如何从Linux中的文本文件获取价值

Question

I have some file xxx.conf in text format. I have some text "disablelog = 1" in this file. When I use

grep -r "disablelog" oscam.conf

output is

disablelog = 1

But i need only value 1. Do you have some idea please?

Answer 1

one way is to use awk to print just the value

grep -r "disablelog" oscam.conf | awk '{print $3}'

you could also use sed to replace diablelog = with empty

grep -r 'disablelog' oscam.conf | sed -e 's/disablelog = //'

If you also want to get the lines with or without space before and after = use

grep -r 'disablelog' oscam.conf | sed 's/disablelog\s*=\s*//'

above command will also match

disablelog=1

Answer 2

尝试：

grep -r "disablelog" oscam.conf | awk -F= '{print $2}'

Answer 3

Assuming you need it as a var in a script:

#!/bin/bash

DISABLELOG=$(awk -F= '/^.*disablelog/{gsub(/ /,"",$2);print $2}' /path/to/oscam.conf)
echo $DISABLELOG

When calling this script, the output should be 1.

Edit: No matter wether there is whitespace or not between the equals sign and the value, the above will handle that. The regex should be anchored in either way to improve performance.

Answer 4

Just for fun a solution without awk

grep -r disablelog | cut -d= -f2 | xargs

xargs is used here to trim the whitespace