首页 文章

如何将android JSON数据转换为PHP数组

提问于
浏览 次
-1

我有一个格式为的JSON数据

[{"product_id":"33","amount":"1"},{"product_id":"34","amount":"3"},{"product_id":"10","amount":"1"},{"username":"test"}]

现在我想在PHP Web系统中获取这些数据 . 我将这些数据从Android应用程序发送到PHP服务器 . 我使用下面的代码将其发送到Web服务器 .

public JSONObject sendAndGetJSONfromURL(String url,List<NameValuePair> params){
        InputStream is = null;
        String result = "";
        JSONObject jArray = null;

        //http post
        try{
                HttpClient httpclient = new DefaultHttpClient();
                HttpPost httppost = new HttpPost(url);
                httppost.setHeader("Content-type", "application/json");
                httppost.setHeader("Accept", "application/json");
                httppost.setEntity(new UrlEncodedFormEntity(params));
                HttpResponse response = httpclient.execute(httppost);
                HttpEntity entity = response.getEntity();
                is = entity.getContent();
.....

在发送之前我打印了

jArray.toString()

并得到输出为

[{"product_id":"33","amount":"1"},{"product_id":"34","amount":"3"},{"product_id":"10","amount":"1"},{"username":"test"}]

我想知道如何从PHP系统中获取这些值 . 任何人都可以帮助我吗？

在通过HTTPRequest发送之前，params变量值的输出如下所示

[cartdata=[{"product_id":"33","amount":"1"},{"product_id":"34","amount":"3"},{"product_id":"10","amount":"1"},{"username":"UWUDAMITH"}]]
php android json httprequest httpresponse

4 回答

  • 0

    这可能不是最好的方法，因为我是Java / Android的新手，但它对我有用:)

    HttpClient httpclient = new DefaultHttpClient();
      HttpPost httppost = new HttpPost("http://www.yourdomain.com/save.php");

      try {
        // Add your data
          List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
          nameValuePairs.add(new BasicNameValuePair("id", "1"));
          nameValuePairs.add(new BasicNameValuePair("json",jArray.toString()));
          httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
          HttpResponse response = httpclient.execute(httppost);

      } catch (ClientProtocolException e) {
          // TODO Auto-generated catch block
      } catch (IOException e) {
          // TODO Auto-generated catch block
      }

    这将向服务器发送一个帖子，其中$ _POST ['id'] = 1，然后$ _POST ['json'] =发送给你的json数据;

    这是save.php . 我实际上把它保存到MySQL，所以你 .

    <?php
    $server = "localhost";
    $username = "user";
    $password = "pass";
    $database = "db";
    $con = mysql_connect($server, $username, $password) or die ("Could not connect: " . mysql_error());
    mysql_select_db($database, $con);

    $id = $_POST["id"];
    $json = $_POST["json"];

    $sql = "INSERT INTO comments (id, json) ";
    $sql .= "VALUES ($id, '$json')";

    if (!mysql_query($sql, $con)) {
        die('Error: ' . mysql_error());
    } else {
        echo "Comment added";
    }
    mysql_close($con);
    echo json_decode($json);
    ?>
    回复于
  • 0

    使用 json_decode()

    http://php.net/manual/en/function.json-decode.php

    用法： json_decode($json, true) //将json解码为关联数组

    回复于
  • 0

    http://php.net/manual/en/function.json-decode.php这应该派上用场了 . 同时检查http://php.net/manual/en/function.json-encode.php是否与函数完全相反 .

    回复于
  • 0

    当您以urlencoded格式发送数据时，这些值将出现在$ _POST PHP的数组中！从哪里可以使用json_decode（）解码它们：

    $json = $_POST['json']; // assumes JSON is posted as "json" POST variable
$data = url_decode($json, true); // gets you associative arrays, not objects

    您需要做的唯一事情是在POST中将JSON字符串作为“json”变量发送，在请求正文中进行正确编码（URL编码） .

    回复于

相关问题