我正在为 HttpURLConnection 和 OutputStreamWriter 苦苦挣扎。
代码实际上到达了服务器,因为我得到了一个有效的错误 回复。发出 POST 请求,但没有收到数据 服务器端。
任何关于正确使用这个东西的提示都非常感谢。
代码在AsyncTask中
protected JSONObject doInBackground(Void... params) {
try {
url = new URL(destination);
client = (HttpURLConnection) url.openConnection();
client.setDoOutput(true);
client.setDoInput(true);
client.setRequestProperty("Content-Type", "application/json; charset=UTF-8");
client.setRequestMethod("POST");
//client.setFixedLengthStreamingMode(request.toString().getBytes("UTF-8").length);
client.connect();
Log.d("doInBackground(Request)", request.toString());
OutputStreamWriter writer = new OutputStreamWriter(client.getOutputStream());
String output = request.toString();
writer.write(output);
writer.flush();
writer.close();
InputStream input = client.getInputStream();
BufferedReader reader = new BufferedReader(new InputStreamReader(input));
StringBuilder result = new StringBuilder();
String line;
while ((line = reader.readLine()) != null) {
result.append(line);
}
Log.d("doInBackground(Resp)", result.toString());
response = new JSONObject(result.toString());
} catch (JSONException e){
this.e = e;
} catch (IOException e) {
this.e = e;
} finally {
client.disconnect();
}
return response;
}
我要发送的 JSON:
JSONObject request = {
"action":"login",
"user":"mogens",
"auth":"b96f704fbe702f5b11a31524bfe5f136efea8bf7",
"location":{
"accuracy":25,
"provider":"network",
"longitude":120.254944,
"latitude":14.847808
}
};
我从服务器得到的响应:
JSONObject response = {
"success":false,
"response":"Unknown or Missing action.",
"request":null
};
我应该得到的回应:
JSONObject response = {
"success":true,
"response":"Welcome Mogens Burapa",
"request":"login"
};
服务器端 PHP 脚本:
<?php
$json = file_get_contents('php://input');
$request = json_decode($json, true);
error_log("JSON: $json");
error_log('DEBUG request.php: ' . implode(', ',$request));
error_log("============ JSON Array ===============");
foreach ($request as $key => $val) {
error_log("$key => $val");
}
switch($request['action'])
{
case "register":
break;
case "login":
$response = array(
'success' => true,
'message' => 'Welcome ' . $request['user'],
'request' => $request['action']
);
break;
case "location":
break;
case "nearby":
break;
default:
$response = array(
'success' => false,
'response' => 'Unknown or Missing action.',
'request' => $request['action']
);
break;
}
echo json_encode($response);
exit;
?>
Android Studio 中的 logcat 输出:
D/doInBackground(Request)﹕ {"action":"login","location":{"accuracy":25,"provider":"network","longitude":120.254944,"latitude":14.847808},"user":"mogens","auth":"b96f704fbe702f5b11a31524bfe5f136efea8bf7"}
D/doInBackground(Resp)﹕ {"success":false,"response":"Unknown or Missing action.","request":null}
如果我将 ?action=login
附加到 URL
,我可以从服务器获得成功响应。但只有 action 参数在服务器端注册。
{"success":true,"message":"Welcome","request":"login"}
结论一定是URLConnection.write(output.getBytes("UTF-8"));
没有传输数据
好吧,数据毕竟传输了。
@greenaps 提供的解决方案可以解决问题:
$json = file_get_contents('php://input');
$request = json_decode($json, true);
上面的 PHP 脚本已更新以显示解决方案。
最佳答案
echo (file_get_contents('php://input'));
将显示 json 文本。像这样使用它:
$jsonString = file_get_contents('php://input');
$jsonObj = json_decode($jsonString, true);
关于java - HttpURLConnection 向 Apache/PHP 发送 JSON POST 请求,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/29680237/