从前两个单词的字符串中获取第一个字母

时间:2017-09-27 09:10:23

标签: ios swift string xcode9

我的字符串如下:

Apple recipe  recapes

Mango Tengaer

Lemone T U

Grapes limoenis  Steyic genteur

所以我尝试的是:

if let bakery = filtered?[indexPath.row]{
   let stringInput = bakery.fruitsname
                let stringInputArr = stringInput.components(separatedBy: " ")
                var stringNeed = ""

                for string in stringInputArr {
                    stringNeed = stringNeed + String(string.characters.first!)
                }

    print(stringNeed)  // have to print first word first letter, second word second letter

}

但是当我为我的第3个Lemone T U字做的时候,我就崩溃了。在这一行:

 stringNeed = stringNeed + String(string.characters.first!)

任何帮助!!

由于

Output i expect as per my above words


AR
MT
LT
GL

8 个答案:

答案 0 :(得分:2)

只需要检查数组计数

var array =  [
"Apple recipe  recapes",
"Mango Tengaer",
"Lemone  T U",
"Grapes limoenis  Steyic genteur"]

for str in array {
    let wordArray = str.split(separator: " ")
    if wordArray.count >= 2 {
        let firstTwoChar = String(wordArray[0].first!)+String(wordArray[1].first!)
        print(firstTwoChar)
    }
}

输出:

Ar
MT
LT
Gl

答案 1 :(得分:2)

试试这个。它将字符串分隔为字符串数组并删除nil。因此,如果字符串有双倍空格,则过滤掉它。确保字符串至少有2个单词。

if let bakery = filtered?[indexPath.row]{
   let stringInput = bakery.fruitsname
   if stringInput.components(separatedBy: " ").count >= 2 {
         let stringNeed = (stringInput.components(separatedBy: " ").map({ $0.characters.first }).flatMap({$0}).reduce("", { String($0) + String($1) }) as NSString).substring(to: 2)
    print(stringNeed)  
  }
}

答案 2 :(得分:0)

检查第一个字母的可选项可能还有额外的空格

if let bakery = filtered?[indexPath.row]{
 let stringInput = bakery.fruitsname
        let stringInputArr = stringInput.components(separatedBy: " ")
        var stringNeed = ""

                for string in stringInputArr {

 //Check the optional
                if let first = string.characters.first{
                    stringNeed = stringNeed + String(first)

                }
            }

}

答案 3 :(得分:0)

试试这个:(我已添加前缀2,因此如果只有单个元素,它将只计算前2个元素,然后不要担心它不会崩溃)

if let bakery = filtered?[indexPath.row]{
     let stringInput = bakery.fruitsname
            let stringInputArr = stringInput.components(separatedBy: " ")
            var stringNeed = ""

            for string in stringInputArr.prefix(2) //returns only first two elements {

             if let strFirst = string.characters.first{
                   stringNeed += String(strFirst)
               }
        }

}

答案 4 :(得分:0)

这是我的解决方案,事实上您已经启动了一个String对象数组。

let myStrings:[String] = ["Apple recipe  recapes","Mango Tengaer", " Lemone T U", "Grapes limoenis", "Steyic genteur", "He"]

var retStr = "";
for (_, str) in myStrings.enumerated()
{
    let stringInOneLine = str.components(separatedBy: CharacterSet.whitespaces).filter({$0.count > 0}).map({String($0.first!).uppercased()}).prefix(2).joined()

    //Separate all components by white spaces
    let c1 = str.components(separatedBy: CharacterSet.whitespaces)
    print("c1: \(c1)")

    //Remove components that are empty: It happens in case there is double spaces for instance
    let c2 = c1.filter({$0.count > 0})
    print("c2: \(c2)")

    //Get only the first letter and in upper case
    let c3 = c2.map({String($0.first!).uppercased()})
    print("c3: \(c3)")

    //Keep only the first two elements (if there is more or less than 2, it's takend care of
    let c4 = c3.prefix(2)
    print("c4: \(c4)")

    //Reform the String with the first letters
    let string = c4.joined()
    print("string: \(string)")



    retStr.append(stringInOneLine)
    retStr.append("\n")
    print("stringInOneLine:\n\(stringInOneLine)")
}

print("retStr:\n\(retStr)")

如果需要可以轻松更改构造retStr的方式(如果你想要一个String数组)。 我通过解释为什么要进行每次调用来分解stringInOnLine构造。 这取决于您的Swift专业水平和全局编程/算法技能,以决定您是否更喜欢所有链接在一行中或在各行中执行。 它有助于说明如何分解链式调用(调试,理解或创建它们)。

答案 5 :(得分:0)

乍一看似乎有些奇怪,但实际上它可以帮到你(我不会说这是最有效的解决方案):

var input: [String] =  ["apple recipe recapes", "Mango Tengaer", "Lemone  T U", "Grapes   limoenis   Steyic genteur", "One", "A B"]
var result = input.filter { $0.contains(" ") && $0.characters.count > 0 }.map { String($0.first!).uppercased() + $0.substring(with: $0.range(of: "( \\w{1})", options: .regularExpression, range: $0.startIndex..<$0.endIndex)!).uppercased().trimmingCharacters(in: CharacterSet(charactersIn: " ")) }

并且实际的result是:

["AR", "MT", "LT", "GL", "AB"]

注意: 它会忽略单个单词或空字符串,但您也可以将它扩展为那些,我认为,如果是要求,它不处理只有空格的输入字符串,正如您在测试中看到的那样。

答案 6 :(得分:0)

如果句子中只有两个单词,这应该可以工作。

String(fullName.components(separatedBy: " ").compactMap { $0.first })

答案 7 :(得分:0)

这应该适用于 Swift 4

String(actualString.components(separatedBy: " ").compactMap { $0.first }).uppercased()
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