PHP Error

[Dentafrice1]

In My Dj Panel My Dj Says is *****ing up

Heres the error:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/habbonew/public_html/dj_staff/says.php on line 3

Heres my page code


<?php
$query = mysql_query("SELECT * FROM users WHERE username = '$logged[username]'");
$row = mysql_fetch_array($query);
echo "<b>".$row["username"]."</b></br>";
echo "<i>Says, ".$row["comment"]."</i><br><br>";
?>

Whats wrong

[timROGERS]

There's a problem with $query

[Dentafrice1]

<?php
$query = mysql_query("SELECT * FROM 'shoutout'");
$row = mysql_fetch_array($query);
echo "<b>".$row["username"]."</b></br>";
echo "<i>Says, ".$row["comment"]."</i><br><br>";
?>

There it is now What is the problem with query?

[craigg.]

<?php
$query = mysql_query("SELECT * FROM users WHERE username = '$logged[username]'");
$row = mysql_fetch_array("$query");
echo "<b>".$row["username"]."</b></br>";
echo "<i>Says, ".$row["comment"]."</i><br><br>";
?>

Also your second post is the same as the first?
Try that what I have ^

- Craig.

[Dentafrice1]

<?php
$query = mysql_query("SELECT * FROM 'shoutout'");
$row = mysql_fetch_array($query);
echo "<b>".$row["username"]."</b></br>";
echo "<i>Says, ".$row["comment"]."</i><br><br>";
?>

I edited it ;) i pasted the wrong thing

But can you help?? Theres something wrong :(

[Eric30]

<?php
$query = mysql_query("SELECT * FROM `shoutout`");
$row = mysql_fetch_array($query);
echo "<b>".$row["username"]."</b></br>";
echo "<i>Says, ".$row["comment"]."</i><br><br>";
?>

There ya go ;)

[nets]

$query = mysql_query('SELECT * FROM users WHERE username = \''.$logged[username].'\'');

[Dentafrice1]

TY Plus rep to both of you

[Splinter]

Wheres the connection? Also just change the $logged['username'] to ".$logged['username']." and it should be ok. Also instead of just using $row = mysql_fetch_array($query);
its best to use:
while($row = mysql_fetch_array($query)) {
echo "RESULTS HERE ;)";
}

[Eric30]

Wheres the connection? Also just change the $logged['username'] to ".$logged['username']." and it should be ok. Also instead of just using $row = mysql_fetch_array($query);
its best to use:
while($row = mysql_fetch_array($query)) {
echo "RESULTS HERE ;)";
}


What would display everything in the table, as this is for a 'DJ Says' he would only need to display the last row.

[Dentafrice1]

TY

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