2013-04-25 52 views
0

我有多EditTexts。 我如何验证EditTextEditText中有“”,如果EditText是“”,那么我希望用户在编辑文本之前必须输入一个数字 这怎么办?验证在onFocusChange

case R.id.P2Throw1Set2: 
    p212r.setText(String.valueOf(Integer.parseInt(p2score.getText().toString()) - Integer.parseInt(p212.getText().toString()))); 
    p2score.setText(String.valueOf(Integer.parseInt(p212r.getText().toString()))); 
    break; 
case R.id.P2Throw2Set2: 
    p222r.setText(String.valueOf(Integer.parseInt(p2score.getText().toString()) - Integer.parseInt(p222.getText().toString()))); 
    p2score.setText(String.valueOf(Integer.parseInt(p222r.getText().toString()))); 
    break; 
case R.id.P2Throw3Set2: 
    p232r.setText(String.valueOf(Integer.parseInt(p2score.getText().toString()) - Integer.parseInt(p232.getText().toString()))); 
    p2score.setText(String.valueOf(Integer.parseInt(p232r.getText().toString()))); 
    break; 
case R.id.P2Throw4Set2: 
    p242r.setText(String.valueOf(Integer.parseInt(p2score.getText().toString()) - Integer.parseInt(p242.getText().toString()))); 
    p2score.setText(String.valueOf(Integer.parseInt(p242r.getText().toString()))); 
    break; 

回答

0

您将不得不禁用除第一个编辑文本以外的所有文本,只在用户提供满意的输入时启用它们。然后换你parseInt函数()调用try/catch块,像这样:

case R.id.P2Throw2Set2: 
    try 
    { 
     p222r.setText(String.valueOf(Integer.parseInt(p2score.getText().toString()) - 
      Integer.parseInt(p222.getText().toString()))); 
     p2score.setText(String.valueOf(Integer.parseInt(p222r.getText().toString()))); 

     // if the previous lines worked, this will work 
     p232r.setEnabled(true); 
    } 
    catch(NumberFormatException e) 
    { 
     // user entered "" or the value was null 
     // in this case, we leave the next edit text disabled 
    } 
    break;