如何限制文本字段只能输入十进制数字?
How to limit the text field to be able to enter only decimal digits?
我有用于 Celectial 导航计算的应用程序,我已将代码 textField.text 转换为 Double,但有时如果用户输入某些字段(如“1.0”和“1”),应用程序会崩溃,结果应用程序崩溃,因为无法扣除 Int 和 Double,以确保我想限制用户仅输入十进制数字“1.0”。对我来说最好的方法是编写一些代码,如果用户在按下完成按钮后自动输入例如“1”,添加“.0”以获得 1.0?
func textField(_ textField: UITextField, shouldChangeCharactersIn range: NSRange, replacementString string: String) -> Bool {
let allowedCharacters = "-1234567890."
let allowedCharacterSet = CharacterSet(charactersIn: allowedCharacters)
let typedCharactersSet = CharacterSet(charactersIn: string)
return allowedCharacterSet.isSuperset(of: typedCharactersSet)
}
func TextField(_ textField: UITextField, shouldChangeCharactersIn range: NSRange, replacementString string: String) -> Bool {
guard let text = latDegTextField.text else { return true }
let count = text.count + string.count - range.length
return count == 2
}
首先使用中的这个方法,即
func textField(_ textField: UITextField, shouldChangeCharactersIn range: NSRange, replacementString string: String) -> Bool {
if textField.text != "" || string != "" {
let res = (textField.text ?? "") + string
return Double(res) != nil
}
return true
}
然后在完成按钮操作中添加:
@IBAction func btnDoneTapped(_ sender: Any) {
print(tf.text)
guard let obj = Double(tf.text!) else { return }
print(obj)
}
并且当您输入 1
并按下完成按钮时 print(tf.text)
将打印 Optional("1")
而 print(obj)
将打印 1.0
使用此代码:-
func textField(_ textField: UITextField, shouldChangeCharactersIn range: NSRange, replacementString string: String) -> Bool {
//Will prevent user from entering space as first character
let enteredCharString = "\(textField.text ?? "")\(string )"
if enteredCharString.trimmingCharacters(in: .whitespaces).count == 0 {
return false
}
switch textField {
case txt_Ammount:
if txt_Ammount.text != "" || string != "" {
let res = (txt_Ammount.text ?? "") + string
return Double(res) != nil
}
default:
true
}
return true
}
我有用于 Celectial 导航计算的应用程序,我已将代码 textField.text 转换为 Double,但有时如果用户输入某些字段(如“1.0”和“1”),应用程序会崩溃,结果应用程序崩溃,因为无法扣除 Int 和 Double,以确保我想限制用户仅输入十进制数字“1.0”。对我来说最好的方法是编写一些代码,如果用户在按下完成按钮后自动输入例如“1”,添加“.0”以获得 1.0?
func textField(_ textField: UITextField, shouldChangeCharactersIn range: NSRange, replacementString string: String) -> Bool {
let allowedCharacters = "-1234567890."
let allowedCharacterSet = CharacterSet(charactersIn: allowedCharacters)
let typedCharactersSet = CharacterSet(charactersIn: string)
return allowedCharacterSet.isSuperset(of: typedCharactersSet)
}
func TextField(_ textField: UITextField, shouldChangeCharactersIn range: NSRange, replacementString string: String) -> Bool {
guard let text = latDegTextField.text else { return true }
let count = text.count + string.count - range.length
return count == 2
}
首先使用
func textField(_ textField: UITextField, shouldChangeCharactersIn range: NSRange, replacementString string: String) -> Bool {
if textField.text != "" || string != "" {
let res = (textField.text ?? "") + string
return Double(res) != nil
}
return true
}
然后在完成按钮操作中添加:
@IBAction func btnDoneTapped(_ sender: Any) {
print(tf.text)
guard let obj = Double(tf.text!) else { return }
print(obj)
}
并且当您输入 1
并按下完成按钮时 print(tf.text)
将打印 Optional("1")
而 print(obj)
将打印 1.0
使用此代码:-
func textField(_ textField: UITextField, shouldChangeCharactersIn range: NSRange, replacementString string: String) -> Bool {
//Will prevent user from entering space as first character
let enteredCharString = "\(textField.text ?? "")\(string )"
if enteredCharString.trimmingCharacters(in: .whitespaces).count == 0 {
return false
}
switch textField {
case txt_Ammount:
if txt_Ammount.text != "" || string != "" {
let res = (txt_Ammount.text ?? "") + string
return Double(res) != nil
}
default:
true
}
return true
}