bash将输出存储为变量

grep -A 26 "some text" somefile.txt |
   awk '/other text/ { gsub(/M/, " "); print $4 }' |
   sort -n -r | uniq | head -1

will return the largest in a list pulled from a large text file, but how do I store the output as a variable?

将返回从大文本文件中提取的列表中的最大值,但如何将输出存储为变量?

3 个解决方案

#1

Use command substitution:

使用命令替换:

my_var=$(grep -A 26 "some text" somefile.txt |
   awk '/other text/ { gsub(/M/, " "); print $4 }' |
   sort -n -r | uniq | head -n1)

Also, for portability, I would suggest always using -n1 for the argument of head. I've come across a couple of incarnations of it where using -1 doesn't work.

另外,为了便携性,我建议总是使用-n1作为head的参数。我遇到过几个版本,其中使用-1不起作用。

#2

For unnested cases back quotes will work too:

对于未经证实的情况,返回引号也会起作用:

variable=`grep -A 26 "some text" somefile.txt |   
awk '/other text/ { gsub(/M/, " "); print $4 }' |  
sort -nru | head -1`

#3

I'd suggest

variable_name=$(grep -A 26 "some text" somefile.txt |
     awk '/other text/ { gsub(/M/, " "); print $4 }' |
     sort -nru | head -1)

#1