[RESOLVED] Cannot modify header information - headers already sent by

[micky007]

Hi,

I'm currently stuck and can not work it out.

<b>Warning</b>: Cannot modify header information - headers already sent by (output started at /home/REMOVED/public_html/REMOVED/track/lead.php:6) in <b>/home/REMOVED/public_html/REMOVED/track/lead.php</b> on line <b>38</b><br />
<br />
<b>Warning</b>: Cannot modify header information - headers already sent by (output started at /home/REMOVED/public_html/REMOVED/track/lead.php:6) in <b>/home/REMOVED/public_html/REMOVED/track/lead.php</b> on line <b>39</b><br />

<br />
<b>Warning</b>: Cannot modify header information - headers already sent by (output started at /home/REMOVED/public_html/REMOVED/track/lead.php:6) in <b>/home/REMOVED/public_html/REMOVED/track/lead.php</b> on line <b>40</b><br />
<iframe src="Array" frameborder="0" height="0" width="0">

Why am i getting this error?

Below is my PHP code:

<?php
// Print a cookie
//echo $_COOKIE["user"];

// A way to view all cookies
//print_r($_COOKIE);


$username="REMOVED";
$password="REMOVED";
$database="REMOVED";
$affid=$_COOKIE["adwarm_lead_affid"];
$subid=$_COOKIE["adwarm_lead_subid"];
$click=$_COOKIE["adwarm_click"];
$leadip=@$REMOTE_ADDR;
$referer=$_SERVER['HTTP_REFERER'];
$status="Pending";
$p=$_GET["p"];
$adid=$_GET["adid"];

mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Oops theres an error, our highly trained monkeys have been notified.");

$query="INSERT INTO affleads VALUES ('','$p','$affid','$subid','$adid','$click','$leadip','$status','$referer')";
mysql_query($query) or dies(mysql_error()); 
mysql_close();

mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Oops theres an error, our highly trained monkeys have been notified.");

$query2="SELECT Pixel FROM affpixels WHERE AffID ='$affid'";
mysql_query($query2) or dies(mysql_error()); 
$result2 = mysql_query($query2);
$pixelurl=mysql_fetch_array($result2, MYSQL_ASSOC);

mysql_close();

setcookie("adwarm_lead_subid", "",1);
setcookie("adwarm_lead_clickid", "",1);
setcookie("adwarm_click", "",1);
?>
<iframe src="<? echo $pixelurl; ?>" frameborder="0" height="0" width="0">

The errors are reporting to the following 3 lines:

setcookie("adwarm_lead_subid", "",1);
setcookie("adwarm_lead_clickid", "",1);
setcookie("adwarm_click", "",1);

Any help would be great.

Thanks

[bradgrafelman]

The errors reference line #6, which in your code snippet above is the print_r() statement that has been commented out.

Are you sure this line is commented out on the server? Did you perhaps forget to save/upload your changes?

[micky007]

Hi,

Once again you help me out, thank you again. I have another problem, for some reason the variable $pixelurl is just outputting text called Array.

mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Oops theres an error, our highly trained monkeys have been notified.");

$query2="SELECT Pixel FROM affpixels WHERE AffID ='$affid'";
mysql_query($query2) or dies(mysql_error()); 
$result2 = mysql_query($query2);
$pixelurl=mysql_fetch_array($result2, MYSQL_ASSOC);

mysql_close();

?>
<? echo $pixelurl; ?>
<iframe src="<? echo $pixelurl; ?>" frameborder="0" height="0" width="0">

The SQL is fine as I've tested it in PHPMYADMIN, but some reason the output is "Array".

Any ideas?

Thanks

[dagon]

Because it is an Array, the function mysql_fetch_array() returns an array, it even says so in its name.

[micky007]

Hi,

Oh sorry, im still learning PHP and have only been learning it for less than 48hrs. Whats the code so its fetches the result of the SQL and stores it in a variable then?

Thanks in advance.

[micky007]

$result2 = mysql_query($query2);

I've tried that but when i echo $result2 it has the output of this:

Resource id #4

How come?

[dagon]

micky007;10976396 wrote:

$result2 = mysql_query($query2);

I've tried that but when i echo $result2 it has the output of this:

Resource id #4

How come?

reading the manual would answer this and the last question, quicker than i can type this

[micky007]

Resolved, i did some googling on that error and have now fixed my problem.

Thanks for the help anyway guys 🙂