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php远程读取json的方法分析

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Jun 08, 2016 pm 05:20 PM

curljsonnbsppostquot

发送json格式的数据有直接使用get url的方式或者post过来的方式，下文我们为各位总结这两种方式的json数据读取方法。

1.直接以文件形式输出的方式

代码如下

复制代码

header("Content-type:text/html;charset=utf-8");
function GetCurl($url){
    $curl = curl_init();
    curl_setopt($curl,CURLOPT_RETURNTRANSFER,1);
    curl_setopt($curl,CURLOPT_URL, $url);
    curl_setopt($curl,CURLOPT_USERAGENT, $_SERVER['HTTP_USER_AGENT']);
    $resp = curl_exec($curl);
    curl_close($curl);
    return $resp;
}
$resp = GetCurl("http://www.111cn.net"); //这个是json数据文件
$resp = json_decode($resp,true);
header('Content-type: application/json');
echo json_encode($resp);

php远程读取json的方法分析

2. post过来的数据接受方式

代码如下	复制代码
$json_string = $_POST["txt_json"]; if(ini_get("magic_quotes_gpc")=="1") { $json_string=stripslashes($json_string); } $user = json_decode($json_string); echo var_dump($user); ?>

在这个文件中，首先得到html文件中POST表单域txt_json的值，放入变量$json_string中，而后判断，如果当前PHP的设定为magic_quotes_gpc=On，即传入的双引号等会被转义，这样json_decode函数无法解析，因此我们要将其反转义化。而后，使用json_decode函数将JSON文本转换为对象，保存在$user变量中，最终用echo var_dump($user);，将该对象dump输出出来

php的HTTP_RAW_POST_DATA

用Content-Type=text/xml 类型，提交一个xml文档内容给了php server,要怎么获得这个POST数据。
The RAW / uninterpreted HTTP POST information can be accessed with: $GLOBALS['HTTP_RAW_POST_DATA'] This is useful in cases where the post Content-Type is not something PHP understands (such as text/xml).

由于PHP默认只识别application/x-www.form-urlencoded标准的数据类型，因此，对型如text/xml的内容无法解析为$_POST数组，故保留原型，交给$GLOBALS['HTTP_RAW_POST_DATA'] 来接收。
另外还有一项 php://input 也可以实现此这个功能

php://input 允许读取 POST 的原始数据。和 $HTTP_RAW_POST_DATA 比起来，它给内存带来的压力较小，并且不需要任何特殊的 php.ini 设置。php://input 不能用于 enctype="multipart/form-data"。

应用

代码如下

复制代码

a.htm

post.php

echo file_get_contents("php://input");?>

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