jQuery的$。阿贾克斯（）与PHP [英] jQuery $.ajax() with PHP

问题描述

我试图使用jQuery $。阿贾克斯（），但我现在面临一些困难。

I'm trying to use jQuery $.ajax() but I'm facing some difficulties.

下面是我想使用POST的文本字段：

Here's the textbox field that I want to use for POST:

<input name="url" class="url" type="text" >

这里的code：

Here's the code:

$.ajax({
        type: "post",
        url: "file.php",
        data: $(this).serialize(),
        success: function(data) { ...............

现在，这是为file.php：

Now this is the file.php:

<?php
if( $_REQUEST['url'] )
{

   $url = $_REQUEST['url'];
   $url = file_get_contents($url);  
   // I would need now to return something here but not sure how??!!
}
?>

现在，我的问题是，如何在这个PHP code返回变量，并利用它们在我的code以上，我的意思是在$阿贾克斯（）成功的一部分。另外，如果我想在$ url变量执行一些其他的东西，怎么办呢？如何回报他们？ ：/

Now, my question is, how to return variables in this PHP code and use them in my code above, I mean in the success part of $.ajax(). Also if I want to perform some additional stuff on the $url variable, how to do it? How to return them? :/

推荐答案

您只需打印/呼应的回归的价值。

You just print/echo your 'return' value.

为file.php

<?php
if( $_REQUEST['url'] )
{

   $url = $_REQUEST['url'];
   $url = file_get_contents($url);  
   // I would need now to return something here but not sure how??!!
   echo "something";
}
?>

然后在您的JS：

Then in your JS:

$.ajax({
    type: "post",
    url: "file.php",
    data: $(this).serialize(),
    success: function(data) {
        console.log(data); // "something"
    }
});

---

作为一个方面说明。您的脚本看起来像它接受任何URL，并获取它。是可能的滥用那样脚本。请确保您知道。

---

As a side note. Your script looks like it accepts any URL and fetches it. It is possible to abuse scripts like that. Make sure you are aware that.

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