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Doc203

08-25-2004 12:50 PM

I am getting a blank page/no error with this code:

I am learning php and I am trying to get this php script to return a result when called by a browser. Can someone look at it and give me a hand?

PHP Code:


		
			
$start_date = "2004-08-01 00:00:00";

$end_date = "2004-08-23 00:00:00";



$result = $DB_site->query("SELECT COUNT(post.postid) AS count, user.username

FROM post

LEFT JOIN user ON (user.userid = post.userid)

AND (UNIX_TIMESTAMP('$start_date')) <= post.dateline

AND post.dateline <= (UNIX_TIMESTAMP('end_date'))

GROUP BY user.userid ORDER BY count DESC, username");

Xenon

08-26-2004 06:38 AM

Code:

(UNIX_TIMESTAMP('end_date'))

you've missed a $ before end_date ;)

Doc203

08-26-2004 04:02 PM

Thanks Xenon, I have added a line also so I could find out a little more and here is what I get now:

"I failed and here is why: 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 #14' at line 1
"

If I run the query directly in Mysql it returns the results correctly. Any ideas? Thanks for the help!

Quote:

<?PHP
require("./global.php");

$start_date = "2004-08-01 00:00:00";
$end_date = "2004-08-23 00:00:00";

$stmt = $DB_site->query("SELECT COUNT(post.postid) AS count, user.username
FROM post
LEFT JOIN user ON (user.userid = post.userid)
AND (UNIX_TIMESTAMP('$start_date')) <= post.dateline
AND post.dateline <= (UNIX_TIMESTAMP('$end_date'))
GROUP BY user.userid ORDER BY count DESC, username");

$sr = mysql_query($stmt) or die ("<p>I failed and here is why: " . mysql_error());

?>

Xenon

08-27-2004 02:01 PM

$DB_site->query already calls mysql_query, so $stmt contains already the mysql result, and therefore you cannot put it as argument of mysql_query again :)

Doc203

08-28-2004 02:26 PM

Thanks Xenon, not only did that help solve my problem, but I understand better as well!

Xenon

08-28-2004 03:17 PM

:) glad i could help

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