Searchbar“if”语句不断显示“else”结果

Question

从我提供的代码中，我想知道我做错了什么？

search.setOnClickListener(new View.OnClickListener() {
    @Override
    public void onClick(View v) {
        String Username = sbar.getText().toString();
        String foundplayer = db.getUsername();
         if (Username.equals(foundplayer)) {
             ResultFrame.setVisibility(View.VISIBLE);
             ResultText.setText(foundplayer);
             Toast.makeText(getApplicationContext(), "Player Found", Toast.LENGTH_LONG).show();                          
         } else {
             Toast.makeText(getApplicationContext(), "Username Not Found", Toast.LENGTH_LONG).show();
         }    
    }

});

public String getUsername() {
    String[] userName = new String[] { "USERNAME" };
    Cursor c = db.query( "NFDB", null, null, null, null, null, null, null);
    String result = "";
    int iName = c.getColumnIndex("USERNAME");
    for (c.moveToFirst(); !c.isAfterLast(); c.moveToNext()) {
        result = c.getString(iName);
    }
    return result;
}

Answer 1

首先，确保您的getUsername（）方法返回值…如果它返回值，那么像这样做。

  String foundplayer = db.getUsername();

这将需要一些时间来处理，因为它涉及到从数据库获取数据，所以编译器不会等待它完成，因为编译器将在完成getUsername（）方法之前转到if条件。因此，您的foundplayer将为空

所以它永远不会满足条件。

  String foundplayer = "";
Thread t=new Thread()
{
public void run()
{
  foundplayer = db.getUsername();
}
};
t.start();
t.join();