Get currently typed word in UITextView

Question

I\'m trying to make a \"tagging window\" much like the one used in Facebook where you type \"@\" and it make suggestions among your friends on which one to tag. I\'d like th

Answer 1

Swift 4:

This solution worked for me inside the method func textView(_ textView: UITextView, shouldChangeTextIn range: NSRange, replacementText text: String) -> Bool of UITextViewDelegate.

  func textView(_ textView: UITextView, shouldChangeTextIn range: NSRange, replacementText text: String) -> Bool {
        guard let textFieldText = textView.text,
              let rangeOfTextToReplace = Range(range, in: textFieldText) else {
                return false
        }

        let currentText = textFieldText.replacingCharacters(in: rangeOfTextToReplace, with: text)
    
        return count <= 150
    }

Answer 2

This works for me.

extension UITextView {

var currentWord : String? {

    let beginning = beginningOfDocument

    if let start = position(from: beginning, offset: selectedRange.location),
        let end = position(from: start, offset: selectedRange.length) {

        let textRange = tokenizer.rangeEnclosingPosition(end, with: .word, inDirection: 1)

        if let textRange = textRange {
            return text(in: textRange)
        }
    }
    return nil
   }
}

Answer 3

As usual I spent a fair amount of time trying to get this to work, and minutes after I posted the question I got it working. So here's the code I used, but I have deprecated call for the let myRange

func textView(textView: UITextView, shouldChangeTextInRange range: NSRange, replacementText text: String) -> Bool {

    let myRange = Range<String.Index>(start: textView.text.startIndex, end: textView.text.startIndex.advancedBy(range.location))
    var subString = textView.text.substringWithRange(myRange)
    subString += text

    let wordArray = subString.componentsSeparatedByString(" ")
    if let wordTyped = wordArray.last {
        currentWord = wordTyped
        print("word typed: " + wordTyped)
    }

    return true
}