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Problems using mysql_num_rows

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boab

Programmer
May 30, 2001
75
GB
Can anyone help here

The code I'm using finds the total number of records

function trawl(){

$qTrawl = "SELECT * FROM feedback ";

$rTrawl = mysql_query($qTrawl);
$num = mysql_num_rows($rTrawl);

return $num;
}

calculates the percentage of a particular response to a question

function perc($criteria,$value,$total){

$qPerC = "SELECT * FROM feedback WHERE $criteria = $value";

$rPerC = mysql_query($qPerC);
$q_Num = mysql_num_rows($rPerC);

if (q_Num<1)
{
percentage = 0;
}else
percentage = ($q_Num/$total)*100
}

return $percentage;
}

these two are called on the page thus:

<?php perC("4a","Very",trawl());?>

when the page is processed I get 3 warnings two relate to the use of mysql_num_rows saying that the supplied argument is not valid the third warning is a divide by zero caused by the previous two errors.

ANy Ideas how to fix this.


THe Start of wisdom is to realise you know nothing. I must be a genius then!
 
A common reason for mysql_num_rows() not working is that mysql_query() barfed for some reason and returned a FALSE rather than a result handle.

Change:
$rTrawl = mysql_query($qTrawl);
and
$rPerC = mysql_query($qPerC);

to read:
$rTrawl = mysql_query($qTrawl) or die(mysql_error());
and
$rPerC = mysql_query($qPerC) or die(mysql_error());

respectively.

Want the best answers? Ask the best questions!

TANSTAAFL!!
 
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