[英]Sending JSON from Java HTTPUrlConnection, but $_POST variable is empty in PHP?
I'm writing an Android app and I want to send some JSON data to a PHP server. 我正在编写一个Android应用程序,并且想将一些JSON数据发送到PHP服务器。 The POST request does go to the server but in my server.php script I check the $_POST variable and it is empty. POST请求确实到达了服务器,但是在我的server.php脚本中,我检查了$ _POST变量,该变量为空。 TCP/IP monitor is not in Eclipse ADT and wireshark doesn't show localhost request so I can't see what is actually being sent. TCP / IP监视器不在Eclipse ADT中,wireshark不显示localhost请求,因此我看不到实际发送的内容。 So does anyone have an idea what is being sent and how I can access it in PHP? 那么,有谁知道发送的内容以及如何在PHP中访问它? Or have I made a mistake in the code somewhere? 还是我在某处的代码中犯了一个错误?
JSONObject json = new JSONObject();
try {
json.put("dog", "cat");
} catch (JSONException e1) {
// TODO Auto-generated catch block
e1.printStackTrace();
}
HttpURLConnection urlConnection = null;
try {
URL url = new URL("http://10.0.2.2/server.php");
urlConnection = (HttpURLConnection)url.openConnection();
urlConnection.setRequestProperty("Content-Type", "application/json");
urlConnection.setRequestProperty("Accept", "application/json");
urlConnection.setRequestMethod("POST");
urlConnection.setDoOutput(true);
OutputStreamWriter os = new OutputStreamWriter(urlConnection.getOutputStream(), "UTF-8");
os.write(json.toString());
os.close();
}
try it. 试试吧。
JSONObject json = new JSONObject();
try {
json.put("dog", "cat");
} catch (JSONException e1) {
// TODO Auto-generated catch block
e1.printStackTrace();
}
HttpClient localDefaultHttpClient=new DefaultHttpClient();
HttpPost lotlisting = new HttpPost("http://10.0.2.2/server.php");
ArrayList localArrayList = new ArrayList();
localArrayList.add(new BasicNameValuePair("json",json.toString()));
try {
lotlisting.setEntity(new UrlEncodedFormEntity(localArrayList));
String str = EntityUtils.toString(localDefaultHttpClient.execute(lotlisting).getEntity());
} catch (Exception e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
you will get the output in str variable; 您将在str变量中获得输出;
I think is missing the flush Try os.flush(); 我认为缺少同花顺试试os.flush(); after os.write(..) 在os.write(..)之后
Don't use $_POST. 不要使用$ _POST。 Do something like this on user server 在用户服务器上执行类似的操作
$json = file_get_contents('php://input');
$animals= json_decode($json,true);
//you can do
echo $animals['dog'];
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