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Mysql Query.


Person

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I need to Query a database with the following.

 

SELECT SUM(1) AS clicks, SUM(`clcpc`), SUM(`chcpc`) FROM `nuke_pnAffiliate_clicktracking` WHERE `pl` = 'rpu' AND  `date` = '20070418'AND `cl` IN (SELECT DISTINCT(cl) FROM rpu_sales WHERE salesman = 'ryan')

 

What would be the best method ?

Thanks in advanced

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ok i have the following now

 

<?php
$host = "localhost";
$user = "user";
$pass = "pass";
$dbname = "user";

$con = mysql_connect($host,$user,$pass) or die (mysql_errno().": ".mysql_error()."<BR>");
mysql_select_db($dbname);

$query = "SELECT SUM(1) AS clicks, SUM(`clcpc`), SUM(`chcpc`) FROM `nuke_pnAffiliate_clicktracking` WHERE `pl` = 'rpu' AND  `date` > '20070228'AND `cl` IN (SELECT DISTINCT(cl) FROM rpu_sales WHERE salesman = 'ryan')";
echo $query

$result= mysql_query($query);
$num_results = mysql_num_rows($result);
    
$row= mysql_fetch_assoc($result);

mysql_free_result($result);

echo $result

?>

 

But this is the error i get

 

<br />

<b>Parse error</b>:  parse error, unexpected T_VARIABLE, expecting ',' or ';' in                        <b>/home/brian/clicktracking.php</b> on line <b>13</b><br />

 

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<?php
$host = "localhost";
$user = "user";
$pass = "pass";
$dbname = "user";

$con = mysql_connect($host,$user,$pass) or die (mysql_errno().": ".mysql_error()."<BR>");
mysql_select_db($dbname);

$query = "SELECT SUM(1) AS clicks, SUM(`clcpc`), SUM(`chcpc`) FROM `nuke_pnAffiliate_clicktracking` WHERE `pl` = 'rpu' AND  `date` > '20070228'AND `cl` IN (SELECT DISTINCT(cl) FROM rpu_sales WHERE salesman = 'ryan')";
echo $query;

$result= mysql_query($query);
$num_results = mysql_num_rows($result);
    
$row= mysql_fetch_assoc($result);

mysql_free_result($result);

echo $result

?>

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now thats fixed ... next error

 

 

<br />

<b>Warning</b>:  mysql_connect(): Can't connect to local MySQL server through socket '/tmp/mysql.sock' (2) in <b>/home/brian/clicktracking.php</b> on line <b>7</b><br />

2002: Can't connect to local MySQL server through socket '/tmp/mysql.sock' (2)<BR>[brian@first brian]$

 

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