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我正在寻找一个现有的 swift2 函数来在空格上拆分字符串输入，同时在带引号的字符串中保留空格。

我已阅读堆栈溢出问题 25678373。我的问题似乎没有重复。

我在 cocoapods 中搜索了类似的功能。我没有找到它。

如果 swift2 中不存在这个 shlex.split 函数，那么完成类似事情的有效替代方法是什么？在保留内部引号字符串中的空格的同时拆分字符串的另一种方法是什么？

这是我在python中的意思的一个例子：

$    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
>>>
4

2 回答 2

2

正如@EricD 在他给你的评论中所写，不存在这样的原生 Swift 函数。但是，您可以很容易地编写自己的这种拆分函数，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 */

请注意，以上假设输入字符串具有匹配的引号分隔符集'。

于 2016-04-04T10:35:40.933 回答
0

“纯”快速（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{" \t\n\r".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
}

用法

let str = "a 2  'b   c'   d  ''"
dump(split(str))
/*
 ▿ 4 elements
   - [0]: a
   - [1]: 2
   - [2]: b   c
   - [3]: d
 */
于 2016-04-04T15:11:34.733 回答