Button &引用;“下一步”;键不';行不通

Button &引用;“下一步”;键不';行不通,button,ios6,sdk,keyboard,uitextfield,Button,Ios6,Sdk,Keyboard,Uitextfield,我有两个文本文件。在第一个字段中,我在keybord上设置了“Next”按钮,但按钮“Done”在第二个字段中起作用。我写了这段代码 - (BOOL)textFieldShouldReturn:(UITextField *)theTextField { if (theTextField == self.loginField) { [theTextField resignFirstResponder]; } else if (theTextField == self.passField)

我有两个文本文件。在第一个字段中,我在keybord上设置了“Next”按钮,但按钮“Done”在第二个字段中起作用。我写了这段代码

- (BOOL)textFieldShouldReturn:(UITextField *)theTextField {
if (theTextField == self.loginField) {
    [theTextField resignFirstResponder];
} else if (theTextField == self.passField) {
    [self.passField becomeFirstResponder];
}
return YES;
}

也许我在某些方面错了。

是否连接xib中的文本字段委托?导入UITEXTFIELDEGRADE。
    -(BOOL)textFieldShouldReturn:(UITextField *)textField
    {
        if(textField == self.loginField)
        {
            [self.passField becomeFirstResponder];
        }
        else if(textField == self.passField)
        {
            [textField resignFirstResponder];
        }
        return YES;
    }

In xib, select LoginField - Return key set as "Next", Select PassField - Return key as "Done"/"Next".