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 
    }

为了完成八月的回答，建议函数的一个可能实现(参见UITextField的委托)。

我没有测试domness代码，但我的代码不会卡住，如果用户达到了限制，它与一个新的字符串兼容，取代一个更小或等于一个。

-(BOOL)textField:(UITextField *)textField shouldChangeCharactersInRange:(NSRange)range replacementString:(NSString *)string {
    //limit the size :
    int limit = 20;
    return !([textField.text length]>limit && [string length] > range.length);
}

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);
}

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 
    }

通常您有多个长度不同的输入字段。

- (BOOL)textField:(UITextField *)textField shouldChangeCharactersInRange:(NSRange)range replacementString:(NSString *)string {
    int allowedLength;
    switch(textField.tag) {
        case 1: 
            allowedLength = MAXLENGTHNAME;      // triggered for input fields with tag = 1
            break;
        case 2:
            allowedLength = MAXLENGTHADDRESS;   // triggered for input fields with tag = 2
            break;
        default:
            allowedLength = MAXLENGTHDEFAULT;   // length default when no tag (=0) value =255
            break;
    }

    if (textField.text.length >= allowedLength && range.length == 0) {
        return NO; // Change not allowed
    } else {
        return YES; // Change allowed
    }
}

这个简单的方法怎么样?它对我来说很好。

extension UITextField {

    func  charactersLimit(to:Int) {

        if (self.text!.count > to) {
            self.deleteBackward()
        }
    }
}

然后:

someTextField.charactersLimit(to:16)