我编写了以下代码来验证文本字段中的文本输入。
else if (textField == txtField_Password)
{
let charSet = CharacterSet(charactersIn: "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789@$&*!")
let charLength = (txtField_Password.text!.count) + (string.count) - range.length
for i in 0..<string.count
{
let c = (string as NSString).character(at: i)
if (!((charSet as NSCharacterSet).characterIsMember(c)))
{
return false
}
}
return (charLength > 20) ? false : true
}
任何人都可以帮助我将character(at :)和characterIsMember()部分转换为上面代码中的swift等价物。
答案 0 :(得分:4)
您只需检查倒置字符集的范围即可简化逻辑。如果字符串仅包含允许的字符,则该函数返回nil
。
else if textField == txtField_Password {
let charLength = txtField_Password.text!.utf8.count + string.utf8.count - range.length
let charSet = CharacterSet(charactersIn: "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789@$&*!")
return string.rangeOfCharacter(from: charSet.inverted) == nil && charLength < 21
}
答案 1 :(得分:1)
请注意,有一种使用正则表达式实现所需内容的简单方法:
let currentText = (textField.text ?? "") as NSString
let newText = currentText.replacingCharacters(in: range, with: string)
let pattern = "^[a-zA-Z0-9@$&*!]{0,20}$"
return newText.range(of: pattern, options: .regularExpression) != nil
答案 2 :(得分:0)
您可以按照以下方式使用某些东西。我赞赏这有点粗糙且可以使用,但应该可以:
charSet = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789@$&*!"
if txtField_Password.text!.count <= 20 {
for i in 0..<str.count
{
let c = Array(str)[i]
let cString = String(c)
if charSet.contains(cString) {
return false
}
}
} else {
return false
}
答案 3 :(得分:0)
使用rangeOfCharacter
:
func textField(_ textField: UITextField, shouldChangeCharactersIn range: NSRange, replacementString string: String) -> Bool {
let specialCharacters = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789@$&*!"
let characterSet = CharacterSet(charactersIn: specialCharacters)
guard let lengh = textfield.text else {return}
if lengh.count >= 20 {
// text exceeded 20 characters. Do something
}
if (string.rangeOfCharacter(from: characterSet) != nil) {
print("matched")
return true
} else {
print("not matched")
}
return true
}