Question

我想从ls输出中提取一组2位数字，并能够从中获取最大和最小的数字，供以后使用。

我这样做： ls 14_data_*.log | egrep -o '[0-9]{2,2}' | head -1}但输出有时只是按升序排列，而head，tail期望返回错误的值。我尝试将它转换为整数但无望。

ls返回了什么以及如何处理？

Answer 1

我不知道你在做什么，但这是使用sort解决问题的方法：

降序：

ls 14_data_*.log | egrep -o '[0-9]{2}' | sort -nr

升序：

ls 14_data_*.log | egrep -o '[0-9]{2}' | sort -n

Answer 2

我真的不认为你需要在这里使用ls。如果您确定您的文件名不包含换行符，则可以使用：

printf '%s\n' 14_data_*.log | grep -Eo '[0-9]{2}'

然后将其传递到sort -n，-r以反转排序顺序。

Answer 3

I believe this is an XY problem.

How I understand what you want: you have some files in the current directory that match the glob 14_data_[0-9][0-9].log. And you want to extract the smallest and highest number from the glob [0-9][0-9].

E.g., given the listing

14_data_1.log
14_data_09.log
14_data_13.log
14_data_87.log
14_data_aa.log

you want the numbers 09 and 87, as min and max respectively.

Here's a simple pure Bash way to achieve this:

shopt -s nullglob

files=( 14_data_[0-9][0-9].log )

fnumbers=( "${files[@]#14_data_}" )
fnumbers=( "${fnumbers[@]%.log}" )

if (( ${#fnumbers[@]} == 0 )); then
    echo "No files found."
else
    printf 'min=%d\nmax=%d\n' "$((10#${numbers[0]}))" "$((10#${numbers[-1]}))"
fi

shopt -s nullglob: make globs expand to nothing if no matches
files=( 14_data_[0-9][0-9].log ): populate an array files with all matching files
fnumbers=( "${files[@]#14_data_}" ): populate an array fnumbers with the fields of files where the leading 14_data_ is removed
fnumbers=( "${fnumbers[@]%.log}" ): and remove the trailing .log from this array

At this point, you have an array fnumbers that contains (in the example case): ( 09 13 87 ).

We finally select the first and last elements of this array to get the lowest and highest number.

将ls输出转换为整数

3 个答案: