命令替换中的Bash引用 [英] Bash quoting within command substitution
问题描述
我很难理解这一点. 以下命令扩展文件:
I am having a hard time understanding this. The following command expand files:
% echo "(echo " * " )"
(echo foo.txt bar.txt )
但这不是
% echo "$(echo " * " )"
*
星号*不加引号.两种情况下都不会发生文件泛滥吗?命令替换如何影响这种情况?
Asterisk * is unquoted. Shouldn't the file globbing happen for both cases? How does command substitution affect this case?
推荐答案
单独查看命令会很有帮助.
It's helpful to look at the commands individually.
% echo "(echo " * " )"
...只有一个引用上下文.在这里，( echo用引号引起来，*不用引号引起来，然后)用引号引起来.
...has only a single quoting context. Here, ( echo is quoted, an * is given unquoted, and then a ) is given quoted.
% echo "$(echo " * " )"
在这里，您正在执行命令替换，而内部命令具有其自己的引用上下文.具体来说，该内部命令是:
Here, you're performing a command substitution, with the inner command having its own quoting context. Specifically, that inner command is:
# the inner command for the substitution is run first
echo " * "
...正如人们所期望的那样，它发出*作为其输出.值得注意的是，由于它具有自己的引用上下文，因此内容开始时未加引号，并且仅在看到"时才被引用.
...which emits * as its output, as one would expect. Notably, because it has its own quoting context, content starts unquoted, and only becomes quoted when the "s are seen.
然后，执行替换； 因为$()在引号内，所以替换过程不会运行字符串拆分和glob扩展(如果遵循这些步骤，则会将内部命令的输出分成多个单词，并且将每个单词扩展为一个单词.
Then, the substitution is performed; because the $() is inside quotes, that substitution process does not run string-splitting and glob expansion (which, if these steps were followed, would break the output of the inner command into multiple words, and expand each word as a glob).
echo "$(echo "*")"
...因此变成...
...thus becomes...
# effectively single-quoted because we already finished running expansions
# so expansions that would be allowed in double-quotes are already finished
echo ' * '
相比之下，如果您未引用:
By contrast, if you didn't quote:
echo $(echo "*")
...将成为...
...would become...
echo *
...其行为将符合您的预期.
...which would have behavior in line with what you anticipated.
或者，反之亦然
echo "$(echo *)"
...将变得类似于
echo 'foo.txt bar.txt'
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