[英]php check if link clicked
The question is that how do i check if a link has been clicked? 问题是如何检查链接是否已单击?
<a href="laggtill.php">+</a>
<a href="laggtill.php">+</a>
(another document)
<?php
session_start();
$_SESSION['car'] = $_SESSION['car'] + 1;
$_SESSION['boat'] = $_SESSION['boat'] + 1;
header("Location: betalning.php");
?>
The first a tag add a car to the cart and the second a tag add a boat to the cart. 第一个标签将汽车添加到购物车,第二个标签将船添加到购物车。 How do i detect which one of the a tag that has been clicked, and if i now click on any of the a tags both a car and a boat will be added to the cart.
我如何检测单击了哪个标签,如果现在单击任何一个标签,则汽车和船都将添加到购物车中。
You can add GET parameters to the links: 您可以将GET参数添加到链接中:
<a href="laggtill.php?add=car">Add car</a>
And then in your PHP document: 然后在您的PHP文档中:
if($_GET['add'] == 'car'){
$_SESSION['car'] += 1;
}
// etc...
This is basically the easiest way to pass data from one page to another using a link. 基本上,这是使用链接将数据从一页传递到另一页的最简单方法。
The concept that you should use. 您应该使用的概念。 is Ajax.
是Ajax。
Every click and others things with the browser only happens on the client ( browser ). 浏览器的每次单击和其他操作仅发生在客户端(browser)上。
Some simple may be: 一些简单的可能是:
// Html // HTML
<a id="linkone" href="laggtill.php">+</a>
<a id="linktwo" href="laggtill.php">+</a>
//Javascript // JavaScript
// Use jquery
$("#linkone").on("click", function(){
//function that send request to server
$.post('someurl.php', {})
.success(function(res){
console.log(res);
// If you stay here, the procces should be ended
})
// if you return false, the click dont redirect other window
// if you return true, the click redirect other window
return false;
});
// php file for first link //第一个链接的php文件
<?php
//capture values
// But only is a clic, then there is not values
session_start();
$_SESSION['car'] = $_SESSION['car'] + 1;
// If you want some simple, one file only works for one link
// For the other link, you should create other file same to this.
header("Location: betalning.php"); // With ajax dont use this line,
?>
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.