Hello can any one help with this one? There is only one line of code that does not work!!!!

PHP Code:

        $sql = "SELECT * FROM adddb WHERE nameof LIKE '$comment' ORDER BY nameof";
        
$result = mysql_query($sql, $conn) or die(mysql_error());
while ($newArray = mysql_fetch_array($result)) {
     
    $nameof = $newArray['nameof'];
    $why = $newArray['why'];
        
    echo "  
    <b>Name:</b><br>             
    $nameof<br><br>
    <b>Why?</b><br>$why<br>";
}
?>



         <?php
    
        [B]$sql = "SELECT * FROM comments WHERE nameid = '$nameof'";[/B] 
// This One
$result = mysql_query($sql, $conn) or die(mysql_error());
while ($newArray = mysql_fetch_array($result)) {
     
    $user = $newArray['user'];
    $comment = $newArray['comment'];

    echo "  
    <b>Name:</b><br>             
    $user<br><br>
    <b>Comments:</b><br>  
    $comment<br><hr>";
}
?>

Care to tell us in what way it doesn't work?

which line? and whats the error?

My quick guess is this one:

PHP Code:

[B]$sql = "SELECT * FROM comments WHERE nameid = '$nameof'";[/B] 


remove the [ b ] and [ / b ]

er, I think they were trying to point out with bbcode that that was the offending line. You can't nest bold tags inside the [php ] tags. Also the comment //this one is a dead giveaway.

I can't see much wrong with the SQL itself - what problem are you having? Does it just not select anything?

Oh yeah I see it now, lol like I said it was a quick guess. Simjay, any information you could give as to what error you're getting would help. The answer you get is only as good as the information you give.

Where you have LIKE '$comment' are you really wanting LIKE '%$comment' or LIKE '$comment%' or even LIKE '%$comment%'. Don't you want a wildcard in there if you're using LIKE?

well all it has to do is display information from the database where nameid = $nameof in the nameid the data would be John Jackson. it only works if nameid = 0.