Error Echoing Database(DB)

[PHP_Person]

Okay, I’ve been making a PHP commenting script, and it seems as though the last thing I need it to echo the database. For some reason, it is giving me this 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 ‘Resource id #6’ at line 1.

My code…

<?php
//conection
mysql_connect("localhost","root","") or die(mysql_error());
echo "Connected!";
mysql_select_db("ycomment") or die(mysql_error());
echo "DBSelected!";

//the real stuff.  oh yeah.
$show = mysql_query("SELECT * FROM comments");
echo "Query made!";
$result = mysql_query($show) or die(mysql_error());
while($show = mysql_fetch_array($result)) {
$name = $show['name'];
$commment = $row['comment'];
	echo "Name: $name";
	echo "Comment: $comment";
}
?>

[aaronszy]

mysql_query() takes a string as a parameter and returns a resource identifier

[php]
// After this call $show contains a resource identifier, not a string
$show = mysql_query(“SELECT * FROM comments”);

// Then here we are passing mysql_query a resource identifier instead of a SQL query
$result = mysql_query($show) or die(mysql_error());
[/php]

I think what you meant to do was something like this:
[php]

<?php //conection mysql_connect("localhost","root","") or die(mysql_error()); echo "Connected!"; mysql_select_db("ycomment") or die(mysql_error()); echo "DBSelected!"; //the real stuff. oh yeah. $result = mysql_query("SELECT * FROM comments"); echo "Query made!"; while($show = mysql_fetch_array($result)) { $name = $show['name']; $commment = $row['comment']; echo "Name: $name"; echo "Comment: $comment"; } ?>

[/php]

[PHP_Person]

I’ve tried that, and it doesn’t work either. Any suggestions?

[aaronszy]

Are you getting the same error message?

[PHP_Person]

I’m not sure what the problem is anymore, it had something to do with using the wrong user and some blah blah simple mistake. ;D