如果 intime 和 outtime 都为空,则Toast或两者都为空,然后传递
private boolean isValidate() {
// TODO Auto-generated method stub
if(sp.getSelectedItem().toString().trim().length() == 0){
Toast.makeText(getApplicationContext(), "Please Enter Name !", Toast.LENGTH_LONG).show();
return false;
}else{
}
if(etDate.getText().toString().trim().length() == 0){
Toast.makeText(getApplicationContext(), "Please Enter Date !", Toast.LENGTH_LONG).show();
return false;
}else{
}
//both intime and out time passed either one of passed here
if(etintime.getText().toString().trim().length() == 0){
Toast.makeText(getApplicationContext(), "Please Enter In Time !", Toast.LENGTH_LONG).show();
return false;
}else{
}
if(etouttime.getText().toString().trim().length() == 0){
Toast.makeText(getApplicationContext(), "Please Enter Out Time !", Toast.LENGTH_LONG).show();
return false;
}else{
}
return true;
}
答案 0 :(得分:0)
使用&&条件运算符执行逻辑和在您的情况下检查两个EditTexts是否为空。
e.g。
if(editText1.getText().toString().equals("") && editText2.getText().toString().equals("")){
//show Toast here.
}