Need Help In Php (p.s. Yarrgh)

Dear yarrgh thankx for your elp before but i caught a ssyntax error in the coding.

The page displays but instead of the info there a long syntax error by mysql? The problem was with the array codin to get the array from mysql!

Heres the problem,

Warning:  mysql_fetch_array(): supplied argument is not a valid MySQL result resource in F:\XAMPP\apachefriends\xampp\htdocs\xampp\all.php on line 137

So you think you can help me on this?

Thankx,

VBLK

tags.

Can you post the code you are using? Make sure you include line 137 and the mysql stuff before that. I need to look to make sure you didn't change it or I didn't post it right.

This error occurs when you attempt to retrieve data from a failed MySQL query. For example, you may have something like:

$result = mysql_query('SELECT X FROM Y');$data = mysql_fetch_array($result);

$result = mysql_query('SELECT X FROM Y');if( $result ) {   $data = mysql_fetch_array($result);}

I've never had that problem. Right now my tables are empty and mysql_fetch_array() doesn't give any errors.

EDIT:

I've just noticed that my results are in loops. For example:

<?php...$a = mysql_query("SELECT * FROM `news`");$b = mysql_num_rows($a);if ($b > 0) {   ...}...?>

<?php...$a = mysql_query("SELECT * FROM `news`");while($blah = mysql_fetch_array($a)) {   ...}?>

Maybe I'm right? Maybe? MAYBE?

Tables being empty doesn't mean that the query will fail. Tables not existing or explictly specified fields not existing within tables will. If you have error reporting turned off, you will not see any errors - even with the method you mentioned, if you had full error reporting turned on, you would see an error at least once (during the first evaluation of the while() loop's condition).

For example:

mysql_query('CREATE TABLE test_table (test_field tinyint(1))');$result = mysql_query('SELECT test_field FROM test_table');$data = mysql_fetch_assoc($result);

Dear Yarrgh,
Hey i read your's and spectres's posts on how to fix it but the unfortunate thing is i don't get a single thing spectre's telling.(maybe it's because i'm dumb ) But hey i've given all my coding below, can you edit that coding and post the corrected coding?

<?php 					//check to see if they are trying to view a certain projectif (isset($_GET["id"])) {	$id = $_GET["id"];	// Connect to the mysql and select the database	mysql_connect ( "localhost" , "" , "" );	mysql_select_db( "projects" );	// Get the data from the database   $a = mysql_query ( "SELECT * from `PROJECTS` WHERE `id`='".$id."'");	// Put the info into an array for easy use	$info = mysql_fetch_array($a);}else {   // Some other code to list the projects}?>

Try this. I know I'm not 'Yarrgh', but I'm the next best thing... honest...

You may want to consider structuring your code a little better, too (I understand that it may have been structured and good old IPB screwed up the indentation - but if not, structure, structure, structure!) - it makes it infinately easier for others to quickly analyze and modify your code.

Also, can you please edit your post and correct the 'code' bbCode tags.

<?php //check to see if they are trying to view a certain projectif (isset($_GET["id"])) {$id = $_GET["id"];// Connect to the mysql and select the databasemysql_connect ( "localhost" , "" , "" );mysql_select_db( "projects" );// Get the data from the database$a = mysql_query("SELECT * from PROJECTS WHERE id = '$id'");// Put the info into an array for easy useif( $a ) {   $info = @mysql_fetch_array($a);} else {   // The script will process this if the query fails.   // If you want, put something here - such as an error message.}}else {// Some other code to list the projects}?>

Sorry it takes me so long to respond. I have school and work. Looks like I was beat to it.

I guess the first time I could've gave you a code that is more user friendly with the error handling just in case users try to view a project that doesn't exist or what ever. I'd say instead of just displaying an error if it doesn't exist to list the projects with an error near the top telling them that the requested project didn't exist.

//check to see if they are trying to view a certain projectif (isset($_GET["id"])) {   $id = $_GET["id"];     // Connect to the mysql and select the database   mysql_connect ( "localhost" , "" , "" );   mysql_select_db( "projects" );   // Get the data from the database   $a = mysql_query("SELECT * from PROJECTS WHERE id = '$id'");   // Put the info into an array for easy use   if( $a ) {	 $info = @mysql_fetch_array($a);   } else {	 list_projects("project doesn't exist");   }}else {   list_projects("");}function list_projects($error) {  //code to list the projects. $error is the variable to display the error.}

Thankx spectre for the help and sorry for disturbing you Yarrgh! well thankx everybidy who helped in!

Its ok. I just noticed I read spectre's first post wrong. lol My bad. I though you had the tables already set up. Sorry I couldn't help this time.