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Accepted
Jerry
Asked: 2024-11-22 02:56:45 +0800 CST

在 Mac 上，这个命令中的 {} 和 // 起什么作用？

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最近，我发现了一个很酷的技巧，可以帮助用另一个子字符串替换环境变量中的子字符串。但我真的不明白它是如何做到的。

TEST_URL="abc/123.com"
NEW_TEST_URL="${TEST_URL/123/123-ro-}"
echo $NEW_TEST_URL

现在 NEW_TEST_URL 是 abc/123-ro-.com

这//看起来与替代命令类似。

我在 Mac 终端上运行了这个

echo $0 返回 -zsh

macos

1 个回答

  1. Best Answer

    您可以在手册页中找到解释man zshexpn：

    ${name/pattern/repl}
${name//pattern/repl}
${name:/pattern/repl}

       Replace the longest possible match of pattern in the
       expansion of parameter name by string repl.  The first form
       replaces just the first occurrence, the second form all
       occurrences, and the third form replaces only if pattern
       matches the entire string.  Both pattern and repl are
       subject to double-quoted substitution, so that expressions
       like ${name/$opat/$npat} will work, but obey the usual rule
       that pattern characters in $opat are not treated specially
       unless either the option GLOB_SUBST is set, or $opat is
       instead substituted as ${~opat}.

       The pattern may begin with a `#', in which case the pattern
       must match at the start of the string, or `%', in which
       case it must match at the end of the string, or `#%' in
       which case the pattern must match the entire string.  The
       repl may be an empty string, in which case the final `/'
       may also be omitted.  To quote the final `/' in other cases
       it should be preceded by a single backslash; this is not
       necessary if the `/' occurs inside a substituted parameter.
       Note also that the `#', `%' and `#% are not active if they
       occur inside a substituted parameter, even at the start.

       If, after quoting rules apply, ${name} expands to an array,
       the replacements act on each element individually.  Note
       also the effect of the I and S parameter expansion flags
       below; however, the flags M, R, B, E and N are not useful.

       For example,

              foo="twinkle twinkle little star" sub="t*e" rep="spy"
              print ${foo//${~sub}/$rep}
              print ${(S)foo//${~sub}/$rep}

       Here, the `~' ensures that the text of $sub is treated as a
       pattern rather than a plain string.  In the first case, the
       longest match for t*e is substituted and the result is `spy
       star', while in the second case, the shortest matches are
       taken and the result is `spy spy lispy star
    • 1

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