shows a blank page

demosthenes705

<?php
session_start();
?>
<html>
<head><title> Edit Sig</title></head>
<?php

$username = $_SESSION['username'];
$image = $_POST['image'];
$font = $_POST['font'];

$sql = mysql_query("UPDATE silent_aggressions SET font='$font', image='$image' WHERE username='$username'") or die mysql_error(;

$sql_doublecheck = mysql_query("SELECT * FROM silent_aggressions WHERE font='$font', image='$image'") or die mysql_error(;

$doublecheck = mysql_num_rows($sql_doublecheck);

if($doublecheck == 0){
echo "<strong><font color=red>Your Sig couldn't be changed!</font></strong>";
} elseif ($doublecheck > 0) {
echo "<strong>Your sig has been changed!!</strong><br />";

}

?>

and all that does is show a blank page

bretticus

try this...

<?php
session_start();
?>
<html>
<head><title> Edit Sig</title></head>
<?php

$username = $_SESSION['username'];
$image = $_POST['image'];
$font = $_POST['font'];

$sql = mysql_query("UPDATE silent_aggressions SET font='$font', image='$image' WHERE username='$username'") 
	or die(mysql_error());

$doublecheck = mysql_affected_rows();

if($doublecheck == 0){
echo "<strong><font color=red>Your Sig couldn't be changed!</font></strong>";
} elseif ($doublecheck > 0) {
echo "<strong>Your sig has been changed!!</strong><br />";

}

?>

bpat1434

Well, if this is a direct copy & paste of your code, you should get a parse error because you're missing a few closing ")".

Now, it seems that you are missing the <body></body> tags in the HTML.

~Brett

demosthenes705

thanks

i remember why i am only getting a blank page

apache2triad puts all errors in the php error log. and i've not found out how to turn it on so it shows on the page

ok i fixed it up and now im getting an sql error

You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near ' image='sig1'' at line 1

(new code with the edits)

<?php
session_start();
?>
<html>
<head><title> Edit Sig</title></head>
<?php
include ('db.php');

$AccountName = $_SESSION['username'];
$image = $_POST['image'];
$font = $_POST['font'];

$sql = mysql_query("UPDATE silent_aggressions SET font='$font', image='$image' WHERE AccountName='$AccountName'") or die (mysql_error());

$sql_doublecheck = mysql_query("SELECT * FROM silent_aggressions WHERE font='$font', image='$image'") or die (mysql_error());

$doublecheck = mysql_num_rows($sql_doublecheck);

if($doublecheck == 0){
echo "<strong><font color=red>Your Sig couldn't be changed!</font></strong>";
} elseif ($doublecheck > 0) {
echo "<strong>Your sig has been changed!!</strong><br />";

}

?>
</body>
</html>

and im pritty sure that there is something wrong with the double check. so if you could tell me how to rewrite that or something please do

bpat1434

Try instead of using the ", image='$image'" use "AND" instead of the "," so your SQL will look like:

SELECT * FROM silent_aggressions WHERE font='$font' AND image='$image'

~Brett

bretticus

If your goal (as it seems from your original feedback text) is to simply tell the user whether or not his or her record was updated or not, just use mysql_query. It will return true on a successful update. If you really do want to "doublecheck" as you say (if you do not trust PHP or something) then mysql_affected_rows will "ask" the database how many records were affected since the last query.