How can i get the value of previous url in php code?
Here is the url page1.php
http://localhost/spk/kelulusan_process.php?lulus=10003
page2.php:
I want to get the parameter from the previous url in page1.php. I use
<?php
$_GET['lulus'];
?>
But the value is null.
For $_GET to work you have to redirect user using form action
Or you can do that by setting session variables and access them any page
You can try $_SERVER["HTTP_REFERER"]
But don't forget to escape $_SERVER["HTTP_REFERER"] since it's common for attacks.
Better is to store the current page in a $_SESSION var.
if (!isset($_SESSION)) {
session_start();
}
$_SESSION['lastpage'] = $_SERVER['REQUEST_URI'];
Then when loading the next page:
if (!isset($_SESSION)) {
session_start();
}
// now you can access the last page
$lastpage = $_SESSION['lastpage'];
page2.php
<?php
echo htmlspecialchars($_GET["lulus"]);
?>
how are you trying to pass from page 1? if through a form set the method to get and action to page2.php
in page1.php make sure you set the name
<form method="get" action="page2.php">
<input type="text" name="lulus">
<input id="submit" name="submit" type="submit" value="Send">
You cannot display $_GET['lulus']; in page2.php unless you write get parameter in page2.php.
Your code is redirect from page1.php to page2.php, You must set $_GET['lulus']; into one session and echo in page2.php
Page1.php
<?
$_SESSION['GET'] = $_GET['lulus'];
?>
Page2.php
<?
echo $_SESSION['GET'];
?>
Instead, echo $_GET['lulus'], you should echo in page2.php $_SESSION. Because $_SESSION variable was overrode by $_GET itself.
Use $_SERVER['HTTP_REFERER'] ... hopefully this will help you. And parse url something like below:
$url = $_SERVER['HTTP_REFERER'] ; //http://localhost/spk/kelulusan_process.php?lulus=10003
$array = explode("?lulus=",$url);
echo $array[1];
echo $_GET['lulus'];