bash 中带有变量、大括号和哈希字符的 ${0##...} 语法的含义是什么？

发布于 2024-08-17 22:50:43 字数 235 浏览 13 评论 0原文

我刚刚在bash中看到一些我不太理解的代码。作为新手 bash 脚本编写者，我不确定发生了什么。

echo ${0##/*}
echo ${0}

我确实没有看到这两个命令的输出有什么区别（打印脚本名称）。 # 只是一条注释吗？ /* 是怎么回事。如果它是注释，为什么它不会干扰右大括号？

谁能给我一些关于这个语法的见解？

原文

I just saw some code in bash that I didn't quite understand. Being the newbie bash scripter, I'm not sure what's going on.

echo ${0##/*}
echo ${0}

I don't really see a difference in output in these two commands (prints the script name). Is that # just a comment? And what's with the /*. If it is a comment, how come it doesn't interfere with the closing } brace?

Can anyone give me some insight into this syntax?

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ぽ尐不点ル 2024-08-24 22:50:43

请参阅子字符串删除部分wiki.bash-hackers.org/syntax/pe" rel="noreferrer">bash-hackers wiki 的参数扩展页面：

${PARAMETER#PATTERN} 和 ${PARAMETER##PATTERN}
这种形式是从字符串的开头删除试图匹配它的所描述的模式。运算符#将尝试删除与模式匹配的最短文本，而##尝试删除最长的文本匹配。
示例字符串（只是大佬的引用）：
MYSTRING="接受的内容要自由，发送的内容要保守”
<表类=“s-表”>
<标题>
语法
结果

<正文>
${MYSTRING#*in}
~~在接受的内容上保持宽松~~，在发送的内容上保持保守。
${MYSTRING##*in}
~~接受的内容要自由，发送的内容要保守。~~

See the section on Substring removal on the parameter expansion page of the bash-hackers' wiki:

${PARAMETER#PATTERN} and ${PARAMETER##PATTERN}
This form is to remove the described pattern trying to match it from the beginning of the string. The operator # will try to remove the shortest text matching the pattern, while ## tries to do it with the longest text matching.
Example string (just a quote from a big man):
MYSTRING="Be liberal in what you accept, and conservative in what you send"
Syntax Result
${MYSTRING#*in} ~~Be liberal in~~ what you accept, and conservative in what you send.
${MYSTRING##*in} ~~Be liberal in what you accept, and conservative in~~ what you send.

Syntax	Result
`${MYSTRING#*in}`	~~Be liberal in~~ what you accept, and conservative in what you send.
`${MYSTRING##*in}`	~~Be liberal in what you accept, and conservative in~~ what you send.

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风铃鹿 2024-08-24 22:50:43

Linux 技巧：Bash 参数和参数扩展

${PARAMETER##WORD}  Results in removal of the longest matching pattern from the beginning rather than the shortest.
for example
[ian@pinguino ~]$ x="a1 b1 c2 d2"
[ian@pinguino ~]$ echo ${x#*1}
b1 c2 d2
[ian@pinguino ~]$ echo ${x##*1}
c2 d2
[ian@pinguino ~]$ echo ${x%1*}
a1 b
[ian@pinguino ~]$ echo ${x%%1*}
a
[ian@pinguino ~]$ echo ${x/1/3}
a3 b1 c2 d2
[ian@pinguino ~]$ echo ${x//1/3}
a3 b3 c2 d2
[ian@pinguino ~]$ echo ${x//?1/z3}
z3 z3 c2 d2

Linux tip: Bash parameters and parameter expansions

${PARAMETER##WORD}  Results in removal of the longest matching pattern from the beginning rather than the shortest.
for example
[ian@pinguino ~]$ x="a1 b1 c2 d2"
[ian@pinguino ~]$ echo ${x#*1}
b1 c2 d2
[ian@pinguino ~]$ echo ${x##*1}
c2 d2
[ian@pinguino ~]$ echo ${x%1*}
a1 b
[ian@pinguino ~]$ echo ${x%%1*}
a
[ian@pinguino ~]$ echo ${x/1/3}
a3 b1 c2 d2
[ian@pinguino ~]$ echo ${x//1/3}
a3 b3 c2 d2
[ian@pinguino ~]$ echo ${x//?1/z3}
z3 z3 c2 d2

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初见终念 2024-08-24 22:50:43

我认为现有的答案（虽然肯定是准确的）忽略了OP问题的实际主旨。

OP 询问：

echo ${0##/*}

我的猜测是，他们在代码中真正看到的是：

echo ${0##*/}

后者本质上意味着“删除最后一个斜杠（如果有）之前的所有内容，包括最后一个斜杠（如果有）”。因此，这是一种获取脚本名称而无需路径的简洁方法，无论脚本是如何调用的。它相当于^*，

basename "$0"

但如果您将其用作变量而不是仅仅将其打印到控制台，则可以说更方便（并且更高效）。（OTOH 基本名称更便于移植，而参数扩展则是一种攻击。）

^{* 或多或少。在某些边缘情况（例如以空格开头的文件名）中，它们不会输出完全相同的内容。}

I think the existing answers (while certainly accurate) miss the practical thrust of the OP's question.

The OP asked about:

echo ${0##/*}

My guess is that what they really saw in the code was:

echo ${0##*/}

The latter essentially means "delete everything up to, and including, the last slash (if any)". So it's a concise way of getting the name of the script without the path, regardless of how the script was called. It's equivalent^* to

basename "$0"

but is arguably handier (and more efficient) if you're using it as a variable rather than just printing it to the console. (OTOH basename is more portable, whereas the parameter expansion is a bashism.)

^{* More or less. There are edge cases (such as file names that start with a space) where they don't output exactly the same thing.}

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您的好友蓝忘机已上羡 2024-08-24 22:50:43

您是指 ##/* 还是 ##*/？

##/*

${0##/*} 有点不寻常 - 它会从 $0< 开头去掉前缀 /... /代码>。

这是一个全有或全无的操作：如果 $0 以斜杠开头（例如 /home/bob/myscript.sh），那么它将删除所有内容并返回一个空细绳。否则（例如./myscript.sh）它将不删除任何内容并返回整个$0。

（双 ## 表示它应该删除最长的匹配项；单个 # 只会删除第一个字符，如果它是斜杠。）

我不知道如何它是有用的。也许它可以用来帮助检测脚本是否是从绝对路径调用的。

##*/

${0##*/} 更常见 - 它会从
从 $0 开始。

例如，如果$0是/home/bob/myscript.sh，它将返回myscript.sh。

## 再次指示它应该删除最长的匹配项，因此它将删除所有斜杠 (.../.../)。

（与#相反，它只会去掉第一个斜杠，例如/home/bob/myscript.sh -> home/bob/myscript。 sh, a/b/myscript.sh -> b/myscript.sh)