给定someletters_12345_moreleters形式的文件名。ext，我想提取5位数字，并将它们放入一个变量。
为了强调这一点，我有一个x个字符的文件名，然后是一个5位数字序列，两边都有一个下划线，然后是另一组x个字符。我想把这个5位数代入一个变量。
我对实现这一目标的多种不同方式非常感兴趣。
当前回答
有点晚了，但我刚刚遇到了这个问题，并发现了以下内容:
host:/tmp$ asd=someletters_12345_moreleters.ext
host:/tmp$ echo `expr $asd : '.*_\(.*\)_'`
12345
host:/tmp$
我用它在一个没有%N日期的嵌入式系统上获得毫秒分辨率:
set `grep "now at" /proc/timer_list`
nano=$3
fraction=`expr $nano : '.*\(...\)......'`
$debug nano is $nano, fraction is $fraction
其他回答
如果有人想要更严格的信息，你也可以像这样在man bash中搜索
$ man bash [press return key]
/substring [press return key]
[press "n" key]
[press "n" key]
[press "n" key]
[press "n" key]
结果:
${parameter:offset}
${parameter:offset:length}
Substring Expansion. Expands to up to length characters of
parameter starting at the character specified by offset. If
length is omitted, expands to the substring of parameter start‐
ing at the character specified by offset. length and offset are
arithmetic expressions (see ARITHMETIC EVALUATION below). If
offset evaluates to a number less than zero, the value is used
as an offset from the end of the value of parameter. Arithmetic
expressions starting with a - must be separated by whitespace
from the preceding : to be distinguished from the Use Default
Values expansion. If length evaluates to a number less than
zero, and parameter is not @ and not an indexed or associative
array, it is interpreted as an offset from the end of the value
of parameter rather than a number of characters, and the expan‐
sion is the characters between the two offsets. If parameter is
@, the result is length positional parameters beginning at off‐
set. If parameter is an indexed array name subscripted by @ or
*, the result is the length members of the array beginning with
${parameter[offset]}. A negative offset is taken relative to
one greater than the maximum index of the specified array. Sub‐
string expansion applied to an associative array produces unde‐
fined results. Note that a negative offset must be separated
from the colon by at least one space to avoid being confused
with the :- expansion. Substring indexing is zero-based unless
the positional parameters are used, in which case the indexing
starts at 1 by default. If offset is 0, and the positional
parameters are used, $0 is prefixed to the list.
基于jor的回答(这对我来说并不适用):
substring=$(expr "$filename" : '.*_\([^_]*\)_.*')
使用sed replace的简单方法:
result=$(echo "someletters_12345_moreleters.ext" | sed 's/.*_\(.*\)_.*/\1/g')
echo $result
有点晚了，但我刚刚遇到了这个问题，并发现了以下内容:
host:/tmp$ asd=someletters_12345_moreleters.ext
host:/tmp$ echo `expr $asd : '.*_\(.*\)_'`
12345
host:/tmp$
我用它在一个没有%N日期的嵌入式系统上获得毫秒分辨率:
set `grep "now at" /proc/timer_list`
nano=$3
fraction=`expr $nano : '.*\(...\)......'`
$debug nano is $nano, fraction is $fraction
很多过时的解决方案都需要管道和子外壳。 自bash版本3(2004年发布)以来，它有一个内置的正则表达式比较操作符=~。
input="someletters_12345_moreleters.ext"
# match: underscore followed by 1 or more digits followed by underscore
[[ $input =~ _([0-9]+)_ ]]
echo ${BASH_REMATCH[1]}
输出:
12345
注意，如果您不是很精通编写RegExp，我建议您阅读精通正则表达式。
如果您只是需要弄清楚如何让RegExp工作，并且它不符合您的想法，请尝试RegEx101.com的在线GUI，并将“Flavor”设置为“PCRE”，以便获得bash使用的[[:digit:]]等POSIX风格的字符类。
