Count = 0 ? Should be 1

[micky007]

Hi Guys,

My count is echoing the $count variable as 0 when it should be 1. When i run the MySQL Query myself in PHPMYADMIN it returns a row. So why when i echo $count is it showing as 0? Any help would be great. Thanks in advance!

<?php

require("./databaseconnection.php"); //Get Database Login Information

//Do ISP Check
mysql_connect(localhost,$username,$password) or die(mysql_error());
@mysql_select_db($database) or die( "Oops theres an error, our highly trained monkeys have been notified.");
$query = "SELECT * FROM `leads` WHERE `IPCheck` = 'NO'";        
$result=mysql_query($query) or die(mysql_error());

while($row = mysql_fetch_array($result))
{
      $ipcountry=$row['IPCountry'];
      $ipcc=$row['IPCC'];
      $ipisp=$row['IPISP'];
      $leadid=$row['LeadID'];
      $ProgramID=$row['ProgramID'];
      
      //Get Valid Consumer ISP
        $query2 = "SELECT * FROM `ValidISP` WHERE `ISP` = '$ipisp'";
        $result2=mysql_query($query2);
        echo $query2."<br/>";
        $count=mysql_num_rows($result2);
        echo $count."<br/>";
        var_dump($result2);
        echo "<br />";
        if($count==1){
            while($row2 = mysql_fetch_array($result2))
            {
                $validisp=$row2['ISP'];
                $validispstatus=$row2['Status'];
            }
        }else {
            $validispstatus="MANUALCHECK";
            }
      //End
      
        if ( $validispstatus == "PASS" ) {
            $query3 = "UPDATE leads SET IPCheck='PASS' WHERE LeadID='$leadid'";
            mysql_query($query3);
        } elseif ( $validispstatus == "FAIL" ) {
                $query3 = "UPDATE leads SET IPCheck='FAIL' WHERE LeadID='$leadid'";
                mysql_query($query3);
        } else {
                $query3 = "UPDATE leads SET IPCheck='MANUAL' WHERE LeadID='$leadid'";
                mysql_query($query3);
               }
}
mysql_close();
//End
echo "ISP Checks completed!";
?>

Example:

http://www.loudnewmedia.com/images/ispcheck1.jpg

http://www.loudnewmedia.com/images/ispcheck2.jpg

Any help would be great please.

[NogDog]

mysql_num_rows() only counts result rows from SELECT queries. For INSERT or UPDATE queries, you want mysql_affected_rows().

[micky007]

mysql_num_rows() only counts result rows from SELECT queries. For INSERT or UPDATE queries, you want mysql_affected_rows().

Hi,

Thanks for your response, but it is a select query (see below)

$query2 = "SELECT * FROM `ValidISP` WHERE `ISP` = '$ipisp'"; 
        $result2=mysql_query($query2); 
        echo $query2."<br/>"; 
        $count=mysql_num_rows($result2); 

I cant work out why its not working, any help would be much appreciated.

Thank you.

[Ashley Sheridan]

Does var_dump($result2); give any output?

[NogDog]

Hi,

Thanks for your response, but it is a select query (see below)
...

Oops, my poor old eyes were playing tricks on me again.

[micky007]

Does var_dump($result2); give any output?

Yea it outputs the following:

resource(4) of type (mysql result)

[NogDog]

Just in case, are you positive you are connecting to the same database in your script versus in phpMyAdmin? (Yeah, I know: but I've made stupider mistakes. )

[traq]

My count [ ... ] should be 1.

are you _sure_? just out of curiosity, try this:

      //Get Valid Consumer ISP 
        $query2 = "SELECT * FROM `ValidISP` WHERE `ISP` = '$ipisp'"; 
        $result2=mysql_query($query2); 
        echo $query2."<br/>"; 
        $count=mysql_num_rows($result2); 
        echo $count."<br/>"; 
        var_dump($result2); 
        echo "<br />"; 

//  comment this out:
//        if($count==1){ 
//  add the following:
        if( true ){

            while($row2 = mysql_fetch_array($result2)) 
            { 
                $validisp=$row2['ISP']; 
                $validispstatus=$row2['Status']; 

//  add the following:
                var_dump( $validisp );
                var_dump( $validispstatus );

            } 
        } 

[Weedpacket]

        echo $query2."<br/>"; 

And see if that is exactly the same as what you're entering into PHPMyAdmin. Try copying and entering it.