HELP with headers already sent

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  • craigtomo
    New Member
    • Feb 2008
    • 8
    #1

    HELP with headers already sent

    I am getting the following error when i try to log on to my data base
    Code:
    Warning: session_register() [function.session-register]: Cannot send session cookie - headers already sent by (output started at D:\virtualservers\182763\root\test\login\checklogin.php:1) in D:\virtualservers\182763\root\test\login\checklogin.php on line 18

Warning: session_register() [function.session-register]: Cannot send session cache limiter - headers already sent (output started at D:\virtualservers\182763\root\test\login\checklogin.php:1) in D:\virtualservers\182763\root\test\login\checklogin.php on line 18

Warning: Cannot modify header information - headers already sent by (output started at D:\virtualservers\182763\root\test\login\checklogin.php:1) in D:\virtualservers\182763\root\test\login\checklogin.php on line 20

Warning: Unknown: Your script possibly relies on a session side-effect which existed until PHP 4.2.3. Please be advised that the session extension does not consider global variables as a source of data, unless register_globals is enabled. You can disable this functionality and this warning by setting session.bug_compat_42 or session.bug_compat_warn to off, respectively. in Unknown on line 0
    the code i have is
    [php]
    <?php

    include('vistac config.php');

    // Define $myusername and $mypassword
    $myusername=$_P OST['myusername'];
    $mypassword=$_P OST['mypassword'];

    $sql="SELECT * FROM $tbl_name WHERE username='$myus ername' and password='$mypa ssword'";
    $result=mysql_q uery($sql);

    // Mysql_num_row is counting table row
    $count=mysql_nu m_rows($result) ;
    // If result matched $myusername and $mypassword, table row must be 1 row

    if($count==1){
    // Register $myusername, $mypassword and redirect to file "login_success. php"
    session_registe r('myusername') ;
    session_registe r('mypassword') ;
    header("locatio n:login_success .php");
    }
    else {
    echo "Wrong Username or Password";
    }

    ob_end_flush();
    ?>
    [/php]

    i cannot find the error and the more i look the more i caannot see

    can any body help me please

    thanks
    Last edited by ronverdonk; Feb 24 '08, 12:36 PM. Reason: code within code tags
    Tags: None
    • ronverdonk
      Recognized Expert Specialist
      • Jul 2006
      • 4259
      #2
      Please enclose any code within the proper code tags. See the Posting Guidelines on how to do that.

      moderator

        Comment

        • ronverdonk
          Recognized Expert Specialist
          • Jul 2006
          • 4259
          #3
          Probably some data was already sent (maybe in your include script). I will show the text of an article by wildteen88, june 2006, who can explain this a lot better then I can.
          The reason why you are getting this message is because you have may have/be:
          • Whitespace before the opening php tag <?php
          • Outputting something to the browser before you use session_start, header, setcookie etc

          session_start, setcookie, header and a few other functions write header information to the web server/browser what to do. Such as when you use session_start it requests the browser to create a cookie which stores the PHPSESSID.

          If you have output before you use these functions then they are unable to send new header information as it has already been sent in the form text/html, and so you get the headers already sent error message.

          You can easily find the source of the problem by looking at the error message. As the answer is actully in there, but most people dont notice it as they may not understand what the error means. So lets take this error message as an example:

          Warning: Cannot modify header information - headers already sent by (output started at C:\server\www\t est.php:6) in C:\server\www\t est.php on line 8
          Now PHP has given us a clue here as it has told use where the output has started! Have look where it says output started at and it gives you the full path to the file and the line number. Now the line number that it stats is not usually where the output has started but where the output has ended, becuase the output could be above that line.
          So lets look at test.php:
          Code:
           1 <html>
 2 <head>
 3 <title>Header test</title>
 4 </head>
 5 <body>
 6 <?php
 7
 8 header("Location: http://www.google.com");
 9
10 ?>
11 </body>
12 </html>
          As you can see line 6 is the opening PHP tag (< ?php) this isn't the output but look above that line you'll notice it has some HTML code. This html code is the cause of the error!

          So when you get this error message again look for the clue where it says output started at and goto the file and line number it says. Look above the line number and find where your output is located to.

          Hope that helps you understand why your are getting this error message. (Thanks to wildteen88)
          Ronald

            Comment

            • craigtomo
              New Member
              • Feb 2008
              • 8
              #4
              i have checked out my files and i cannot fine the error
              as per the post the error says it is on line 1 which is the start of my file
              it still has me stumped

              i am a newbie to php and all i am trying to do is create a login to a database

              can you rhelp me with the scritpt i need
              i thought the script i had would work but it keeps gieving me the errors stated above
              and i also apolosgise for not postin gin the correct format

              i will try and do it correctly form now on

                Comment

                • ronverdonk
                  Recognized Expert Specialist
                  • Jul 2006
                  • 4259
                  #5
                  Are you sure you do not send anything out in the first line? Implicitly, maybe a blank character right after the <?php, because that is also a char out.

                  Ronald

                    Comment

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