使用ajaxRequest.open将变量发送到php [英] Using an ajaxRequest.open to send a variable to php

问题描述

我有一个脚本，该脚本使用ajaxRequest.open来调用php脚本.我想知道如何将变量发送到php文件.更具体地说，我有一个表单文本字段，希望能够发送到php文件.

I have a script which uses ajaxRequest.open to call a php script. I'm wondering how I could send a variable to the php file. More specifically, I've got a form text field I'd like to be able to send to the php file.

ajaxRequest工作正常，我只需要知道如何发送变量以及php如何读取它.

The ajaxRequest is working fine, I just need to know how I'd send a variable and how php could read it.

这是(非常简短的)脚本，正在调用我的php文件...

Here is the (very abbreviated) script which is calling my php file...

}
  ajaxRequest.open("GET", "../../ajaxphp.php", true);
  ajaxRequest.send(null); 
}

推荐答案

首先，您需要获取要发送到* .php的变量witch的值.您可以通过以下方式做到这一点:

First you need to obtain the value of variable witch you are going to send to *.php You can do it by this ways:

var value = document.getElementById("someID").value;

或jQuery方式:

var value = $("#someID").val();

然后，您需要将变量放入ajax请求中:

Then, you need to put the variable to your ajax request:

ajaxRequest.open("GET", "../../ajaxphp.php?variable="+value, true);
//The last argument in this line (witch is set a "true") want to say, that is a  asynchronous request 
ajaxRequest.send(null);
//null want to say, that you not sending any parameter, really is automatically sent in the first line of code

然后，当您在代码*中拾取变量的值时.php，可以做下一个:

Then, when you pick up the value of the variable in the code *. php, can do the next:

<?php
$myVariable = $_GET["variable"];
echo $myVariable;
?>

或者像这样:

<?
$myVariable = $_REQUEST["variable"];
echo $$myVariable;
?>

这篇关于使用ajaxRequest.open将变量发送到php的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！