Question

对于拆分字符串，Java中是否有默认/简单方法，但是要使用引号或其他符号？

例如，鉴于此文：

There's "a man" that live next door 'in my neighborhood', "and he gets me down..."

获得：

There's
a man
that
live
next
door
in my neighborhood
and he gets me down

Answer 1

这样的东西适用于您的输入：

    String text = "There's \"a man\" that live next door "
        + "'in my neighborhood', \"and he gets me down...\"";

    Scanner sc = new Scanner(text);
    Pattern pattern = Pattern.compile(
        "\"[^\"]*\"" +
        "|'[^']*'" +
        "|[A-Za-z']+"
    );
    String token;
    while ((token = sc.findInLine(pattern)) != null) {
        System.out.println("[" + token + "]");
    }

以上打印（as seen on ideone.com）：

[There's]
["a man"]
[that]
[live]
[next]
[door]
['in my neighborhood']
["and he gets me down..."]

它使用Scanner.findInLine，正则表达式模式是以下之一：

"[^"]*"      # double quoted token
'[^']*'      # single quoted token
[A-Za-z']+   # everything else

毫无疑问，这并不总是100％有效;引用可以嵌套等的情况将是棘手的。

参考

regular-expressions.info/Character class

Answer 2

根据您的逻辑令人怀疑，您可以区分撇号和单引号，即There's和in my neighborhood

如果你想要上面的东西，你必须开发某种配对逻辑。我正在考虑正则表达式。或者某种两部分解析。

拆分/标记/扫描一个知道引号的字符串

2 个答案:

参考