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如何在Swift中设置选择器视图的起始行？我看到有一个Objective C的代码,但我不明白.如果有人能解释我如何将Objective C代码翻译成Swift,这也很棒!

谢谢

下面的代码将更改所有3个组件的选择器视图的字体颜色.然而,当我试图旋转车轮时,它会崩溃.我认为它与didSelectRow函数有关.也许这两个函数必须以某种方式嵌套？任何的想法？

    func pickerView(pickerView: UIPickerView, attributedTitleForRow row: Int, forComponent component: Int) -> NSAttributedString? {
    var attributedString: NSAttributedString!
    if component == 0 {
        attributedString = NSAttributedString(string: a.text!, attributes: [NSForegroundColorAttributeName : UIColor.redColor()])
    }
    if component == 1 {
        attributedString = NSAttributedString(string: b.text!, attributes: [NSForegroundColorAttributeName : UIColor.redColor()])
    }
    if component == 2 {
        attributedString = NSAttributedString(string: c.text!, attributes: [NSForegroundColorAttributeName : UIColor.redColor()])
    }
    return attributedString
}


func pickerView(pickerView: UIPickerView, didSelectRow row: Int, inComponent component: Int){

    switch component {
    case 0:
        aOutput.text = a[row]      -->  **Code …

我通过创建渐变图层创建了一个用于tableView的进度条.它完美地运作.

iphone 5:

为了在多个设备上使用该应用程序,我在Storyboard中创建了UIView,对其进行了标记并添加了约束.但是,当我在iPhone 6上使用该应用程序时,CALayer不会调整大小.

iPhone 6:

我发现这非常愚蠢,但没关系.我环顾四周,试图了解如何解决这个问题几个月,但我做得很短.是否任何人知道如何使CALayers与UIView的调整？任何帮助将非常感谢!谢谢.

        progressBar =  cell.contentView.viewWithTag(3) as UIView!
        progressBar.layer.cornerRadius = 4
        progressBar.layer.masksToBounds = true


        // create gradient layer
        let gradient : CAGradientLayer = CAGradientLayer()

        // create color array
        let arrayColors: [AnyObject] = [
            UIColor (red: 255/255, green: 138/255, blue: 1/255, alpha: 1).CGColor,
            UIColor (red: 110/255, green: 110/255, blue: 118/255, alpha: 1).CGColor]

        // set gradient frame bounds to match progressBar bounds
        gradient.frame = progressBar.bounds

        // set gradient's color array
        gradient.colors = arrayColors

        //Set progress(progressBar) …

我有一个有几千行的应用程序,在该代码中有很多println()命令.这会减慢应用程序的速度吗？它显然是在模拟器中执行的,但是当你从app store/TestFlight存档,提交和下载应用程序时会发生什么.这段代码是否仍然"活跃",那么"注释掉"的代码呢？

它是字面上永远不会读或我应该在提交测试航班/应用程序商店时删除注释掉的代码吗？

我已经在我的应用程序中实现了文本到语音,它可以正常使用我目前使用的代码.基本上算法创建文本,然后如果用户点击UIButton,则说出文本.

挑战:我想启用相同的UIButton来暂停合成器,如果按钮已经被轻敲(即当前正在说话文本),然后如果再次点击按钮则继续说话停止.

我知道AVFoundation Reference中有一些函数,但我无法正确实现它们.

有谁知道如何在Swift中这样做？

import UIKit
import AVFoundation


    @IBOutlet var generatedText: UILabel!

@IBAction func buttonSpeakClicked(sender: UIButton){
    var mySpeechSynthesizer:AVSpeechSynthesizer = AVSpeechSynthesizer()
    var mySpeechUtterance:AVSpeechUtterance = AVSpeechUtterance(string:generatedText.text)
    mySpeechUtterance.rate = 0.075

mySpeechSynthesizer .speakUtterance(mySpeechUtterance)
}

我的应用程序已注册为后台模式:IP语音,后台提取和远程通知.

我使用下面的代码成功接收了voip推送通知,但我无法启动应用程序来接听电话.收到voipPush后,如何在没有任何用户互动的情况下启动应用程序？

import UIKit
import PushKit

@UIApplicationMain
class AppDelegate: UIResponder, UIApplicationDelegate, PKPushRegistryDelegate {

var voipRegistry:PKPushRegistry!
var window: UIWindow?

func application(application: UIApplication, didFinishLaunchingWithOptions launchOptions: [NSObject: AnyObject]?) -> Bool {
    voipRegistry = PKPushRegistry(queue: dispatch_get_main_queue())
    voipRegistry.delegate = self
    voipRegistry.desiredPushTypes = Set([PKPushTypeVoIP])
    print("VOIP Push registered")

    return true
}

func pushRegistry(registry: PKPushRegistry!, didUpdatePushCredentials credentials: PKPushCredentials!, forType type: String!) {
    let voipToken: String! = credentials.token.description
    print("\n\n##### PushKit credential: \n\n", voipToken)
}

func pushRegistry(registry: PKPushRegistry!, didReceiveIncomingPushWithPayload payload: PKPushPayload!, forType type: String!) {
    let data = payload.dictionaryPayload
    print("\n\n##### …

如何使用switch case语句创建在switch-case语句之外有效的变量/常量.如果没有办法做到这一点,我还能做些什么来达到相同的效果,即创建受条件限制的变量,并使其在"全局"或更高范围内可访问？

var dogInfo = (3, "Fido")

switch dogInfo {

case(var age, "wooff"):
    println("My dog Fido is \(age) years old")

case (3, "Fido"):
    var matchtrue = 10           --> 10
    matchtrue                    -->10

default:
    "No match"
}

matchtrue                       --> Error: Use of unresolved identifier 'matchtrue'

这是我解决的问题:

var randomNumberOne = 0, randomNumberTwo = 0, randomNumberThree = 0

func chosen (#a: Int, #b: Int) -> (randomNumberOne: Int, randomNumberTwo: Int, randomNumberThree: Int){

if a > 0 {
    let count1 = UInt32(stringArray1.count)-1
    let randomNumberOne = Int(arc4random_uniform(count1))+1 …

我确定之前已经问过这个问题,但我找不到一个能够解决嵌套if-else和switch-case逻辑问题的答案.
我有UITableView两个部分,每个部分有两个自定义单元格.就是这样.4个细胞.但无论我做什么,我都会得到"在预期返回的功能中失去回归UITableViewCell"

问题如何更改此设置以便在底部获得一个满足快速逻辑的else语句？

任何帮助将非常感谢

override func tableView(tableView: UITableView, cellForRowAtIndexPath indexPath: NSIndexPath) -> UITableViewCell {

    if indexPath.section == 0{

        switch (indexPath.row) {
        case 0:
            let cell0: SettingsCell! = tableView.dequeueReusableCellWithIdentifier("cell0", forIndexPath: indexPath) as! SettingsCell
        cell0.backgroundColor = UIColor.redColor()
        break

        case 1:
            let cell1: SettingsCell! = tableView.dequeueReusableCellWithIdentifier("cell1", forIndexPath: indexPath) as! SettingsCell
        cell1.backgroundColor = UIColor.whiteColor()
         break

        default:
            break
        }
    }

    if indexPath.section == 1{

        switch (indexPath.row) {
        case 0:
            let cell10: SettingsCell! = tableView.dequeueReusableCellWithIdentifier("cell10", forIndexPath: indexPath) as! SettingsCell
        cell10.backgroundColor = …

我试图索引到一个函数数组但我得到错误:"表达式解析为未使用的l值".我试图谷歌这意味着什么,但信息是稀缺的,我发现似乎无关.有谁知道我在做错了什么？任何帮助将非常感激 !谢谢.

 override func tableView(tableView: UITableView, cellForRowAtIndexPath indexPath: NSIndexPath) -> UITableViewCell {

    var cell:UITableViewCell = myTableView.dequeueReusableCellWithIdentifier("cell") as UITableViewCell
    var chart =  cell.contentView.viewWithTag(42) as TKChart
    chart.delegate = self
    chart.removeAllData(); //needed because of cell recycling

    var userDef1 = 1
    var userDef2 = 1

    func lineChart() -> TKChart {...}
    func columnChart() -> TKChart {...}

    var chartsArray = [AnyObject]()


    if userDef1 == 1{
        chartsArray.append(lineChart())
    }


    if userDef2 == 1{
        chartsArray.append(columnChart())
    }


     if indexPath.row == 0{
        chartsArray[0]  **error: Expression resolves to an unused l-value**
    } …

我正在尝试创建一个字典,我可以将其制作成JSON格式的对象并发送到服务器.

例:

    var users = [
[
"First": "Albert", 
"Last": "Einstein", 
    "Address":[
        "Street": "112 Mercer Street",
        "City": "Princeton"]
],
[
"First": "Marie", 
"Last": "Curie", 
    "Address":[
        "Street": "108 boulevard Kellermann",
        "City": "Paris"]]
]

我用这个功能

func nsobjectToJSON(swiftObject: NSObject) -> NSString {
    var jsonCreationError: NSError?
    let jsonData: NSData = NSJSONSerialization.dataWithJSONObject(swiftObject, options: NSJSONWritingOptions.PrettyPrinted, error: &jsonCreationError)!
    var strJSON = NSString()

    if jsonCreationError != nil {
        println("Errors: \(jsonCreationError)")
    }
    else {
        // everything is fine and we have our json stored as an NSData object. We …