Question

我正在尝试实现一个简单的步进器，它将其值反映到标签上。使用＆＃34; Int（sender.value）＆＃34;我收到错误＆＃34;模糊地使用价值＆＃34;

//  ViewController.swift
//  Stepper
//
//  Created by Prabhu Konchada on 19/06/15.
//  Copyright (c) 2015 Prabhu. All rights reserved.
//

import UIKit

class ViewController: UIViewController {

    @IBOutlet weak var StepperValue: UILabel!

    @IBOutlet weak var OutputLabel: UILabel!

    override func viewDidLoad() {
        super.viewDidLoad()
        // Do any additional setup after loading the view, typically from a nib.
    }

    override func didReceiveMemoryWarning() {
        super.didReceiveMemoryWarning()
        // Dispose of any resources that can be recreated.
    }


    @IBAction func StepperTap(sender: AnyObject) {

        self.OutputLabel.text = String(Int(sender.value).description)
    }
}

Answer 1

你可以这样做：

@IBOutlet var lblStep: UILabel!
@IBAction func stepPressed(sender: UIStepper) {

    lblStep.text = sender.value.description

}

否则您必须将AnyObject转换为UIStepper 像是

var stepControl : UIStepper = sender as! UIStepper

就像：

@IBAction func StepperTap(sender: AnyObject) {
    var stepControl : UIStepper = sender as! UIStepper
    self.OutputLabel.text = stepControl.value.description
}

整个代码为6.1版，你必须改为as！对于6.3.2：

import UIKit

class ViewController: UIViewController {

    @IBOutlet var lblStep: UILabel!
    override func viewDidLoad() {
        super.viewDidLoad()
        // Do any additional setup after loading the view, typically from a nib.
    }

    override func didReceiveMemoryWarning() {
        super.didReceiveMemoryWarning()
        // Dispose of any resources that can be recreated.
    }


    @IBAction func stepPressed(sender: AnyObject) {

        var step : UIStepper = sender as UIStepper

        lblStep.text = step.value.description

    }
}

Answer 2

尝试修改您的代码：

Select * from t1,t2,t3 
where t1.id = 1 or t2.id = 1 0r t3.id = 1

模糊地使用＆＃34;值＆＃34;同时实现UIStepper

2 个答案: