如何在php中创建一个search.json类似Twitter的内容 [英] How to create in php a search.json twitter-like

问题描述

我在我的网站上制作了一个search.php文件，该文件生成一个JSON字符串，可以帮助我在自己的应用程序中使用实时ajax.

但是现在，我想将其作为对其他人的API来打开，但是我发现$ .get $ .getJSON $ .ajax不允许使用其他服务器/域中的search.php文件. /p>

如何将我的php搜索转换为search.json，就像Twitter一样，将参数传递给它.

Thx

解决方案

getJSON受浏览器的安全限制所限制，该安全限制可锁定非起源域.为了进行跨域，您必须使用JSONP，这需要将数据包装在由回调变量(例如$ _GET ['jsonp_callback'])定义的函数中.例如

Search.php

<?php
    echo $_GET['jsonp_callback'] . '(' . json_encode($data). ');'
    // prints: jsonp123({"search" : "value", etc. });
?>

jQuery

$.ajax({
  dataType: 'jsonp',
  data: 'search=value',
  jsonp: 'jsonp_callback',
  url: 'http://yourserver.com/search.php',
  success: function () {
    // do stuff
  },
});

只需确保您在php脚本中定义的回调变量与通过.ajax查询调用的jsonp值匹配(或者默认为回调").

I've made on my website a search.php file that produce a JSON string, helping me to use real-time ajax for my apps.

But now, I'd like to open it as an API to others, but I discovered that $.get $.getJSON $.ajax doesn't allow to use my search.php file from other servers/domains.

How can I do to transform my php search into a search.json, exactly like Twitter, passing parameters to it.

Thx

解决方案

getJSON is limited by your browser's security restrictions that lock down non-origin domains. In order to do cross-domain, you have to use JSONP, which requires you wrap the data in a function that is defined by the callback variable (e.g. $_GET['jsonp_callback']). e.g.

Search.php

<?php
    echo $_GET['jsonp_callback'] . '(' . json_encode($data). ');'
    // prints: jsonp123({"search" : "value", etc. });
?>

jQuery

$.ajax({
  dataType: 'jsonp',
  data: 'search=value',
  jsonp: 'jsonp_callback',
  url: 'http://yourserver.com/search.php',
  success: function () {
    // do stuff
  },
});

Just make sure that the callback variable that you define in your php script matches the jsonp value that you call through the .ajax query (or it defaults to "callback").

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