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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"},{"用户名":"test"}]

现在我想在我的 PHP Web 系统中获取这些数据。我正在将这些数据从 android 应用程序发送到 PHP 服务器。我正在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

jArray.toString()
并得到输出

[{"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。谁能帮帮我吗？

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"}]]
4

4 回答 4

0

采用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

于 2012-09-23T06:52:06.160 回答
0

http://php.net/manual/en/function.json-decode.php这应该会派上用场。还要检查http://php.net/manual/en/function.json-encode.php以获取与该功能完全相反的内容。

于 2012-09-23T06:52:08.377 回答
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！从那里您将能够使用 json_decode() 对它们进行解码：

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

您唯一需要做的就是在 POST 中将 JSON 字符串作为“json”变量发送，并在请求正文中进行正确编码（URL 编码）。

于 2012-09-23T06:54:02.023 回答
0

这可能不是最好的方法，因为我是 Java/Android 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
      }

这将使用 $_POST['id'] = 1 将帖子发送到服务器，然后将 $_POST['json'] = 发送到您的 json 数据；

这是save.php。我实际上将它保存到 MySQL，所以是的。

    <?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);
    ?>
于 2012-09-23T08:06:30.873 回答