[SOLVED] how to trigger execution of two other php files if a condition is satisfied

[phplearner2008]

Hi,

 

I have expressinterest.php file where I insert values FROM, TO to database

 

$sql = "INSERT INTO interests_sent(FROM,TO) values($FROM,$TO)";

$result = mysql_query($sql) or die("Query: " . $sql . "\n\n" . mysql_error());

$sql1 = "INSERT INTO interests_received(FROM,TO) values($FROM,$TO)";

$result1 = mysql_query($sql1) or die("Query: " . $sql1 . "\n\n" . mysql_error());

 

If $result is successful, I want another php file called InterestSent.php file to  execute a SELECT query & echo result in that page.

 

Similarly, if $result1 is successful, I want another php file called InterestReceived.php file to execute another SELECT query & echo result in that page.

 

How can the files InterestSent.php and InterestReceived.php be made known that $result and $result1 were successful?

I cannot use "header" tag as it will direct the page to another php file, whereas I want two other php files to execute separate queries based on the results of this page.

 

Anyone who can help me with this?

Thanks.

[JasonLewis]

if($result){
include("InterestSent.php");
}

if($result1)}
include("InterestReceived.php");
}

 

?