Question

我有UITextField用于输入搜索字符串，UITableView用于输入结果。我想要的是当用户输入超过3个字母时运行搜索功能，并且自从最后一个符号添加到UITextView后至少经过了0.5秒。

我找到了（Detect when user stopped / paused typing in Swift）函数，我把它添加到我的ViewController中，它有类SearchVC和方法server_search

func textField(_ textField: UITextField, shouldChangeCharactersIn range: NSRange, replacementString string: String) -> Bool {
    NSObject.cancelPreviousPerformRequests(
        withTarget: self,
        selector: #selector(SearchVC.server_search),
        object: textField)
    self.perform(
        #selector(SearchVC.server_search),
        with: textField,
        afterDelay: 0.5)
    return true
}

但没有任何反应。

Answer 1

您可以使用Timer ...

来实现这一目标

var timer: Timer?

func textField(_ textField: UITextField, shouldChangeCharactersIn range: NSRange, replacementString string: String) -> Bool {

    timer?.invalidate()  // Cancel any previous timer

    // If the textField contains at least 3 characters…
    let currentText = textField.text ?? ""
    if (currentText as NSString).replacingCharacters(in: range, with: string).characters.count >= 3 {

         // …schedule a timer for 0.5 seconds
         timer = Timer.scheduledTimer(timeInterval: 0.5, target: self, selector: #selector(performSearch()), userInfo: nil, repeats: false)
    }

    return true
}

func performSearch() {

}

并且不要忘记将视图控制器设置为UITextField delegate

Answer 2

定时器的使用有一些优点。根据您的实施，您将取消所有为你的对象执行的操作，这可能会失控。

相反，计时器可让您对其进行细粒度控制。请参阅以下实现：

var searchTimer: Timer?
func textField(_ textField: UITextField, shouldChangeCharactersIn range: NSRange, replacementString string: String) -> Bool {
    // Invalidate remove the previous timer, since it's optional if the previous timer is already cancelled or nil, it won't affect execution
    searchTimer?.invalidate()
    searchTimer = Timer.scheduledTimer(withTimeInterval: 0.5, repeats: false, block: { (timer) in
        //do Something crazy
        self.server_search()
    })
    return true
}

Answer 3

虽然使用import numpy as np def moving_average(a, n=3) : ret = np.cumsum(a, dtype=float) ret[n:] = ret[n:] - ret[:-n] return ret[n - 1:] / n仍然是有效的答案，但是从Swift 3开始，您可以使用Timer…

DispatchWorkItem

在swift iOS应用中延迟搜索

3 个答案: