string 删除变量开头和结尾的子串匹配模式
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Remove substring matching pattern both in the beginning and the end of the variable
提问by bobylapointe
As the title says, I'm looking for a way to remove a defined pattern both at the beginning of a variable and at the end. I know I have to use #and %but I don't know the correct syntax.
正如标题所说，我正在寻找一种方法来删除变量开头和结尾的已定义模式。我知道我必须使用#，%但我不知道正确的语法。
In this case, I want to remove http://at the beginning, and /score/at the end of the variable $linewhich is read from file.txt.
在这种情况下，我想http://在开头和/score/结尾处删除$line从file.txt.
回答by
Well, you can't nest ${var%}/${var#}operations, so you'll have to use temporary variable.
好吧，您不能嵌套${var%}/${var#}操作，因此您必须使用临时变量。
Like here:
像这儿：
var="http://whatever/score/"
temp_var="${var#http://}"
echo "${temp_var%/score/}"
Alternatively, you can use regular expressions with (for example) sed:
或者，您可以将正则表达式与（例如）sed 一起使用：
some_variable="$( echo "$var" | sed -e 's#^http://##; s#/score/$##' )"
回答by jaypal singh
$ var='https://www.google.com/keep/score'
$ var=${var#*//} #removes stuff upto // from begining
$ var=${var%/*} #removes stuff from / all the way to end
$ echo $var
www.google.com/keep
回答by Charles Duffy
There IS a way to do it one step using only built-in bash functionality (no running external programs such as sed) -- with BASH_REMATCH:
有一种方法可以仅使用内置的 bash 功能（不运行诸如 的外部程序sed）来一步完成它——使用BASH_REMATCH：
url=http://whatever/score/
re='https?://(.*)/score/'
[[ $url =~ $re ]] && printf '%s\n' "${BASH_REMATCH[1]}"
This matches against the regular expression on the right-hand side of the =~test, and puts the groups into the BASH_REMATCHarray.
这与=~测试右侧的正则表达式匹配，并将组放入BASH_REMATCH数组中。
That said, it's more conventional to use two PE expressions and a temporary variable:
也就是说，使用两个 PE 表达式和一个临时变量更为传统：
shopt -s extglob
url=http://whatever/score/
val=${url#http?(s)://}; val=${val%/score/}
printf '%s\n' "$val"
...in the above example, the extgloboption is used to allow the shell to recognized "extglobs" -- bash's extensions to glob syntax (making glob-style patterns similar in power to regular expressions), among which ?(foo)means that foois optional.
...在上面的示例中，该extglob选项用于允许 shell 识别“extglobs”——bash 对 glob 语法的扩展（使 glob 样式模式在功能上类似于正则表达式），其中?(foo)意味着这foo是可选的。
By the way, I'm using printfrather than echoin these examples because many of echo's behaviors are implementation-defined -- for instance, consider the case where the variable's contents are -eor -n.
顺便说一下，我在这些示例中使用printf而不是echo，因为 的许多echo行为是实现定义的——例如，考虑变量的内容是-e或 的情况-n。
回答by Gilles Quenot
You have to do it in 2 steps :
您必须分两步完成：
$ string="fooSTUFFfoo"
$ string="${string%foo}"
$ string="${string#foo}"
$ echo "$string"
STUFF
回答by michael501
how about
怎么样
export x='https://www.google.com/keep/score';
var=$(perl -e 'if ( $ENV{x} =~ /(https:\/\/)(.+)(\/score)/ ) { print qq();}')
