当我加载一个UIView时,我如何在iPhone SDK上设置UITextField的最大字符数?


当前回答

最好的方法是设置文本更改通知。在视图控制器方法的-awakeFromNib中,你需要:

[[NSNotificationCenter defaultCenter] addObserver:self selector:@selector(limitTextField:) name:@"UITextFieldTextDidChangeNotification" object:myTextField];

然后在同一个类中添加:

- (void)limitTextField:(NSNotification *)note {
    int limit = 20;
    if ([[myTextField stringValue] length] > limit) {
        [myTextField setStringValue:[[myTextField stringValue] substringToIndex:limit]];
    }
}

然后链接出口myTextField到你的UITextField,它将不让你添加任何字符后,你达到限制。一定要把这个添加到你的dealloc方法中:

[[NSNotificationCenter defaultCenter] removeObserver:self name:@"UITextFieldTextDidChangeNotification" object:myTextField];

其他回答

我想补充一下@sickp给出的答案。

在他的Swift代码中有一个问题,任何多字节文本(例如表情符号)都会出现。NSRange和String在Swift中是不兼容的,所以委托类将它们结合在一起是令人沮丧的。技巧是简单地转换String对象到NSString正确的解决方案,基于什么@sickp写,实际上是这样的:

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

    let currentText = (textField.text as NSString?) ?? NSString()
    let currentCharacterCount = currentText.length
    if range.length + range.location > currentCharacterCount {
        return false
    }
    let newLength = currentText.replacingCharacters(in: range, with: string).characters.count
    return newLength <= 25
}

适用于Swift 3.1或更高版本

首先添加协议UITextFieldDelegate

如:-

class PinCodeViewController: UIViewController, UITextFieldDelegate { 
.....
.....
.....

}

之后创建你的UITextField并设置委托

完成经验:-

import UIKit

class PinCodeViewController: UIViewController, UITextFieldDelegate {

let pinCodetextField: UITextField = {
    let tf = UITextField()
    tf.placeholder = "please enter your pincode"
    tf.font = UIFont.systemFont(ofSize: 15)
    tf.borderStyle = UITextBorderStyle.roundedRect
    tf.autocorrectionType = UITextAutocorrectionType.no
    tf.keyboardType = UIKeyboardType.numberPad
    tf.clearButtonMode = UITextFieldViewMode.whileEditing;
    tf.contentVerticalAlignment = UIControlContentVerticalAlignment.center
 return tf
  }()


 override func viewDidLoad() {
   super.viewDidLoad()
   view.addSubview(pinCodetextField)
    //----- setup your textfield anchor or position where you want to show it----- 


    // after that 
pinCodetextField.delegate = self // setting the delegate

 }
func textField(_ textField: UITextField, shouldChangeCharactersIn range: NSRange, replacementString string: String) -> Bool {
    return !(textField.text?.characters.count == 6 && string != "")
     } // this is return the maximum characters in textfield 
    }

这应该足以解决问题(用你想要的极限替换4)。只要确保在IB中添加委托。

- (BOOL)textField:(UITextField *)textField shouldChangeCharactersInRange:(NSRange)range replacementString:(NSString *)string
{
     NSString *newString = [textField.text stringByReplacingCharactersInRange:range withString:string];
     return (newString.length<=4);
}

有通用的解决方案设置最大长度在Swift。 通过IBInspectable,你可以在Xcode属性检查器中添加新的属性。

import UIKit
private var maxLengths = [UITextField: Int]()
extension UITextField {

    @IBInspectable var maxLength: Int {
        get {
            guard let length = maxLengths[self]
            else {
                return Int.max
            }
            return length
        }
        set {
            maxLengths[self] = newValue
            addTarget(
                self,
                action: Selector("limitLength:"),
                forControlEvents: UIControlEvents.EditingChanged
            )
        }
    }

    func limitLength(textField: UITextField) {
        guard let prospectiveText = textField.text
            where prospectiveText.characters.count > maxLength else {
                return
        }
        let selection = selectedTextRange
        text = prospectiveText.substringWithRange(
            Range<String.Index>(prospectiveText.startIndex ..< prospectiveText.startIndex.advancedBy(maxLength))
        )
        selectedTextRange = selection
    }

}

你不能直接这样做- UITextField没有maxLength属性,但你可以设置UITextField的委托,然后使用:

- (BOOL)textField:(UITextField *)textField shouldChangeCharactersInRange:(NSRange)range replacementString:(NSString *)string