Bash：别名中嵌套反引号会导致问题
我正在尝试编写一个别名，该别名将跳转到包含指定文件的 cwd 的后代目录（或第一个 find 找到的此类文件名的出现）：
以下命令组合可实现所需的结果：
cd `dirname \`find -name 'MyFile.txt' | sed -n 1p\``
但是，我似乎无法以正确的方式转义它来创建工作别名：
alias jump="cd \`dirname \\\`find -name '$1' | sed -n 1p\\\`\`"
输出：
/*
dirname: missing operand
Try `dirname --help' for more information.
bash: cd: find: No such file or directory
我的逻辑是反引号需要在带有单个 \ 的双引号字符串中转义，我可以'要做\\ 被转换为字符串中的单个反斜杠，因此第二个嵌套反引号需要 1+2=3。
有什么建议吗？
I'm trying to write an alias which will jump to the descendant directory of cwd which contains a specified file (or the first find found occurrence of such a filename):
The following command combination achieves the desired result:
cd `dirname \`find -name 'MyFile.txt' | sed -n 1p\``
However, I can't seem to escape this in the correct way to create a working alias:
alias jump="cd \`dirname \\\`find -name '$1' | sed -n 1p\\\`\`"
Output:
/*
dirname: missing operand
Try `dirname --help' for more information.
bash: cd: find: No such file or directory
My logic is that backticks need escaping in a double quoted string with a single \ and I can't do \\ gets translated to a single backslash within a string, so the second nested backtick requires 1+2=3.
Any suggestions?
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评论（3）
别名不能接受像
$1这样的参数。使用函数代替。还可以使用
$(command)进行命令替换，而不是使用反引号，因为这样更容易嵌套。该函数将是：
An alias cannot take an argument like
$1. Use a function instead.Also use
$(command)for command substitution instead of backticks, as it is easier to nest.The function would be:
反引号是命令替换的旧形式，你可以'不要轻易地筑巢。但是，新的
$()表单 确实很容易嵌套：Backticks are the old form of command substitution, and you can't nest them easily. However, the new
$()form does nest easily:Backticks不提供嵌套。尝试使用命令替换，其语法为$(..)在您的情况下，它将是
Backticksdoesn't offer nesting. Try usingcommand substitutionwhich has the syntax$(..)In your case it will be