bash 中带有变量、大括号和哈希字符的 ${0##...} 语法是什么意思? [英] What is the meaning of the ${0##...} syntax with variable, braces and hash character in bash?

问题描述

我刚刚在 bash 中看到了一些我不太明白的代码.作为新手 bash 脚本编写者，我不确定发生了什么.

echo ${0##/*}回声 ${0}

我真的看不出这两个命令的输出有什么不同(打印脚本名称).那 # 只是一个评论吗?和 /* 有什么关系.如果是注释，它怎么不干扰结束 } 大括号?

谁能让我深入了解一下这种语法?

解决方案

见高级 Bash 脚本指南中的子字符串删除‡:

<块引用>

${string#substring}

从$string前面删除substring的最短匹配.

${string##substring}

从$string前面删除substring的最长匹配.

子串可能包含通配符*，匹配所有内容.表达式 ${0##/*} 打印 $0 的值，除非它以正斜杠开头，在这种情况下它什么都不打印.

_{‡ 截至 2019 年 3 月 7 日，该指南错误地声称匹配是 $substring，就好像 substring 是多变的.不是:substring 只是一个模式.}

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?

解决方案

See the section on Substring removal in the Advanced Bash-Scripting Guide‡:

${string#substring}

Deletes shortest match of substring from front of $string.

${string##substring}

Deletes longest match of substring from front of $string.

The substring may include a wildcard *, matching everything. The expression ${0##/*} prints the value of $0 unless it starts with a forward slash, in which case it prints nothing.

_{‡ The guide, as of 3/7/2019, mistakenly claims that the match is of $substring, as if substring was the name of a variable. It's not: substring is just a pattern.}

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