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grlayouts

Wrong parameter count for mysql_result

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ok what i have is[code]
$turnupdate = mysql_result("SELECT lastran FROM cronjobs WHERE cronjob='turns';");[/code] when i try to take a result from my data base to display on my php page in getting [code]Warning: Wrong parameter count for mysql_result() in /home/pimpdomi/public_html/header.php on line 17[/code]

any idea's?

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mysql_result() is used to pull a specific value from a query. you're trying to use it to actually run your query. for instance:
[code]
<?php
// run a mysql query:
$turnupdate = mysql_query("SELECT lastran FROM cronjobs WHERE cronjob = 'turns'");

// get your result from that query:
$turn = mysql_result($turnupdate, 0);
?>
[/code]

hope this helps. look in the manual for a full description of the required arguments (as well as the optional ones) for mysql_result()

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