Question

我尝试构建一个Android应用程序，将参数发送到php代码作为where

在那个php代码中查询条件，我应该在log.i中查询查询结果但是

我收到以下错误

org.json.JSONException：java.lang.String类型的值john无法转换为JSONObject

MainActivity

 public class MainActivity extends Activity {

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);

        try {
            JSONObject toSend = new JSONObject();
            toSend.put("msg", "3");

            JSONTransmitter transmitter = new JSONTransmitter();
            transmitter.execute(new JSONObject[] {toSend});

        } catch (JSONException e) {
            e.printStackTrace();
        }

    }

}

JSONTransmitter类。

public class JSONTransmitter extends AsyncTask<JSONObject, JSONObject, JSONObject> {

    String url = "http://192.168.1.10:89/b.php";


    protected JSONObject doInBackground(JSONObject... data) {
        JSONObject json = data[0];
        HttpClient client = new DefaultHttpClient();
        HttpConnectionParams.setConnectionTimeout(client.getParams(), 100000);
        StrictMode.setThreadPolicy(new StrictMode.ThreadPolicy.Builder().permitNetwork().build());
        JSONObject jsonResponse = null;

        HttpPost post = new HttpPost(url);
        try {
            StringEntity se = new StringEntity("json="+json.toString());
            post.addHeader("content-type", "application/x-www-form-urlencoded");
            post.setEntity(se);

            HttpResponse response;
            response = client.execute(post);
            String resFromServer = org.apache.http.util.EntityUtils.toString(response.getEntity());

            jsonResponse=new JSONObject(resFromServer);
            Log.i("Response from server", jsonResponse.getString("msg"));
            Toast.makeText(null, resFromServer, Toast.LENGTH_LONG);
        } catch (Exception e) { e.printStackTrace();}

        return jsonResponse;
    }

}

php code

<?php
 mysql_connect("localhost", "root", "password") 
or die(mysql_error());
mysql_select_db("test") or die(mysql_error());


$result = mysql_query(" select  part_name from  Services_parts where  part_id=  1 ") 
or die(mysql_error());  

$row = mysql_fetch_array( $result );


$j_out = new stdClass();
$j_out->part_name= $row['part_name'];

echo json_encode($j_out);
?>

Answer 1

执行与此等效的代码：

JSONObject obj = new JSONObject("john");

显然会导致异常

这意味着来自服务器的json数据格式错误。问题可能在你的php代码里面，用于返回json中的项目数组，适应php + java代码：

$res_items = array();
for ( ... )  {
    array_push($res_items, $some_item);
}
$js_result = get_json_encode(array(
                "results" => $res_items
            ));
echo $js_result;

然后在java代码中：

JSONObject jsResp = new JSONObject(serverJsonResult);
JSONArray results = jsResp.getJSONArray("results");
for (int n = 0; n < results.length(); ++n) {
   JSONObject js_row = results.getJSONObject(n);
   // ...
}

java.lang.String类型的值john无法转换为JSONObject

1 个答案: