是否斯威夫特有一个修剪方法的字符串?例如:
let result = " abc ".trim()
// result == "abc"
是否斯威夫特有一个修剪方法的字符串?例如:
let result = " abc ".trim()
// result == "abc"
当前回答
在Swift 3.0中
extension String
{
func trim() -> String
{
return self.trimmingCharacters(in: CharacterSet.whitespaces)
}
}
你可以打电话
let result = " Hello World ".trim() /* result = "Hello World" */
其他回答
Swift 5和4.2
let trimmedString = " abc ".trimmingCharacters(in: .whitespaces)
//trimmedString == "abc"
我创建了这个函数,允许输入字符串并返回由任意字符修剪的字符串列表
func Trim(input:String, character:Character)-> [String]
{
var collection:[String] = [String]()
var index = 0
var copy = input
let iterable = input
var trim = input.startIndex.advancedBy(index)
for i in iterable.characters
{
if (i == character)
{
trim = input.startIndex.advancedBy(index)
// apennding to the list
collection.append(copy.substringToIndex(trim))
//cut the input
index += 1
trim = input.startIndex.advancedBy(index)
copy = copy.substringFromIndex(trim)
index = 0
}
else
{
index += 1
}
}
collection.append(copy)
return collection
}
As在swift中没有找到这样做的方法(在swift 2.0中编译和工作完美)
**Swift 5**
extension String {
func trimAllSpace() -> String {
return components(separatedBy: .whitespacesAndNewlines).joined()
}
func trimSpace() -> String {
return self.trimmingCharacters(in: .whitespacesAndNewlines)
}
}
**Use:**
let result = " abc ".trimAllSpace()
// result == "abc"
let ex = " abc cd ".trimSpace()
// ex == "abc cd"
您还可以发送您想要修剪的字符
extension String {
func trim() -> String {
return self.trimmingCharacters(in: .whitespacesAndNewlines)
}
func trim(characterSet:CharacterSet) -> String {
return self.trimmingCharacters(in: characterSet)
}
}
validationMessage = validationMessage.trim(characterSet: CharacterSet(charactersIn: ","))
不要忘记导入Foundation或UIKit。
import Foundation
let trimmedString = " aaa "".trimmingCharacters(in: .whitespaces)
print(trimmedString)
结果:
"aaa"
否则你会得到:
error: value of type 'String' has no member 'trimmingCharacters'
return self.trimmingCharacters(in: .whitespaces)