如何将NSRange转换为Range<String。索引>在Swift?
我想使用以下UITextFieldDelegate方法:
func textField(textField: UITextField!,
shouldChangeCharactersInRange range: NSRange,
replacementString string: String!) -> Bool {
textField.text.stringByReplacingCharactersInRange(???, withString: string)
当前回答
这与Emilie的回答相似，但是，因为你特别问如何将NSRange转换为Range<String。索引>，你会这样做:
func textField(textField: UITextField, shouldChangeCharactersInRange range: NSRange, replacementString string: String) -> Bool {
let start = advance(textField.text.startIndex, range.location)
let end = advance(start, range.length)
let swiftRange = Range<String.Index>(start: start, end: end)
...
}
其他回答
Swift 5解决方案
简短的回答带有主要的扩展
extension NSRange {
public init(range: Range<String.Index>,
originalText: String) {
self.init(location: range.lowerBound.utf16Offset(in: originalText),
length: range.upperBound.utf16Offset(in: originalText) - range.lowerBound.utf16Offset(in: originalText))
}
}
详情请看这里
func textField(_ textField: UITextField, shouldChangeCharactersIn range: NSRange, replacementString string: String) -> Bool {
guard let current = textField.text, let r = Range(range, in: current) else {
return false
}
let text = current.replacingCharacters(in: r, with: string)
// ...
return true
}
在Swift 2.0中假设func textField(textField: UITextField, shouldChangeCharactersInRange range: NSRange, replacementString string: string) -> Bool {:
var oldString = textfield.text!
let newRange = oldString.startIndex.advancedBy(range.location)..<oldString.startIndex.advancedBy(range.location + range.length)
let newString = oldString.stringByReplacingCharactersInRange(newRange, withString: string)
func textField(textField: UITextField, shouldChangeCharactersInRange range: NSRange, replacementString string: String) -> Bool {
let strString = ((textField.text)! as NSString).stringByReplacingCharactersInRange(range, withString: string)
}
extension StringProtocol where Index == String.Index {
func nsRange(of string: String) -> NSRange? {
guard let range = self.range(of: string) else { return nil }
return NSRange(range, in: self)
}
}
