android - android中获取方法问题

Question

我在 php 中创建了类似 get 的参数并且我正在使用 Get 方法提交到链接。例如:“http://example.com/subscribe/?act=SubscribeForEmail&EmailAddress= ”。

如果用户输入 edittext(例如 aa@gmail.com)并提交，它将提交为

http://example.com/subscribe/?act=SubscribeForEmail&EmailAddress=aa@gmail.com .怎么样了

在安卓中可能吗？

我尝试通过示例使用它。但它强制关闭

代码如下:

EditText Ename;
Button btncreate;
String n = null;
String contentOfMyInputStream1;
String output = null;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);
    Ename = (EditText)findViewById(R.id.msgTextField);
    btncreate = (Button)findViewById(R.id.sendButton);
    btncreate.setOnClickListener(this);
}

@Override
public boolean onCreateOptionsMenu(Menu menu) {
    getMenuInflater().inflate(R.menu.main, menu);
    return true;
}

@Override
public void onClick(View v) {   
    String st1;
    st1 = Ename.getText().toString();
    try {
        output = "http://example.com/subscribe/?act=SubscribeForEmail&EmailAddress="+st1;
        downloadUrl(output);//request been send
    } 
    catch (IOException e) {
        e.printStackTrace();
    }
    if (output != null) {
        Toast.makeText(this, output, Toast.LENGTH_SHORT).show();
    }
}

public String downloadUrl(String url) throws IOException {
    HttpClient httpclient = new DefaultHttpClient();
    HttpRequestBase httpRequest = null;
    HttpResponse httpResponse = null;
    InputStream inputStream = null;
    String response = "";
    StringBuffer buffer = new StringBuffer();
    httpRequest = new HttpGet(url); 
    httpResponse = httpclient.execute(httpRequest);
    inputStream = httpResponse.getEntity().getContent();
    int contentLength = (int) httpResponse.getEntity().getContentLength();
    if (contentLength < 0)  {
        // Log.e(TAG, "The HTTP response is too long.");
    }
    byte[] data = new byte[256];
    int len = 0;
    while (-1 != (len = inputStream.read(data)) ) {
        buffer.append(new String(data, 0, len));
    }
    inputStream.close();
    response = buffer.toString();
    return response;
}

我得到的错误是

Fatal Exception:main
android.os.NetworkOnMainTreadException
at android.os.StrictMode$AndroidBlockGuardPolicy.onNetwork(StrictMode.java:1117)
at java.net.InetAddress.lookHostByName(InetAddress.java:385)
at java.net.InetAddress.getAllByNameImpl(InetAddress.java:385)
at org.apache.http.impl.conn.DefaultClientConnectionOperator.openConnection(DefaultClientConnectionOperator.java:137)

请帮我解决这个问题。

提前致谢。

Answer 1

就像 Exception 告诉您的那样简单。您在主 Thread 上运行网络内容。使用 AsyncTask 或 new Thread 来执行您的 downloadUrl 方法。

使用此示例代码，取自 here :

private class LongOperation extends AsyncTask<String, Void, String> {

      @Override
      protected String doInBackground(String... params) {
            //Here you put your downloadUrl() method.
            //because this method does stuff in background
            return downloadUrl(params[0])
      }      

      @Override
      protected void onPostExecute(String output) {
          //after downloading is finished, toast it.
          if (output != null)
              Toast.makeText(this, output, Toast.LENGTH_SHORT).show();
      }

      @Override
      protected void onPreExecute() {
      }

      @Override
      protected void onProgressUpdate(Void... values) {
      }
}   

通过这两行调用它:

LongOperation downloadTask = new LongOperation();
downloadTask.execute(output);

This site也可能有所帮助。