php mysql query works in phpmyadmin but not on webpage

Question

i wrote the following query

SELECT COUNT(userID) From statistics WHERE userID = ""

this query displays the number of unathunticated visit to the website.

the query works in phpmyadmin when i use double quotes however it doesnt when i use single quotes like below it just gives me the number of record stored in the table

 $queryB = "SELECT COUNT(userID) From statistics WHERE userID = ''";
 $resultB =mysql_query($queryA, $con) or die(mysql_error());
 $authB = mysql_result($resultB, "COUNT(userID)");

 echo "the number of authenticated visits were $authB<br />\n";

i've no idea why it breaks, any ideas?

Luca Rainone · Accepted Answer · 2012-04-11 14:01:28Z

4

you store your query in $queryB but you use $queryA

answered Apr 11, 2012 at 14:01

Luca Rainone

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1 Comment

Michael Berkowski Over a year ago

+1 nice catch.I would bet that $queryA was a test with no WHERE clause.

xholicka · Accepted Answer · 2012-04-11 14:03:18Z

0

not sure if it will work...its just first think that came to mind: how about when u use escaped double quotes?

$queryB = "SELECT COUNT(userID) From statistics WHERE userID = \"\""

answered Apr 11, 2012 at 14:03

xholicka

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web-nomad · Accepted Answer · 2012-04-11 14:01:57Z

0

Try this:

 $queryB = "SELECT COUNT(userID) AS total From statistics WHERE userID = ''";
 $resultB =mysql_query($queryB, $con) or die(mysql_error());
$authB = mysql_fetch_assoc($resultB);
 echo "the number of authenticated visits were ".$authB['total']."<br />\n";

answered Apr 11, 2012 at 14:01

web-nomad

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dnagirl · Accepted Answer · 2012-04-11 14:02:10Z

0

Does userID have a default value? If the default value is NULL, then change your query to

$queryB = "SELECT COUNT(userID) From statistics WHERE userID IS NULL";

answered Apr 11, 2012 at 14:02

dnagirl

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IT Advanture · Accepted Answer · 2012-04-11 14:03:46Z

0

you should change a little to your code

$queryB = "SELECT COUNT(userID) From statistics WHERE userID = ''";
$resultB =mysql_query($queryB, $con) or die(mysql_error());
$authB = mysql_result($resultB, 0, 0);

echo "the number of authenticated visits were $authB<br />\n";

answered Apr 11, 2012 at 14:03

IT Advanture

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