我想知道当用户打开活动时是否有办法不获取相同的消息,在我的活动打开活动时,如果用户返回我想要的同一活动,您将看到一条带有吐司的消息提出代码的另一条消息:
@Override
protected void onCreate(Bundle savedInstanceState) {
// TODO Auto-generated method stub
super.onCreate(savedInstanceState);
setContentView(R.layout.q_050);
//--> Type mismatch: cannot convert from String to int
Bundle value = getIntent().getExtras();
int num = value.getString("sent").toString();
if(num){
Toast.makeText(getApplicationContext(), "next message", Toast.LENGTH_SHORT).show();
}
else{
Toast.makeText(getApplicationContext(), getApplicationContext().getString(R.string.checkpoint), Toast.LENGTH_SHORT).show();
}
答案 0 :(得分:1)
这是你的第一个活动:
@Override
protected void onCreate(Bundle savedInstanceState) {
// TODO Auto-generated method stub
super.onCreate(savedInstanceState);
setContentView(R.layout.q_050);
Bundle value = getIntent().getExtras();
String str= value.getString("sent").toString();
if(str =="activity_two"){
Toast.makeText(getApplicationContext(), "next message", Toast.LENGTH_SHORT).show();
}
else{
Toast.makeText(getApplicationContext(), getApplicationContext().getString(R.string.checkpoint), Toast.LENGTH_SHORT).show();
}
}
这是你的第二项活动:
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.second);
returnButton = (Button) findViewById(R.id.returnButton);//button to return to previous activity
returnButton.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View arg0) {
Intent intent = new Intent(SecondActivity.this,
FirstActivity.class);
intent.putExtra("sent", "activity_two"); //here 1 is a number used in first activity for if else
startActivity(intent);
}
});
}
答案 1 :(得分:0)
在活动A:
上创建一个字符串String test = "blahblahblah";
创建一个包:
Bundle b = new Bundle();
然后,将上面的字符串放到包中:
b.putString("testing", test);
定义您的活动B:
Intent i = new Intent(this, YourActivityB.class);
把你的额外包装:
i.putExtras(b);
开始活动:
startActivity(i);