bash - 使用括号对表达式进行分组以防万一

Question

我想在 case 中用 () 对表达式进行分组，如下所示:

case a in
'*(a|b)*') printf '%s\n' 'something something';;
esac

但这并没有取得任何成功。我也试过:

*(a|b)**'('a|b')'*我都没有成功。

Answer 1

这将是特定于 Bash 的:

您需要启用 extglob 并使用此特定语法

#!/usr/bin/env bash

shopt -s extglob

case "$1" in
*@(a|b)*) printf '%s\n' 'something something';;
esac

参见man bash:

If the extglob shell option is enabled using the shopt builtin, several extended pattern matching operators are recognized. In the following description, a pattern-list is a list of one or more patterns separated by a |. Composite patterns may be formed using one or more of the following sub-patterns:

• ?(pattern-list)
  Matches zero or one occurrence of the given patterns
• *(pattern-list)
  Matches zero or more occurrences of the given patterns
• +(pattern-list)
  Matches one or more occurrences of the given patterns
• @(pattern-list)
  Matches one of the given patterns
• !(pattern-list)
  Matches anything except one of the given patterns

或者，您可以使用命令组末尾的特殊 ;& 标记让 case 继续执行下一个模式的命令组。

这是 NOT POSIX，但由 bash、ksh 或 ksh 处理strong>zsh 仍然:

#!/usr/bin/env ksh

case "$1" in
  *a*) ;& # Continue to next commands group
  *b*) printf '%s\n' 'something something';;
esac

现在，that other guy在评论中指出。

POSIX 方式:

#!/usr/bin/env sh

case "$1" in
  *a*|*b*) printf '%s\n' 'something something';;
esac