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session_start(): Cannot send session cookie - headers already sent by
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hi
thrr is no one who can help me this is the prog which when i run faces problem of session.
session_start(): Cannot send session cookie - headers already sent
pls plssssssssss help me to solve the problem below is the code
any one this is simple prog but i can't run
if i can't run thn my boss will kick me out of the office.
so pls hlp me.
[php]<?session_start();?>
<?php
include"db_config.php";
$connect = mysql_connect("$db_host", "$db_user", "$db_password");
mysql_select_db("$db_name", $connect) or die(mysql_error());
$f_name= $_POST['fname'];
$l_name= $_POST['lname'];
$pass= $_POST['password'];
$query = "SELECT * from helpagent WHERE first='$f_name' and last='$l_name' and password='$pass' ";
$result = mysql_query($query);
for ($i = 0; $i < mysql_num_rows($result); $i++)
{
$id = mysql_result($result, $i, "agentid");
$fname = mysql_result($result, $i, "first");
$lname = mysql_result($result, $i, "last");
$pas = mysql_result($result, $i, "password");
if (("$f_name" == "$fname") and ("$l_name" == "$lname") and ("$pass" == "$pas"))
{
$_SESSION['sfname'] = $fname;
$_SESSION['slname'] = $lname;
$_SESSION['sid'] = $id;
echo("<script>document.location.href='user_page.php';</script>");
exit;
}//if loop
}//for loop
?>[/php]
thrr is no one who can help me this is the prog which when i run faces problem of session.
session_start(): Cannot send session cookie - headers already sent
pls plssssssssss help me to solve the problem below is the code
any one this is simple prog but i can't run
if i can't run thn my boss will kick me out of the office.
so pls hlp me.
[php]<?session_start();?>
<?php
include"db_config.php";
$connect = mysql_connect("$db_host", "$db_user", "$db_password");
mysql_select_db("$db_name", $connect) or die(mysql_error());
$f_name= $_POST['fname'];
$l_name= $_POST['lname'];
$pass= $_POST['password'];
$query = "SELECT * from helpagent WHERE first='$f_name' and last='$l_name' and password='$pass' ";
$result = mysql_query($query);
for ($i = 0; $i < mysql_num_rows($result); $i++)
{
$id = mysql_result($result, $i, "agentid");
$fname = mysql_result($result, $i, "first");
$lname = mysql_result($result, $i, "last");
$pas = mysql_result($result, $i, "password");
if (("$f_name" == "$fname") and ("$l_name" == "$lname") and ("$pass" == "$pas"))
{
$_SESSION['sfname'] = $fname;
$_SESSION['slname'] = $lname;
$_SESSION['sid'] = $id;
echo("<script>document.location.href='user_page.php';</script>");
exit;
}//if loop
}//for loop
?>[/php]
Last edited by cscgal; Jun 30th, 2006 at 7:42 pm.
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Join Date: Jul 2004
Posts: 494
Reputation: 
Solved Threads: 21
Posting Pro in Training
I suggest you read the other threads you started, as well as the manual. There are also some very good how-tos out there to teach you how to properly do what you want to.
www.uncreativelabs.net
Old computers are getting to be a lost art. Here at Uncreative Labs, we still enjoy using the old computers. Sometimes we want to see how far a particular system can go, other times we use a stock system to remind ourselves of what we once had.
Old computers are getting to be a lost art. Here at Uncreative Labs, we still enjoy using the old computers. Sometimes we want to see how far a particular system can go, other times we use a stock system to remind ourselves of what we once had.
Check the manual... if that script is included check that the file that it is actually including it doesn't have any print or echo. session_start() should be at the very beginning of the code.
Also change this:
[php]
<?session_start();?>
<?php
include"db_config.php";
[/php]
to this
[php]
<?php
session_start();
include"db_config.php";
[/php]
There is no need to break the PHP tags there... and use <?php also instead of <? it will make the script work in other servers. Or maybe your server doesn;t support short php tags.
Also change this:
[php]
<?session_start();?>
<?php
include"db_config.php";
[/php]
to this
[php]
<?php
session_start();
include"db_config.php";
[/php]
There is no need to break the PHP tags there... and use <?php also instead of <? it will make the script work in other servers. Or maybe your server doesn;t support short php tags.
I'm assuming that your including this file into another file, and in that other file you have already echoed something when this file is included.
Otherwise I cannot see what is wrong, as suggested the manual always helps.
Otherwise I cannot see what is wrong, as suggested the manual always helps.
-1
#5 27 Days Ago
Be sure that your SafeMode CLOSED. Ask your hosting company t switch it to CLOSED. The problem will be solved
•
•
•
•
Originally Posted by manish812 
hi
thrr is no one who can help me this is the prog which when i run faces problem of session.
session_start(): Cannot send session cookie - headers already sent
pls plssssssssss help me to solve the problem below is the code
any one this is simple prog but i can't run
if i can't run thn my boss will kick me out of the office.
so pls hlp me.
[php]<?session_start();?>
<?php
include"db_config.php";
$connect = mysql_connect("$db_host", "$db_user", "$db_password");
mysql_select_db("$db_name", $connect) or die(mysql_error());
$f_name= $_POST['fname'];
$l_name= $_POST['lname'];
$pass= $_POST['password'];
$query = "SELECT * from helpagent WHERE first='$f_name' and last='$l_name' and password='$pass' ";
$result = mysql_query($query);
for ($i = 0; $i < mysql_num_rows($result); $i++)
{
$id = mysql_result($result, $i, "agentid");
$fname = mysql_result($result, $i, "first");
$lname = mysql_result($result, $i, "last");
$pas = mysql_result($result, $i, "password");
if (("$f_name" == "$fname") and ("$l_name" == "$lname") and ("$pass" == "$pas"))
{
$_SESSION['sfname'] = $fname;
$_SESSION['slname'] = $lname;
$_SESSION['sid'] = $id;
echo("<script>document.location.href='user_page.php';</script>");
exit;
}//if loop
}//for loop
?>[/php]
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Join Date: Oct 2009
Posts: 15
Reputation:
Solved Threads: 3
Newbie Poster
0
#6 27 Days Ago
PHP Syntax (Toggle Plain Text)
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