Setcookie () &quot;The Hell&quot;

Setcookie () "The Hell"

Anon

Hi all,
writing cookies in php is really a tough task. As I checked all possible "whitespaces" "echo" "header" before setting cookie but the Setcookie() function seems stubborn.. where i am wrong..Is there any solution for this warning..

Warning: Oops, php3_SetCookie called after header has been sent in /home/avenuesw/avenuesworldwide-www/abc/tabish/editpages.php3 on line 45

can I write the Database connectivity before setsookie() ??

Pulling my hair 🙁
Tabish

kpmonroe

hmm,
this error is obviously showing that a header has already been sent.

are you calling any kind of include prior to the setcookie?
if so, does that include send any headers?
ex.
include(global.php);
where global.php sends a header.

would it be possible to see the code?

Anon

That warning definitely is due to you sending output to browser before setting the cookie.

You can set the cookie after db connection, but be SURE that all of that comes before any output.

SO you want:

<?php

// do the database stuff

setcookie("thecookie", "whatever", $expire);
// of course set it to what you want

<BODY>

etc. . .

REMEMBER the cookie is set before the tag is set, before headers sent, etc.

Warning: Oops, php3_SetCookie called after header has been sent in /home/avenuesw/avenuesworldwide-www/abc/tabish/editpages.php3 on line 45

Anon

oops, thought these were posted as plain text.

I wanted to have an HTML tag before the BODY tag, but after the "?>", in the post above.
But it didn't show up, darn it.

kpmonroe

what could be happening is that your database calls are producing an error, and then sending them out as standard output, thus sending header information prior to the set cookie. I didnt see any error trapping in the code supplied in the other thread.

ex.
$link=mysql_connect($hostName, $userName, $passWord);
$db = mysql_select_db($databaseName, $link);
$statement = "SELECT * FROM $tableName";
if (!($result = mysql_query($statement, $link))){
printError(sprintf("Error in executing %s statement", $statement));
printError(sprintf("Error:%d %s", mysql_errno($link), mysql_error($link)));
$num = mysql_num_rows ($result);