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我不知道为什么我不能从我的 PHP 代码中得到正确的响应。对于我的 PHP 代码，它真的很简单。它应该只返回 POST 参数。但在我的 Android 项目中，它为所有参数返回 null。我试图在我的 iOS 项目中测试 PHP 代码。一切都很完美。它返回["target": zh, "arrayString": <__NSSingleObjectArrayI 0x6000000192a0>(hello world), "source": en]所以，我想也许我的 Android 项目有一些问题，然后我尝试创建一个新项目并再次执行，但仍然遇到同样的问题。以下是我的新 Android 项目的步骤。

构建 Gradle

dependencies {
    compile 'com.android.volley:volley:1.1.0'
}

AndroidManifest

<uses-permission android:name="android.permission.INTERNET" />

主要活动

import android.os.Bundle;
import android.support.v7.app.AppCompatActivity;
import android.util.Log;

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.Volley;

import org.json.JSONArray;
import org.json.JSONException;
import org.json.JSONObject;

public class MainActivity extends AppCompatActivity {

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

        String requestString = "http://ddoommaaiinn.com/testPost.php";
        JSONObject parameters = new JSONObject();
        try {
            parameters.put("source", "en");
            parameters.put("target", "zh");
            JSONArray jsonArray = new JSONArray();
            jsonArray.put("hello world");
            parameters.put("stringArray", jsonArray);
        } catch (JSONException e) {
            e.printStackTrace();
        }

        Log.d("JSONParams", parameters.toString());

        JsonObjectRequest jsonRequest = new JsonObjectRequest(Request.Method.POST, requestString, parameters, new Response.Listener<JSONObject>() {
            @Override
            public void onResponse(JSONObject response) {
                //TODO: handle success
                Log.d("successful", response.toString());
            }
        }, new Response.ErrorListener() {
            @Override
            public void onErrorResponse(VolleyError error) {
                error.printStackTrace();
                //TODO: handle failure
            }
        });

        Volley.newRequestQueue(this).add(jsonRequest);
    }
}

调试

12-27 08:47:19.994 4363-4363/com.pakhocheung.testjsonobjectrequest D/JSONParams: {"source":"en","target":"zh","stringArray":["hello world"]}
12-27 08:47:20.050 4363-4395/com.pakhocheung.testjsonobjectrequest D/NetworkSecurityConfig: No Network Security Config specified, using platform default
12-27 08:47:20.079 4363-4397/com.pakhocheung.testjsonobjectrequest D/OpenGLRenderer: HWUI GL Pipeline
12-27 08:47:20.885 4363-4397/com.pakhocheung.testjsonobjectrequest I/zygote: android::hardware::configstore::V1_0::ISurfaceFlingerConfigs::hasWideColorDisplay retrieved: 0
12-27 08:47:20.886 4363-4397/com.pakhocheung.testjsonobjectrequest I/OpenGLRenderer: Initialized EGL, version 1.4
12-27 08:47:20.886 4363-4397/com.pakhocheung.testjsonobjectrequest D/OpenGLRenderer: Swap behavior 1
12-27 08:47:20.886 4363-4397/com.pakhocheung.testjsonobjectrequest W/OpenGLRenderer: Failed to choose config with EGL_SWAP_BEHAVIOR_PRESERVED, retrying without...
12-27 08:47:20.886 4363-4397/com.pakhocheung.testjsonobjectrequest D/OpenGLRenderer: Swap behavior 0
12-27 08:47:20.892 4363-4397/com.pakhocheung.testjsonobjectrequest D/EGL_emulation: eglCreateContext: 0xa6f859e0: maj 2 min 0 rcv 2
12-27 08:47:20.895 4363-4397/com.pakhocheung.testjsonobjectrequest D/EGL_emulation: eglMakeCurrent: 0xa6f859e0: ver 2 0 (tinfo 0xb050d880)
12-27 08:47:20.998 4363-4397/com.pakhocheung.testjsonobjectrequest D/EGL_emulation: eglMakeCurrent: 0xa6f859e0: ver 2 0 (tinfo 0xb050d880)
12-27 08:47:21.279 4363-4363/com.pakhocheung.testjsonobjectrequest D/successful: {"source":null,"target":null,"arrayString":null}

PHP 代码

<?php
$source = $_POST['source'];
$target = $_POST['target'];
$arrayString = $_POST['stringArray'];
echo json_encode(array('source'=>$source,'target'=>$target,'arrayString'=>$arrayString));
?>
4

2 回答 2

1

当使用JsonObjectRequest所需的数据时，将在php://input(json) 字符串中找到。要获取该数据，请执行以下操作：

<?php
$postArray = json_decode(file_get_contents('php://input'), true);

// now you access the values nearly as before:
$source = $postArray['source'];
$target = $postArray['target'];
$arrayString = $postArray['arrayString'];

echo json_encode(array('source'=>$source,'target'=>$target,'arrayString'=>$arrayString));
?>
于 2017-12-27T15:47:56.647 回答
0

看起来你 $_POST 不工作

用于测试尝试将硬编码值

$source = 'test';
$target = 'test-target';
$arrayString = 'test-arrayString';

echo json_encode(array('source'=>$source,'target'=>$target,'arrayString'=>$arrayString));

它应该给出以下json

{"source":"test","target":"test-target","arrayString":"test-arrayString"}

如果有什么遗漏请补充

于 2017-12-27T14:58:07.047 回答