Swift2 相当于 Python 的 shlex.split，以保留带引号的字符串中的空格

Question

我正在寻找一个现有的swift2函数来分割空白上的字符串输入，同时在带引号的字符串中保留空白。

我已经阅读了堆栈溢出问题25678373。我的问题似乎没有重复。

我在cocoapods中搜索了类似的功能。我没有找到。

如果这个shlex.split函数在swift2中不存在，那么有什么有效的替代方法可以实现类似的功能呢？在内部带引号的字符串中保留空白的同时，拆分字符串的另一种方法是什么？

以下是我在python中的意思：

$    python
Python 2.7.6 (default, Jun 22 2015, 18:00:18) 
[GCC 4.8.2] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import shlex
>>> input=""" alpha 2 'chicken with teeth' 4 'cat with wings' 6 turkey"""
>>> results = shlex.split(input)
>>> type(results)
<type 'list'>
>>> results[0]
'alpha'
>>> results[2]
'chicken with teeth'
>>> for term in results:
...     print(term)
... 
alpha
2
chicken with teeth
4
cat with wings
6
turkey
>>> 

Answer 1

正如@EricD在给您的评论中所写，不存在这样的原生Swift函数。然而，你可以很容易地编写自己的这种分割函数，例如

extension String {
    func shlexSplit() -> [String] {
        /* separate words by spaces */
        var bar = self.componentsSeparatedByString(" ")
        /* identify array idx ranges of quoted (') sets of words */
        var accumulating = false
        var from = 0
        var joinSegments : [(Int, Int)] = []
        for (i,str) in bar.enumerate() {
            if str.containsString("'") {
                if accumulating { joinSegments.append((from, i)) }
                else { from = i }
                accumulating = !accumulating
            }
        }
        /* join matching word ranges with " " */
        for (from, through) in joinSegments.reverse() {
            bar.replaceRange(from...through, 
                with: [bar[from...through].joinWithSeparator(" ")])
        }
        return bar
    }
}

使用示例

/* exampe usage */
let foo = "alpha 2 'chicken with teeth' 4 'cat with wings' 6 turkey"
let bar = foo.shlexSplit()
bar.forEach{ print($0) }
/* alpha
 2
 'chicken with teeth'
 4
 'cat with wings'
 6
 turkey */

注意，上面假设输入字符串具有匹配的引号分隔符集合'。

Answer 2

"ure"swift（无基础）示例

extension String {
    // split by first incidence of character 
    func split(c: Character)->(String,String) {
        var head: String = "", tail: String = ""
        if let i = characters.indexOf(c) {
            let j = startIndex.distanceTo(i)
            head = String(characters.prefix(j))
            tail = String(characters.dropFirst(j + 1))
        } else {
            head = self
        }
        return (head, tail)
    }
}
// what you are looking for
func split(str: String)->[String] {
    // internal state
    var state:((String,String), [String], Bool) = (str.split("'"), [], false)
    repeat {
        if !state.2 {
            // you can define more whitespace characters
            state.1
                .appendContentsOf(state.0.0.characters.split{" tnr".characters.contains($0)}
                    .map(String.init))
            state.2 = true
        } else {
            state.1.append(state.0.0)
            state.2 = false
        }
        state.0 = state.0.1.split("'")
    } while !state.0.0.isEmpty
    return state.1
}

使用

let str = "a 2  'b   c'   d  ''"
dump(split(str))
/*
 ▿ 4 elements
   - [0]: a
   - [1]: 2
   - [2]: b   c
   - [3]: d
 */