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)

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" */

下面介绍如何从字符串的开头和结尾删除所有空白。

(用Swift 2.0测试的示例。)

let myString = "  \t\t  Let's trim all the whitespace  \n \t  \n  "
let trimmedString = myString.stringByTrimmingCharactersInSet(
    NSCharacterSet.whitespaceAndNewlineCharacterSet()
)
// Returns "Let's trim all the whitespace"

(用Swift 3+测试的示例。)

let myString = "  \t\t  Let's trim all the whitespace  \n \t  \n  "
let trimmedString = myString.trimmingCharacters(in: .whitespacesAndNewlines)
// Returns "Let's trim all the whitespace"

另一种类似的方式:

extension String {
    var trimmed:String {
        return self.trimmingCharacters(in: CharacterSet.whitespaces)
    }
}

使用:

let trimmedString = "myString ".trimmed

斯威夫特3

let result = " abc ".trimmingCharacters(in: .whitespacesAndNewlines)

Swift 5和4.2

let trimmedString = " abc ".trimmingCharacters(in: .whitespaces)
 //trimmedString == "abc"