使用通配符扩展来回显zsh中的所有变量 [英] Use wildcard expansion to echo all variables in zsh

问题描述

在以相同模式开头的多个变量中，可以使用通配符来回显所有匹配的模式吗?

With multiple variables beginning with the same pattern can a wildcard be used to echo all matched patterns?

zzz1=test1; zzz_A=test2; zzza=test3

匹配以zzz开头的所有变量的最佳方法是什么. echo $zzz*或for i in $zzz*; do echo $i; done之类的内容将输出:

What is the best way to match all variables starting with zzz. Where something like echo $zzz* or for i in $zzz*; do echo $i; done would output:

test1
test2
test3

推荐答案

因此，要根据上面的评论直接回答...否，zsh无法使用通配符扩展和回显变量，但是typeset可以提供所需的结果.

So to directly answer based on comments above... No, zsh cannot expand and echo variables using a wildcard, but typeset can provide the desired result.

typeset -m 'zzz*'输出:

zzz_A=test2
zzz1=test1
zzza=test3

或更准确地得到我想要的输出，如此处所述:

or more accurately to get my desired output as explained here:

for i in `typeset +m 'zzz*'`; do echo "${i}:  ${(P)i}"; done

zzz1:  test1
zzz_A:  test2
zzza:  test3

或者只是...

for i in `typeset +m 'zzz*'`; do echo "${(P)i}"; done

test1
test2
test3

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