problem
How about SELECT distinct badge.id, convert( varchar,eventime,110) as date Shows all the badge numbers IN on that date. If the badge number is not there, they didn't check in at ALL that day. This seems too easy to not be able to do it with SQL. There must be something we're missing in the query. Try this: select distinct badge.id,convert( varchar,eventime,110) as date,count(*) from events, badge where events.cardnum = badge.id and devid = '1' and convert( varchar, events.eventime, 110) BETWEEN '$startdate' and 'enddate' and type = '1' group by date In MySQL you don't need to specify a field for count(). And you should have to convert the eventime field twice, you should be able to reference the calc field in the group by. On Apr 21, 2004, at 2:19 PM, Adam Williams wrote: Yeah I basically had that with my previous SQL statement, I was grouping by event.cardnum instead of counting the cardnums by date. I think what I'm trying to do is beyond the scope of SQL and I'll have to write some PHP to take the SQL statement results and feed them into an array and count the distinct cardnums for each date and then spit it all into an html table. thanks On Wed, 21 Apr 2004, Daniel Clark wrote: AND: any count =1 shows they came in that day. How about: SELECT convert( varchar,eventime,110) as date from events, badge, count(convert( varchar,eventime, 110)) as count WHERE events.cardnum = badge.id and devid = '1' and convert( varchar, events.eventime, 110) BETWEEN '$startdate' and 'enddate' AND type = '1' GROUP BY convert( varchar,events.eventime, 110), badge I'm using some proprietary software/hardware where when a visitor swipes their entry card, it is recorded in a mssql 2000 server. My boss wants a count of unique vistors for a range of dates. So, I need to have it give a count of unique vistors, meaning that I need to count all vists for a day as one visit (because if they go outside to smoke and come back and swipe their card again to get in, each one is a separate visit, but i need to count all visits by each person as one visit since i just want to know if they came at all each day, not how many times they came in). This is my SQL statement: select distinct count(convert( varchar,eventime, 110)) as count, convert( varchar,eventime,110) as date from events, badge wher events.cardnum = badge.id and devid = '1' and convert( varchar, events.eventime, 110) BETWEEN '$startdate' and 'enddate' and type = '1' group by convert( varchar,events.eventime, 110) for reference, devid = '1' is the hardware device, where everytime it triggers, it means someone swiped their card to get in, and type = '1' means patron (because we have a type = 2 that is for staff and we jsut want to know how many patrons visited) When I execute this statement, its returning the result for each date of the total number of card swipes (so if a person comes in twice on a date, its recording it as 2 swipes, but I just need to know that they came to the building at all on this date, so I just need it to register that there was a count of atleast one for this card that was swiped) any suggestions? thanks -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] How to Erase MySQL table.
TRUNCATE table xxx will delete all the rows of data. DROP table xxx will delete the table. Hi everyone, I can not find in my manual a MySQL command which erase permanently a table and it?s content from a MySQL database. Thanks in advance Charalambos _ MSN 8 helps eliminate e-mail viruses. Get 2 months FREE*. http://join.msn.com/?page=features/virus -- 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] get rid of the HTML tags
Try strip_tags() http://www.phpbuilder.com/manual/function.strip-tags.php I have this string: $my_string = bhello world/ba href='teste.php'this is a test for a link with a image img src='test.jpg'/a Is there any function that retrieves only the string without the html tags?! -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] [PMX:55%] Re: [PHP-DB] get rid of the HTML tags
I've been getting the same errors, sometimes days latter. Why are my posts not getting through? I got 3 Delivery Report Failures on Thursday all due to Diagnostic was Unable to transfer, Message timed out Information Message timed out -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Session_Start won't work Help?
In PHP.INI there is a variable session.save_path. Make sure it's uncommented and has a directory to store the temp session files on the web server. I'm running Apache/2.0.49 and PHP/4.3.5 on windows XP. I get warnings: WARNING: session_start(): open/tmp\sess_242f6f668d85d512f9f4379ffa1e1, O_DWR failed: No such file or directory (2) in c:\program files\apache group\apache2\htdocs\register_new.php on line 10 I also get another similar warning for seesion start that says cannot send seesion cookie and another that says cannot send session cache. Is this some sort of configuratrion problem in my httpd.cong file or my php.ini file? HELP? -- 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] password input type
I'm pretty sure it just shields the on looker from viewing. View source in the browser still shows what the VALUE is, if any. Does using a pasword input type input name=password type=password make the transfer more secure from someone sniffing my connection or does it only shield an onlooker from seeing what the user enters? - Matt -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Php datetime
I've heard you want to use a TIMESTAMP field for that. i have created a mysql table where i would like to store the date and time whenever a user logs into a web site. the login is done through php/mysql as well but i'm not sure what the php code is to add the current date/time into a mysql database. the field in the db is created as a 'datetime' field. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] mysql_num_rows
$query =SELECT username FROM users WHERE username=('$username');
// $result = mysql_query($query, $db);
Above line should be uncommented.
if (!$query)
And the if checking the number of rows returned from the $result set.
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Re: [PHP-DB] INSERT dynamic values problem - Need help urgently!
Try:
OPTION VALUE=\$row['class_code']\ .$row[class_code]. /option;
while ($row = mysql_fetch_array($sql))
{
print OPTION VALUE=\$class_code\ .$row[class_code]. /option;
}
$result = $db-query($sql);
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RE: [PHP-DB] What's wrong with this query?
What about the single quotes? Might try this. $sql = INSERT INTO log SET term=\'$search\', returns=\'$arrayword\', time=CURTIME(), date=CURDATE(), ip=\'$ip\'; $sql = INSERT INTO log SET term='$search', returns='$arrayword', time=CURTIME(), date=CURDATE(), ip='$ip'; echo $sql; $logit = mysql_query($sql) or die(mysql_error()); -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Oracle and upper case field name
I might try having a common PHP function that converts all field names to lower case, and call that function with every field name. Or call each field by column number, $row[1] . When Oracle returns the result set, the field names are all in upper case. However, I have an application that needs to work on both MSSQL and Oracle. I don?t want to keep 2 copies. One has upper case field names and the other lower case. How could I make PHP or ADODB to work with lower case field names even the result set is upper case? -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Oracle and upper case field name
Yeah, Looks good to me :-) Like http://us2.php.net/manual/en/function.array-change-key-case.php ?? When Oracle returns the result set, the field names are all in upper case. However, I have an application that needs to work on both MSSQL and Oracle. I don?t want to keep 2 copies. One has upper case field names and the other lower case. How could I make PHP or ADODB to work with lower case field names even the result set is upper case? ---John Holmes... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
