session start and two files

Bob_PHP_Builder

Basically i want to be able to send a number to mysql database once and only once. When i open detail.php in a web browser it works and will only create one number once only. In the detail.php i am using session start.

When i try and add detail.php to index.php i get the following error.

I get the following errors:

Warning: session_start() [function.session-start]: Cannot send session cookie - headers already sent by (output started at E:\WEBSITES\photoorder\index.php:9) in E:\WEBSITES\photoorder\detail.php on line 4

Warning: session_start() [function.session-start]: Cannot send session cache limiter - headers already sent (output started at E:\WEBSITES\photoorder\index.php:9) in E:\WEBSITES\photoorder\detail.php on line 4

Below is the header for index.php

<!--BEGIN-->
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
	<title>Aurigma Image Uploader</title>
	<meta http-equiv="MSThemeCompatible" content="yes" />
	<link href="style.css" type="text/css" rel="stylesheet" />

<script src="../iuembed.js" type="text/javascript"></script>

<script type="text/javascript">




//<![CDATA[		
//Separator for ordered formats in description field of an item.
var separator = "...\n\r";    

var maxFormatCount = 3;  

var imageUploader1 = null;



function fullPageLoad(){	
	imageUploader1=getImageUploader("ImageUploader1");
}

//]]>
	</script>

</head>

<body>
include ("detail.php");
</body

Many Thanks

Bob_PHP_Builder

here is the details.php file


<?php 

// start session, and see if has been run before - redirect to index.php 
session_start(); 
if( isset($_SESSION['been_run']) && $_SESSION['been_run'] == 1 ){ 

exit; 
} 


/* Connect to database */ 

$db= mysql_connect("localhost", "user", "pass"); 
mysql_select_db("table", $db); 

$lastrow = mysql_query("SELECT id 
FROM current 
ORDER BY o_id DESC LIMIT 1"); 

while($row= mysql_fetch_array($lastrow)) 
{ 
    $JOB= "BOB"; 
    $newlastrow= ($row["o_id"]+1); 
    printf($BOB); 
    printf($newlastrow); 

} 
$today = date("F j, Y, g:i a");                // Date to Print 
echo "\n $today";   

$date = date("Ymd");                  // Date to Database                              

$sql=     "INSERT INTO current
    VALUES ('$newlastrow', NULL, '$date', '1', '1')"; 
    mysql_query($sql); 

// set 'been run' flag variable 

$_SESSION['been_run'] = 1; 

?>

halojoy

you can remove session_start() from details.php
and make this
first line in your index.php

<?php session_start(); ?>
<!--BEGIN-->
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" 
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>

Bob_PHP_Builder

thank you very much..

I have one last problem.

how can i have the exit below turn into the index.php with the same info as before?

if( isset($_SESSION['been_run']) && $_SESSION['been_run'] == 1 ){ 

exit; 
}

halojoy

maybe you should change
to this in details.php
if is only INSERT you do not want to run several times

<?php 

/* Connect to database */ 
$db= mysql_connect("localhost", "user", "pass"); 
mysql_select_db("table", $db); 

$lastrow = mysql_query("SELECT id FROM current 
ORDER BY o_id DESC LIMIT 1"); 

while($row= mysql_fetch_array($lastrow)) 
{ 
    $JOB= "BOB"; 
    $newlastrow= ($row["o_id"]+1); 
    printf($BOB); 
    printf($newlastrow); 

} 

$today = date("F j, Y, g:i a");   // Date to Print 
echo "\n $today";   

// if not been_run == 1
if( !(isset($_SESSION['been_run']) && $_SESSION['been_run'] == 1) ){ 

$date = date("Ymd");  // Date to Database 
$sql="INSERT INTO current VALUES ('$newlastrow', NULL, '$date', '1', '1')"; 
mysql_query($sql); 

// set 'been run' flag variable 
$_SESSION['been_run'] = 1; 

}

?>

also you have $JOB, but print $BOB

    $JOB= "BOB";
    $newlastrow= ($row["o_id"]+1);
    printf($BOB);

Bob_PHP_Builder

Your a champion Halojoy!! Thank you