我对PHP、JavaScript和许多其他脚本语言很有经验,但我对Java或Android没有太多经验。
我正在寻找一种方法将POST数据发送到PHP脚本并显示结果。
我对PHP、JavaScript和许多其他脚本语言很有经验,但我对Java或Android没有太多经验。
我正在寻找一种方法将POST数据发送到PHP脚本并显示结果。
当前回答
您可以使用WebServer类POST一个HttpRequest,并在其侦听器接口中跟踪响应。
WebServer server=new WebServer(getApplicationContext());
server.setOnServerStatusListner(new WebServer.OnServerStatusListner() {
@Override
public void onServerResponded(String responce) {
}
@Override
public void onServerRevoked() {
}
});
现在创建一个DataRack来绑定数据
List<DataRack> racks=new ArrayList<DataRack>();
racks.add(new DataRack("name","Simon"));
racks.add(new DataRack("age","40"));
racks.add(new DataRack("location","Canada"));
现在只需发送带有该机架的POST请求
server.connectWithPOST(MainActivity.this,"http://sangeethnandakumar.esy.es/PROJECTS/PUBLIC_SERVICE/posttest.php",racks);
你需要包括我的库。文件在这里
其他回答
通过这种方式,我们可以用http post方法发送数据并获得结果
public class MyHttpPostProjectActivity extends Activity implements OnClickListener {
private EditText usernameEditText;
private EditText passwordEditText;
private Button sendPostReqButton;
private Button clearButton;
/** Called when the activity is first created. */
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.login);
usernameEditText = (EditText) findViewById(R.id.login_username_editText);
passwordEditText = (EditText) findViewById(R.id.login_password_editText);
sendPostReqButton = (Button) findViewById(R.id.login_sendPostReq_button);
sendPostReqButton.setOnClickListener(this);
clearButton = (Button) findViewById(R.id.login_clear_button);
clearButton.setOnClickListener(this);
}
@Override
public void onClick(View v) {
if(v.getId() == R.id.login_clear_button){
usernameEditText.setText("");
passwordEditText.setText("");
passwordEditText.setCursorVisible(false);
passwordEditText.setFocusable(false);
usernameEditText.setCursorVisible(true);
passwordEditText.setFocusable(true);
}else if(v.getId() == R.id.login_sendPostReq_button){
String givenUsername = usernameEditText.getEditableText().toString();
String givenPassword = passwordEditText.getEditableText().toString();
System.out.println("Given username :" + givenUsername + " Given password :" + givenPassword);
sendPostRequest(givenUsername, givenPassword);
}
}
private void sendPostRequest(String givenUsername, String givenPassword) {
class SendPostReqAsyncTask extends AsyncTask<String, Void, String>{
@Override
protected String doInBackground(String... params) {
String paramUsername = params[0];
String paramPassword = params[1];
System.out.println("*** doInBackground ** paramUsername " + paramUsername + " paramPassword :" + paramPassword);
HttpClient httpClient = new DefaultHttpClient();
// In a POST request, we don't pass the values in the URL.
//Therefore we use only the web page URL as the parameter of the HttpPost argument
HttpPost httpPost = new HttpPost("http://www.nirmana.lk/hec/android/postLogin.php");
// Because we are not passing values over the URL, we should have a mechanism to pass the values that can be
//uniquely separate by the other end.
//To achieve that we use BasicNameValuePair
//Things we need to pass with the POST request
BasicNameValuePair usernameBasicNameValuePair = new BasicNameValuePair("paramUsername", paramUsername);
BasicNameValuePair passwordBasicNameValuePAir = new BasicNameValuePair("paramPassword", paramPassword);
// We add the content that we want to pass with the POST request to as name-value pairs
//Now we put those sending details to an ArrayList with type safe of NameValuePair
List<NameValuePair> nameValuePairList = new ArrayList<NameValuePair>();
nameValuePairList.add(usernameBasicNameValuePair);
nameValuePairList.add(passwordBasicNameValuePAir);
try {
// UrlEncodedFormEntity is an entity composed of a list of url-encoded pairs.
//This is typically useful while sending an HTTP POST request.
UrlEncodedFormEntity urlEncodedFormEntity = new UrlEncodedFormEntity(nameValuePairList);
// setEntity() hands the entity (here it is urlEncodedFormEntity) to the request.
httpPost.setEntity(urlEncodedFormEntity);
try {
// HttpResponse is an interface just like HttpPost.
//Therefore we can't initialize them
HttpResponse httpResponse = httpClient.execute(httpPost);
// According to the JAVA API, InputStream constructor do nothing.
//So we can't initialize InputStream although it is not an interface
InputStream inputStream = httpResponse.getEntity().getContent();
InputStreamReader inputStreamReader = new InputStreamReader(inputStream);
BufferedReader bufferedReader = new BufferedReader(inputStreamReader);
StringBuilder stringBuilder = new StringBuilder();
String bufferedStrChunk = null;
while((bufferedStrChunk = bufferedReader.readLine()) != null){
stringBuilder.append(bufferedStrChunk);
}
return stringBuilder.toString();
} catch (ClientProtocolException cpe) {
System.out.println("First Exception caz of HttpResponese :" + cpe);
cpe.printStackTrace();
} catch (IOException ioe) {
System.out.println("Second Exception caz of HttpResponse :" + ioe);
ioe.printStackTrace();
}
} catch (UnsupportedEncodingException uee) {
System.out.println("An Exception given because of UrlEncodedFormEntity argument :" + uee);
uee.printStackTrace();
}
return null;
}
@Override
protected void onPostExecute(String result) {
super.onPostExecute(result);
if(result.equals("working")){
Toast.makeText(getApplicationContext(), "HTTP POST is working...", Toast.LENGTH_LONG).show();
}else{
Toast.makeText(getApplicationContext(), "Invalid POST req...", Toast.LENGTH_LONG).show();
}
}
}
SendPostReqAsyncTask sendPostReqAsyncTask = new SendPostReqAsyncTask();
sendPostReqAsyncTask.execute(givenUsername, givenPassword);
}
}
您可以使用WebServer类POST一个HttpRequest,并在其侦听器接口中跟踪响应。
WebServer server=new WebServer(getApplicationContext());
server.setOnServerStatusListner(new WebServer.OnServerStatusListner() {
@Override
public void onServerResponded(String responce) {
}
@Override
public void onServerRevoked() {
}
});
现在创建一个DataRack来绑定数据
List<DataRack> racks=new ArrayList<DataRack>();
racks.add(new DataRack("name","Simon"));
racks.add(new DataRack("age","40"));
racks.add(new DataRack("location","Canada"));
现在只需发送带有该机架的POST请求
server.connectWithPOST(MainActivity.this,"http://sangeethnandakumar.esy.es/PROJECTS/PUBLIC_SERVICE/posttest.php",racks);
你需要包括我的库。文件在这里
在新版本的Android中,你必须把所有的web I/O请求放到一个新的线程中。AsyncTask最适合小请求。
下面是一个完整的解决方案,它运行在后台线程中,向Web API发送一个HTTPS POST多部分请求。实际测试和工作代码。注意:“\n”字符需要在顶部的“最终字符串”中进行正确的请求格式化。
我有困难要么理解,转换,或完成以上解决方案,我的多部分POST需求。
@Override
protected Integer doInBackground(String... files) {
final String MULTIPART_BOUNDARY = "xxYYzzSEPARATORzzYYxx";
final String MULTIPART_SEPARATOR = "--" + MULTIPART_BOUNDARY + "\n";
final String MULTIPART_FORM_DATA = "Content-Disposition: form-data; name=\"%s\"\n\n";
final String FORM_DATA_FILE1 = "file1";
final String FORM_DATA_FILE2 = "file2";
OutputStream outputStream;
Integer responseCode = 0;
try {
URL url = new URL("https://www.example.com/api/endpoint?n1=v1&n2=v2");
HttpsURLConnection urlConnection = (HttpsURLConnection) url.openConnection();
urlConnection.setRequestProperty("Content-Type", "multipart/form-data; boundary=" + MULTIPART_BOUNDARY);
urlConnection.setConnectTimeout(6000);
urlConnection.setRequestMethod("POST");
urlConnection.setDoOutput(true);
outputStream = new BufferedOutputStream(urlConnection.getOutputStream());
BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(outputStream, StandardCharsets.UTF_8));
writer.write(MULTIPART_SEPARATOR);
writer.write(String.format(MULTIPART_FORM_DATA, FORM_DATA_FILE1));
writer.write(files[0]);
writer.write(MULTIPART_SEPARATOR);
writer.write(String.format(MULTIPART_FORM_DATA, FORM_DATA_FILE2));
writer.write(files[1]);
writer.write(MULTIPART_SEPARATOR);
writer.flush();
writer.close();
outputStream.close();
urlConnection.connect();
responseCode = urlConnection.getResponseCode();
Log.d("ResponseCode:", String.valueOf(responseCode));
urlConnection.disconnect();
} catch (IOException e) {
e.printStackTrace();
}
return responseCode;
}
你可以使用URLConnection setDoOutput(true), getOutputStream()(用于发送数据),和getInputStream()(用于接收)。Sun在这方面有一个例子。