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我只是想用JsonRequestObject到值發送到PHP腳本和接收JSON數據，但下面的代碼無法正常工作通行證的JSONObject作爲請求參數使用POST方法凌空
package com.demo.volleyjsondemo;
import android.content.Context;
import android.support.v7.app.AppCompatActivity;
import android.os.Bundle;
import android.util.Log;
import android.widget.TextView;
import com.android.volley.AuthFailureError;
import com.android.volley.Request;
import com.android.volley.Response;
import com.android.volley.VolleyError;
import com.android.volley.toolbox.JsonObjectRequest;
import com.android.volley.toolbox.StringRequest;
import com.demo.volleyjsondemo.Utils.Constants;
import com.demo.volleyjsondemo.Utils.RequestSingleTone;
import org.json.JSONException;
import org.json.JSONObject;
import java.util.HashMap;
import java.util.Map;
public class MainActivity extends AppCompatActivity {
JsonObjectRequest jsonObjectRequest;
TextView txtName, txtEmail;
JSONObject parameters;
Context context;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
init();
try {
sendReq();
} catch (JSONException e) {
e.printStackTrace();
}
}
private void init() {
context = MainActivity.this;
txtName = (TextView) findViewById(R.id.txtName);
txtEmail = (TextView) findViewById(R.id.txtEmail);
}
private void sendReq() throws JSONException {
//just to demonstrate how to send parameters with json request
parameters = new JSONObject();
try {
parameters.put(Constants.NAME, "jack");
Log.e("paramter",parameters.toString());
} catch (JSONException e) {
e.printStackTrace();
}
//jsonobjectrequest to send request and get response in json
jsonObjectRequest = new JsonObjectRequest(
Request.Method.POST //request method
, Constants.BASE_URL + Constants.GET_PERSON_DETAILS_URL //URL of php file
, new JSONObject(parameters.toString()) //parameters to send to server
, new Response.Listener<JSONObject>() { //response will come here in case of success
@Override
public void onResponse(JSONObject response) {
try {
Log.e("response", response.getString(Constants.NAME));
} catch (JSONException e) {
e.printStackTrace();
}
}
},
new Response.ErrorListener() { //response will come here in case of error
@Override
public void onErrorResponse(VolleyError error) {
error.printStackTrace();
}
});
//add request to singletone
RequestSingleTone.getInstance(context).addRequest(jsonObjectRequest);
}
}
PHP代碼
<?php
require_once 'dbconfig.php';
$name = $_POST['name'];
//$name="jack";
$singlePersonInfoQuery = "SELECT * FROM test WHERE name='".$name."'";
$result = mysqli_query($con,$singlePersonInfoQuery);
if(mysqli_num_rows($result) > 0){
$raw = mysqli_fetch_assoc($result);
echo json_encode(array("name"=>$raw['name'],"email"=>$raw['email']));
}else{
echo json_encode(array("name"=>$name,"email"=>"blank"));
}
?>
我得到以下錯誤
10-06 14:36:34.943 5256-5256/com.demo.volleyjsondemo W/System.err: com.android.volley.ParseError: org.json.JSONException: Value perfect<br of type java.lang.String cannot be converted to JSONObject
10-06 14:36:34.943 5256-5256/com.demo.volleyjsondemo W/System.err: at com.android.volley.toolbox.JsonObjectRequest.parseNetworkResponse(JsonObjectRequest.java:73)
10-06 14:36:34.943 5256-5256/com.demo.volleyjsondemo W/System.err: at com.android.volley.NetworkDispatcher.run(NetworkDispatcher.java:123)
10-06 14:36:34.943 5256-5256/com.demo.volleyjsondemo W/System.err: Caused by: org.json.JSONException: Value perfect<br of type java.lang.String cannot be converted to JSONObject
10-06 14:36:34.943 5256-5256/com.demo.volleyjsondemo W/System.err: at org.json.JSON.typeMismatch(JSON.java:111)
10-06 14:36:34.943 5256-5256/com.demo.volleyjsondemo W/System.err: at org.json.JSONObject.<init>(JSONObject.java:160)
10-06 14:36:34.943 5256-5256/com.demo.volleyjsondemo W/System.err: at org.json.JSONObject.<init>(JSONObject.java:173)
10-06 14:36:34.943 5256-5256/com.demo.volleyjsondemo W/System.err: at com.android.volley.toolbox.JsonObjectRequest.parseNetworkResponse(JsonObjectRequest.java:68)
10-06 14:36:34.943 5256-5256/com.demo.volleyjsondemo W/System.err: ... 1 more
我猜想上面的警告是因爲錯誤在PHP代碼script..Above接收參數，做工精細，如果我不傳遞任何參數或使用stringrequest，但我想用jsonrequestobject..I無法理解什麼是錯
我們真的不能只從代碼說，你需要調試PHP腳本，並確保它總是輸出形成正確的JSON。在Java錯誤在PHP響應什麼不對勁的提示。 –
當我正常使用$ NAME =「傑克」，它的工作原理fine..It是給我正確的JSON數據，但我很困惑，向Android傳遞參數和php.That裝置接收名稱不寫的長度適當 –
傳遞在文件接收參數，我發現我無法接收參數，因爲它顯示0 length..Please幫助我，我怎麼能傳遞請求參數，並用它在PHP腳本 –