如何将android JSON数据转换为PHP数组

我有格式的JSON数据

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

现在我想在我的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它发送到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，我会打印

jArray.toString()

并得到了输出

[{"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"}]]

---

这可能不是最好的方法，因为我是Java / Android的新手，但它适用于我:)

      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
      }

这将发送一个帖子到服务器$ _POST ['id'] = 1，然后$ _POST ['json'] =到您的json数据;

这里是save.php。 我实际上保存到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);
    ?>

---

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

---

http://php.net/manual/en/function.json-decode.php这应该派上用场。 也请查看http://php.net/manual/en/function.json-encode.php，以获取与该功能完全相反的内容。

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