基于jor的回答(这对我来说并不适用):

substring=$(expr "$filename" : '.*_\([^_]*\)_.*')

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

以下是我的做法:

FN=someletters_12345_moreleters.ext
[[ ${FN} =~ _([[:digit:]]{5})_ ]] && NUM=${BASH_REMATCH[1]}

解释:

Bash-specific:

[[]]为条件表达式 =~表示条件为正则表达式 如果前一个命令成功，&&将链接这些命令

正则表达式(RE): _([[:digit:]]{5})_

_是字面量，用于为被匹配的字符串划分/锚定匹配边界 ()创建捕获组 [[:digit:]]是一个字符类，我认为它不言自明 {5}表示前面的字符中的恰好五个，类(如本例中所示)或组必须匹配

In english, you can think of it behaving like this: the FN string is iterated character by character until we see an _ at which point the capture group is opened and we attempt to match five digits. If that matching is successful to this point, the capture group saves the five digits traversed. If the next character is an _, the condition is successful, the capture group is made available in BASH_REMATCH, and the next NUM= statement can execute. If any part of the matching fails, saved details are disposed of and character by character processing continues after the _. e.g. if FN where _1 _12 _123 _1234 _12345_, there would be four false starts before it found a match.

Inklusive end，类似于JS和Java实现。删除+1如果你不想这样做。

function substring() {
    local str="$1" start="${2}" end="${3}"
    
    if [[ "$start" == "" ]]; then start="0"; fi
    if [[ "$end"   == "" ]]; then end="${#str}"; fi
    
    local length="((${end}-${start}+1))"
    
    echo "${str:${start}:${length}}"
} 

例子:

    substring 01234 0
    01234
    substring 012345 0
    012345
    substring 012345 0 0
    0
    substring 012345 1 1
    1
    substring 012345 1 2
    12
    substring 012345 0 1
    01
    substring 012345 0 2
    012
    substring 012345 0 3
    0123
    substring 012345 0 4
    01234
    substring 012345 0 5
    012345

更多示例调用:

    substring 012345 0
    012345
    substring 012345 1
    12345
    substring 012345 2
    2345
    substring 012345 3
    345
    substring 012345 4
    45
    substring 012345 5
    5
    substring 012345 6
    
    substring 012345 3 5
    345
    substring 012345 3 4
    34
    substring 012345 2 4
    234
    substring 012345 1 3
    123

使用sed replace的简单方法:

result=$(echo "someletters_12345_moreleters.ext" | sed 's/.*_\(.*\)_.*/\1/g')
echo $result

通用解决方案，其中数字可以在文件名中的任何位置，使用这样的序列中的第一个:

number=$(echo $filename | egrep -o '[[:digit:]]{5}' | head -n1)

另一个精确提取变量一部分的解决方案:

number=${filename:offset:length}

如果你的文件名总是使用stuff_digits_…你可以使用awk:

number=$(echo $filename | awk -F _ '{ print $2 }')

还有一种方法可以删除除数字以外的所有内容，使用

number=$(echo $filename | tr -cd '[[:digit:]]')