Question

想要将人名带到下一个活动，如果登录不正确，这将显示吐司不起作用，只显示吐司。应该在下一个活动中添加什么，这个代码有什么问题

       @Override
         protected void onCreate(Bundle savedInstanceState) {

         super.onCreate(savedInstanceState);

         setContentView(R.layout.activity_main);



          final EditText eText1 = (EditText)findViewById(R.id.editText1);

         final EditText eText2 = (EditText)findViewById(R.id.editText2);

          final Button btn1=(Button)findViewById(R.id.button1);

          btn1.setOnClickListener( new View.OnClickListener(){



        public void onClick(View v) {

        String str = eText1.getText().toString();

        String log= "malaika" ;

       if(str == log){

       Intent intent=new Intent(getApplicationContext(),FirstButton.class);

        intent.putExtra("welcome", log);
                    }
        else {

       Toast tos =Toast.makeText(getApplication(),"wrong",Toast.LENGTH_SHORT);

        tos.show();
                }

          }});

Answer 1

首先在equals的地方使用==并开始新的活动

if(str.equals(log))
{
       Intent intent=new Intent(getApplicationContext(),FirstButton.class);
       intent.putExtra("welcome", log);

       /*'welcome' is acting as a key which will carry  your
       name from this activity to the FirstButton ACTIVITY
          I suggest changing it
        */

      startActivity(intent);//for starting new activity
}
else 
{
       Toast tos =Toast.makeText(getApplication(),"wrong",Toast.LENGTH_SHORT);
       tos.show();
}

以及您可以通过

获取姓名的FirstButton活动

Intent i=getIntent();
String name=i.getStringExtra("welcome");

我想在主要活动中添加意图

1 个答案: