小编use*_*343的帖子

我正在尝试创建一个向translate.google.com发送单词的应用,获取翻译结果并将其显示给用户.我编写了URL,但我不知道如何从网页中提取单词/短语.

示例伪:en是英语的代码,es是西班牙语的代码

String from = "en";
String to = "es";
String word = "hello"; //this will be user input text really
String URL = "http://www.translate.google.com/#" + from + "/" + to + "/" + word;

因此,请求网址将类似于http://www.translate.google.com/#en/es/hello

我现在需要能够从结果框中检索信息并将其放在String中,以便我可以将其显示给用户.

此刻我无法找出问题所在.我不断收到以下代码的相同错误消息:

package com.example.test1;

import android.app.Activity;
import android.os.Bundle;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;
import android.widget.Toast;
import android.util.Log;
import com.memetix.mst.language.Language;
import com.memetix.mst.translate.Translate;

public class MainActivity extends Activity {
  private EditText text;
  private EditText text1;
  //private String translatedText;


  @Override
   public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
    Translate.setClientId("myId");
Translate.setClientSecret("mySecret");
    text = (EditText) findViewById(R.id.editText1);
    text1 = (EditText) findViewById(R.id.editText2);

    Button Trans1 = (Button)findViewById(R.id.button1);
    Trans1.setOnClickListener(new View.OnClickListener(){
    public void onClick(View view) {
        String word = text.getText().toString();
        try {
            String translatedText = Translate.execute(word, Language.ENGLISH, Language.GERMAN);
            text1.setText(translatedText);
        } catch …