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$_POST to $HTTP_POST_VARS...

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  #1  
Old February 22nd, 2004, 06:02 PM
zzz zzz is offline
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$_POST to $HTTP_POST_VARS...
if i'm using earlier version of php, such as 4.0.5
and what i know is i need to change $_POST TO $HTTP_POST_VARS..
my question is...how about $_REQUEST and $_COOKIE
do i have to change to another$.... or still can remain the same..something like
Code:
$sql = mysql_query("SELECT * FROM users WHERE username = '$_COOKIE[username]'") OR DIE("Sorry Theres a mysql error.");

and
Code:

$userid = $_REQUEST['id'];
$code = $_REQUEST['code'];
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  #2  
Old February 22nd, 2004, 07:28 PM
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bluephoenix bluephoenix is offline
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RE: $_POST to $HTTP_POST_VARS...
The following comes from php.net/manual/en/reserved.variables.php

Server variables: $_SERVER
Note: Introduced in 4.1.0. In earlier versions, use $HTTP_SERVER_VARS.

Environment variables: $_ENV
Note: Introduced in 4.1.0. In earlier versions, use $HTTP_ENV_VARS.

HTTP Cookies: $_COOKIE
Note: Introduced in 4.1.0. In earlier versions, use $HTTP_COOKIE_VARS.

HTTP GET variables: $_GET
Note: Introduced in 4.1.0. In earlier versions, use $HTTP_GET_VARS.

HTTP POST variables: $_POST
Note: Introduced in 4.1.0. In earlier versions, use $HTTP_POST_VARS.

HTTP File upload variables: $_FILES
Note: Introduced in 4.1.0. In earlier versions, use $HTTP_POST_FILES.

Request variables: $_REQUEST
Note: Introduced in 4.1.0. There is no equivalent array in earlier versions.

Note: Prior to PHP 4.3.0, $_FILES information was also included into $_REQUEST.

Session variables: $_SESSION
Note: Introduced in 4.1.0. In earlier versions, use $HTTP_SESSION_VARS.

Global variables: $GLOBALS
Note: $GLOBALS has been available since PHP 3.0.0.
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  #3  
Old February 22nd, 2004, 07:48 PM
zzz zzz is offline
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RE: RE: $_POST to $HTTP_POST_VARS...
woooops..oh no..can you tell me what can i do with this code
php Code:
Original - php Code 
  1.  
  2. <?
  3.  
  4. include("config.php");
  5. include("functions.php");
  6. include("user_check.php");
  7.  
  8. $userid = $_REQUEST['id'];
  9. $code = $_REQUEST['code'];
  10.  
  11. $sql = mysql_query("UPDATE users SET activated='1' WHERE userid='$userid' AND password='$code'");
  12.  
  13. $sql_doublecheck = mysql_query("SELECT * FROM users WHERE userid='$userid' AND password='$code' AND activated='1'");
  14. $doublecheck = mysql_num_rows($sql_doublecheck);
  15.  
  16. if($doublecheck == 0){
  17.     echo "<strong><font color=red>Sorry,<BR>your account could not be activated!</font></strong>";
  18. } elseif ($doublecheck > 0) {
  19.     
  20.     echo "<strong>Your account has been activated!</strong> You may login below!<br />";
  21.     login_form();
  22. }
  23.  
  24. ?>
  25.  

if i'm using php4.0.5, can i still use this code by changing the code, which will be doing the same thing like above...?or i have to get a newer version of php..if i have to, that means i have to edit all my previous code??coz previous codes is using HTTP_POST_VARS and so on....will it affect my current coding progress...?i need ot know..if anybody can tell me ...pls!:
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  #4  
Old February 23rd, 2004, 01:08 AM
System System is offline
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Message Moved
Thread moved from 'Database Help' to 'PHP Installation' by sliver.

Reason: This is more a php question
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  #5  
Old February 23rd, 2004, 01:10 AM
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sliver sliver is offline
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RE: $_POST to $HTTP_POST_VARS...
php Code:
Original - php Code 
  1.  
  2. if($HTTP_POST_VARS['id']) {
  3.   $userid=$HTTP_POST_VARS['id']
  4. } elseif($HTTP_GET_VARS['id']) {
  5.   $userid=$HTTP_GET_VARS['id'];
  6. }
  7. if($HTTP_POST_VARS['code']) {
  8.   $code=$HTTP_POST_VARS['code'];
  9. } elseif($HTTP_GET_VARS['code']) {
  10.   $code=$HTTP_GET_VARS['code'];
  11. }
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  #6  
Old February 23rd, 2004, 05:33 AM
zzz zzz is offline
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RE: RE: $_POST to $HTTP_POST_VARS...

Quote:
this
php Code:
Original - php Code 
  1.  
  2. if($HTTP_POST_VARS['id']) {
  3.   $userid=$HTTP_POST_VARS['id']
  4. } elseif($HTTP_GET_VARS['id']) {
  5.   $userid=$HTTP_GET_VARS['id'];
  6. }
  7. if($HTTP_POST_VARS['code']) {
  8.   $code=$HTTP_POST_VARS['code'];
  9. } elseif($HTTP_GET_VARS['code']) {
  10.   $code=$HTTP_GET_VARS['code'];
  11. }

to replace this right??
Code:
$userid = $_REQUEST['id'];
$code = $_REQUEST['code'];

but it's not woring at all...still error message :
Parse error: parse error in c:program filesapache groupapachehtdocsloginactivate.php on line 10

line 10 is this.. $userid=$HTTP_GET_VARS['id'];
the whole code look like this
php Code:
Original - php Code 
  1. <?
  2.  
  3. include("config.php");
  4. include("functions.php");
  5. include("user_check.php");
  6.  
  7.  
  8. if($HTTP_POST_VARS['id']) {
  9.   $userid=$HTTP_POST_VARS['id']
  10. } elseif($HTTP_GET_VARS['id']) {
  11.   $userid=$HTTP_GET_VARS['id'];
  12. }
  13. if($HTTP_POST_VARS['code']) {
  14.   $code=$HTTP_POST_VARS['code'];
  15. } elseif($HTTP_GET_VARS['code']) {
  16.   $code=$HTTP_GET_VARS['code'];
  17. }
  18.  
  19.  
  20. $sql = mysql_query("UPDATE users SET activated='1' WHERE userid='$userid' AND password='$code'");
  21.  
  22. $sql_doublecheck = mysql_query("SELECT * FROM users WHERE userid='$userid' AND password='$code' AND activated='1'");
  23. $doublecheck = mysql_num_rows($sql_doublecheck);
  24.  
  25. if($doublecheck == 0){
  26.     echo "<strong><font color=red>Sorry,<BR>your account could not be activated!</font></strong>";
  27. } elseif ($doublecheck > 0) {
  28.     
  29.     echo "<strong>Your account has been activated!</strong> You may login below!<br />";
  30.     login_form();
  31. }
  32.  
  33. ?>

thanksss alot...please help...HELP~!!@~!~
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  #7  
Old February 23rd, 2004, 05:39 AM
nawlej nawlej is offline
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RE: $_POST to $HTTP_POST_VARS...
you are getting the parse error because you forgot to terminate the variable statement with a semicolon ;
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