用PHP解析JSON POST请求 [英] JSON POST request parsing in PHP

问题描述

我在java中生成了一个包含JSON对象的HTMLPost请求，并希望用PHP解析它。

I've generate an HTMLPost Request containing a JSON object in java and would like to parse it in PHP.

public static String transferJSON(JSONObject j) {
    HttpClient httpclient= new DefaultHttpClient();
    HttpResponse response;
    HttpPost httppost= new HttpPost(SERVERURL);
    List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);  
    nameValuePairs.add(new BasicNameValuePair("json", j.toString()));

    httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
    response = httpclient.execute(httppost);  
}

并且在服务器上

<?php

if ($_SERVER['REQUEST_METHOD'] === 'POST') {

  // input = "json=%7B%22locations%22%3A%5B%7B%22..."
  $input = file_get_contents('php://input');

  // jsonObj is empty, not working
  $jsonObj = json_decode($input, true);

我想这是因为JSON特殊字符是编码的。

I guess this is because the JSON special characters are encoded.

json_decode返回空响应

The json_decode return empty response

知道为什么吗？

推荐答案

您实际上是在发布一个HTTP表单实体（ application / x-www-form-），而不是POST一个 application / json 实体。 urlencoded ），单值对json =（编码json）。

Instead of POSTing an application/json entity, you are actually posting an HTTP form entity (application/x-www-form-urlencoded) with a single value pair json=(encoded json).

而不是

List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);  
nameValuePairs.add(new BasicNameValuePair("json", j.toString()));
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

尝试

 httppost.setEntity(new StringEntity(j.toString(),"application/json","UTF-8"));

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