[SOLVED] Strange problem.. can't get mysql_fetch_array() to work?

[kernelgpf]

Here's my entire script- it's worked previously, and is suddenly not working.. no errors, nothing being printed. That's the problem.

 

<?php
include "header.php";
print "Searching usernames is searching all player's BASIC names, not their display names.<p>";
if($_GET['action'] == "search"){
$name=$_POST['username'];
$idnumber=$_POST['idnumber'];
$acctkind=$_POST['acctkind'];
$onstatus=$_POST['onstatus'];

if($acctkind == "Premium"){
$acctkind=1;
}
if($acctkind == "normal"){
$acctkind=0;
}
if($acctkind == "Doesn\'t Matter"){
$acctkind="";
}

if($onstatus == "Doesn\'t Matter"){
$query=mysql_query("select displayname,id from users where basicname LIKE '$name%' AND id LIKE '$idnumber%' AND acctkind LIKE '$acctkind%' ORDER BY id ASC")or die(mysql_error());
}

if($onstatus == "Online"){
$query=mysql_query("select displayname,id from users where basicname LIKE '$name%' AND id LIKE '$idnumber%' AND acctkind LIKE '$acctkind%' AND lli > (NOW() - INTERVAL 2 MINUTE)")or die(mysql_error());
}

if($onstatus == "Offline"){
$query=mysql_query("select displayname,id from users where basicname LIKE '$name%' AND id LIKE '$idnumber%' AND acctkind LIKE '$acctkind%' AND lli < (NOW() - INTERVAL 2 MINUTE)")or die(mysql_error());
}


$num=mysql_num_rows($query);
if($num == "0"){
print "No matches to your search query.";
include "footer.php";
exit;
}

while($thing=mysql_fetch_array($query)){
print "<a href='userprofiles.php?id=$thing[id]'>$thing[displayname]</a><br>";
}
include "footer.php";
exit;
}
?>





<form method=post action=searchusers.php?action=search>
Username: <input type=text name=username><br>
Player ID#: <input type=text name=idnumber><br>
Account Type: <select name=acctkind>
<option>Doesn't Matter</option>
<option>Premium</option>
<option>normal</option></select><br>
Online/Offline: <select name=onstatus>
<option>Doesn't Matter</option>
<option>Online</option>
<option>Offline</option>
</select><br>
<input type=submit name=submit value='Search Users'>
</form>



<?php
include "footer.php";
?>

[darkfreaks]

try mysql_fetch_assoc

[kernelgpf]

Tried- didn't work.

[darkfreaks]

try

 

<?php

print '<a href=userprofiles.php?id='.$thing[id].'>'.$thing[displayname].'</a><br>';?>

[pocobueno1388]

Have you tried to echo some text between the

if ($_GET['action'] == "search")

brackets?

 

Maybe your script isn't even reaching that part of the code when you press search.

[darkfreaks]

no poc its bad dynamic linkcode 

[kernelgpf]

Doesn't work either, darkfreaks.

 

And Colin- it's getting there, I've printed some stuff out.

[kernelgpf]

As an extra helpful note, it's printing out "<a href='userprofiles.php?id='></a>".. no variables. =[

[darkfreaks]

<?php

print '<a href=userprofiles.php?id='. $thing['id'] .'>'. $thing['displayname'] .'</a><br>';?>

 

 

[kernelgpf]

Caused a parse error.

[darkfreaks]

<?php

print '<a href=userprofiles.php?id='. $thing['id'] .'>'. $thing['displayname'] .'</a><br>';?>

[pocobueno1388]

Okay, try this:

 

EDIT Don't try it yet...hold on.

EDIT AGAIN Okay, try it now.

 

<?php
include "header.php";

print "Searching usernames is searching all player's BASIC names, not their display names.<p>";

if ($_GET['action'] == "search") {
    $name=$_POST['username'];
    $idnumber=$_POST['idnumber'];
    $acctkind=$_POST['acctkind'];
    $onstatus=$_POST['onstatus'];
    
    if ($acctkind == "Premium") $acctkind=1;
    if ($acctkind == "normal") $acctkind=0;
    if ($acctkind == "Doesn\'t Matter") $acctkind="";
    
    if ($onstatus == "Doesn\'t Matter")
        $query= "select displayname,id from users where basicname LIKE '%$name%' AND id LIKE '%$idnumber%' 
        AND acctkind LIKE '%$acctkind%' ORDER BY id ASC";
    
    else if ($onstatus == "Online")
        $query= "select displayname,id from users where basicname LIKE '%$name%' AND id LIKE '%$idnumber%' 
        AND acctkind LIKE '%$acctkind%' AND lli > (NOW() - INTERVAL 2 MINUTE)";
    
    else if ($onstatus == "Offline")
        $query= "select displayname,id from users where basicname LIKE '%$name%' AND id LIKE '%$idnumber%' 
        AND acctkind LIKE '%$acctkind%' AND lli < (NOW() - INTERVAL 2 MINUTE)";
        
     $result = mysql_query($query)or die(mysql_error());
   
     
    $num = mysql_num_rows($result);
    
    if ($num == "0") {
        print "No matches to your search query.";
        include "footer.php";
        exit;
    }
    
    while ($thing = mysql_fetch_assoc($result)) {
        print "<a href='userprofiles.php?id={$thing['id']}'>{$thing['displayname']}</a><br>";
    }
    
    include "footer.php";
    exit;
}
?>


<form method="post" action="searchusers.php?action=search">
Username: <input type="text" name="username"><br>
Player ID#: <input type="text" name="idnumber"><br>
Account Type: <select name="acctkind">
<option>Doesn't Matter</option>
<option>Premium</option>
<option>normal</option></select><br>
Online/Offline: <select name="onstatus">
<option>Doesn't Matter</option>
<option>Online</option>
<option>Offline</option>
</select><br>
<input type="submit" name="submit" value="Search Users">
</form>



<?php
include "footer.php";
?>

[kernelgpf]

Parse error, line 22... the first couple queries you have just the query's guts in the var. $query, but in others, you have the guts and the "mysql_query("");".. which is meant to be?

[pocobueno1388]

You must have missed my edit? Try it again, it's updated.

[kernelgpf]

-nods-

 

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 'Resource id #6' at line 1

[pocobueno1388]

Okay, change this line

$result = mysql_query($query)or die(mysql_error());

 

To

$result = mysql_query($query)or die(mysql_error().$query);

 

Then copy and paste what it prints.

 

In all your queries you had

LIKE '$var%'

 

I looked up the syntax for LIKE and they had a percent sign on both sides of the variable...so I changed that. You might be able to do it with the one as well...you might want to change it to one just to make sure, and see if it works.

[kernelgpf]

It works that way too. Two percent signs makes it so it searches all user IDs with a "1" in them, while my way only brings up account IDs that START with 1.

 

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 'Resource id #6' at line 1Resource id #6

 

[pocobueno1388]

It should have printed out the darn query...

 

Try putting this:

echo $query.'<p>';   
$result = mysql_query($query)or die(mysql_error().' ERROR: '.$query);

[kernelgpf]

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 'Resource id #6' at line 1ERROR:Resource id #6

 

What the heck is "resource ID #6"? O_o

[pocobueno1388]

I wonder why the echo $query.'<p>';  didn't show up 0_o

 

I'm sure it has something to do with all those LIKE statements in there. I would post your query and error in the MySQL forum. Fenway, Barand or some other super guru will come by and probably correct it in no time.

[kernelgpf]

I didn't paste that part, sorry.

 

I don't have time.. I'll keep messing with it. Thanks for the help anywho.. =S

[kernelgpf]

I figured it out- when I switched servers, the $onstatus variable didn't have the backslash in it. =S

 

Thanks for your help though. =]