<I>迅速</i>将字符串转换为范围<I>错误</i>

Question

 func setupAvatarColor(name: String, incoming: Bool) {
        let diameter = incoming ? UInt(collectionView.collectionViewLayout.incomingAvatarViewSize.width) : UInt(collectionView.collectionViewLayout.outgoingAvatarViewSize.width)

        let rgbValue = name.hash
        let r = CGFloat(Float((rgbValue & 0xFF0000) >> 16)/255.0)
        let g = CGFloat(Float((rgbValue & 0xFF00) >> 8)/255.0)
        let b = CGFloat(Float(rgbValue & 0xFF)/255.0)
        let color = UIColor(red: r, green: g, blue: b, alpha: 0.5)

        **let nameLength = count(name);**
        let initials : String? = name.substringToIndex(advance(sender.startIndex, min(3, nameLength)))
        let userImage = JSQMessagesAvatarFactory.avatarWithUserInitials(initials, backgroundColor: color, textColor: UIColor.blackColor(), font: UIFont.systemFontOfSize(CGFloat(13)), diameter: diameter)

        avatars[name] = userImage
    }


I receiving the error '`String is not convertible to Range<I>'in the highlighted code (10th line)`. AnyOne?

Answer 1

In Swift 2 Apple removed alot of global functions.

In your case to get the length of a String do:

str.characters.count

Answer 2

You can't use count anymore in Swift 2.0 for getting the string length.

So instead of:

let nameLength = count(name);

Use:

let nameLength = name.endIndex;

or

let nameLength = name.characters.count;

Answer 3

Swift 2.0 lets you call advance on endIndex and startIndex which is what you need here I guess. Ps, avataWithUserInitials takes NSString as first argument and substringToIndex returns String so I don't see any reason to define it as String optional.

let initials = name.substringToIndex((name.characters.count < 3) ? name.endIndex:name.startIndex.advancedBy(3))