Java/Kotlin：标记字符串忽略嵌套引号的内容

Question

我想用空格分割一个字符，但将空格保留在引号内（以及引号本身）。问题是，引号可以嵌套，而且我需要对单引号和双引号都这样做。所以，从this "'"is a possible option"'" and ""so is this"" and '''this one too''' and even ""mismatched quotes"我想得到的行[this, "'"is a possible option"'", and, ""so is this"", and, '''this one too''', and, even, ""mismatched quotes"]。

这个问题已经被问过了，但不是我要问的确切问题。这里有几个解决方案：一个使用匹配器（在这种情况下"""x"""将被拆分为[""", x"""]，所以这不是我需要的）和 Apache Commons （它可以使用"""x"""但不能使用""x""，因为它需要前两个双引号并留下最后两个与x）。也有人建议手动编写一个函数，但这是最后的手段。

Answer 1

您可以使用以下正则表达式来实现：["']+[^"']+?["']+. 使用该模式，您可以检索要拆分的索引，如下所示：val indices = Regex(pattern).findAll(this).map{ listOf(it.range.start, it.range.endInclusive) }.flatten().toMutableList()剩下的就是用子字符串构建列表。这里完整的功能：fun String.splitByPattern(pattern: String): List<String> {&nbsp; &nbsp; val indices = Regex(pattern).findAll(this).map{ listOf(it.range.start, it.range.endInclusive) }.flatten().toMutableList()&nbsp; &nbsp; var lastIndex = 0&nbsp; &nbsp; return indices.mapIndexed { i, ele ->&nbsp; &nbsp; &nbsp; &nbsp; val end = if(i % 2 == 0) ele else ele + 1 // magic&nbsp; &nbsp; &nbsp; &nbsp; substring(lastIndex, end).apply {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; lastIndex = end&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; }}用法：val str = """this "'"is a possible option"'" and ""so is this"" and '''this one too''' and even ""mismatched quotes"""".trim()println(str.splitByPattern("""["']+[^"']+?["']+"""))输出：[this , "'" is a possible option"'", and , ""so is this"", and , '''this one too''', and even , ""mismatched quotes"]在Kotlin 的操场上试试吧！