[RESOLVED] a REALLY weird problem

thepeccavi

I have the STRANGEST problem...

I finished creating a page to edit a table in my database....... Tesed it.... OK! right! now, I want to create one that edits the SAME FIELDS (by name) just on another table....

Ctrl+A... Ctrl+N... Ctrl+V! now find and replace the table names....... done!#

Testing...... UUUUH???!?!?!?! Error:

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /<path>/bsc/admin/editpage.php on line 77

WTF!!!??!?!?!

This is my code... last time I had not selected the DB (dohhh!) At the top of the page I connect to the DB (this page handles a form TWICE!!!):

// Require config
require('../config.php');

// If config has no password
if(empty($mysqlPassword)){
	// Connect to DB without password
	$mysqlConnect = mysql_connect($mysqlHost, $mysqlUser);
}

// If config HAS a password
else{
	// Connect to DB with password
	$mysqlConnect = mysql_connect($mysqlHost, $mysqlUser, $mysqlPassword);
}

// Select DB
mysql_select_db($mysqlDB);

// If a form has been submitted
if(isset($_POST[submit])){

// Get form data
$teamname = $_POST[team];
$newContent = $_POST[content];
$newDisplayName = $_POST[displayname];
}

....Processes one form (sending stuff to the D😎....

else{
// Create form
// Prepare query
$getContent = 'SELECT displayname, content FROM ' . $mysqlTablePrefix . 'content WHERE teamname = \'' . $teamname . '\'';

// Run query
$getContentResults = mysql_query($getContent);

// Print into form
while($content = mysql_fetch_assoc($getContentResults)){ // LINE 77
	printf("Display Name: <input type=\"text\" name=\"displayname\" value=\"%s\" /><br />\nInfo:<br />\n<textarea name=\"content\" rows=\"10\" cols=\"40\">%s</textarea><br /><br />", $content[displayname], $content[content]);
}
echo '<input type="hidden" name="team" value="' . $teamname . '" />' . "\n";
echo '<input type="submit" name="submit" value="Finish" />';
}

UUUHH?!??!?!?!! WTF is wrong????

This is SO weird because the same script (well.. different table name) STILL WORKS!!!

thorpe

For starters. Learn to debug, test your query actually works before trying to use the results. Try...

$getContentResults = mysql_query($getContent) or die(mysql_error()."sql = ".$getContent);

Your query looks very dodgy.

thepeccavi

Thank you... I have been SUCH a ****ing git!!! that is UNBELIVABLE!!! HOW could I not check that all the fields exist??? oooh.. dear... no field "displayname"

oops!

Thank you again

hagen00

dude! You get even more worked up than me when you code. And that tells you something!!!

thepeccavi

OK... this is odd too...

I am connected to the DB (see origional post) but now this is wrong... how should I debug it?

// If $newContent is not set yet...
if(!isset($newContent)){

// GET THE USER TO SET IT!!!
// Create form
// Prepare query
$getContent = 'SELECT content FROM ' . $mysqlTablePrefix . 'content WHERE teamname = \'' . $teamname . '\'';

// Run query
$getContentResults = mysql_query($getContent) or die(mysql_error()."sql = ".$getContent);

// Print into form
while($content = mysql_fetch_assoc($getContentResults)){
	printf("<p>Content:</p>\n<textarea name=\"content\" rows=\"40\" cols=\"80\">%s</textarea><br /><br />", $content[content]);
}
echo '<input type="hidden" name="team" value="' . $teamname . '" />' . "\n";
echo '<input type="submit" name="submit" value="Finish" />';
}
?>

I get my

echo '<input type="hidden" name="team" value="' . $teamname . '" />' . "\n";
echo '<input type="submit" name="submit" value="Finish" />';

nicely, but the rest of the code does not do anything! not even an error!

What is up now?

thepeccavi

hi hagen... wel.. im not normally THAT worked up.. I just get really annoyed with myself when i do something THAT obvious wrong...lol

hagen00

hehe, ok.

In terms of debugging your code. Just echo out your variables all over the page and see if they say what they should. Put your SQL statement into your SQL manager (phpmyadmin?) and test if it works.

I found the best way to debug is just to echo stuff out on the page. Go throught the lines step by step until you see something you don't expect.

We can't debug your code for you because we don't know what the values of the variables should be.

Good luck

thepeccavi

Right i have added

echo '$content[content]: ' . $content[content];

to my while() statment.... nothing appears... it seems that the while statment does not execute??!!! I a now OFFICIALLY confused!

thepeccavi

This is now fixed... as the while() statement was only going to print one thing (heh heh) I have now got rid of the while() and just got it to print the results:

Old code:

while($content = mysql_fetch_assoc($getContentResults)){
printf("something here that i can't remember off-hand...", $content[content]);
}

New code:

// Print into form
// Make $content an associative array (only containin content... lol)
$content = mysql_fetch_assoc($getContentResults);

// Echo a label for and a textarea containing the current code...
echo '<p>Content:</p>' . "\n" . '<textarea name="content" rows="40" cols="80">' . $content[content] . '</textarea><br /><br />';

Sorry about all this messing around

hagen00

No, that's good! That's how you debug! Well done.