如何在bash中按名称打印文件？

Question

how i can grep or cat file and print values by it names ? for example: file os-release - i want print only NAME and VERSION

NAME="Ubuntu"
VERSION="14.04.3 LTS, Trusty Tahr"
ID=ubuntu
ID_LIKE=debian  
PRETTY_NAME="Ubuntu 14.04.3 LTS"
VERSION_ID="14.04"
HOME_URL="http://www.ubuntu.com/"
SUPPORT_URL="http://help.ubuntu.com/"
BUG_REPORT_URL="http://bugs.launchpad.net/ubuntu/"

Answer 1

If you want to print the lines containing NAME and VERSION entirely, you can use grep :

$ grep "^NAME=\|^VERSION=" /etc/*-release
NAME="Ubuntu"
VERSION="14.04.3 LTS, Trusty Tahr"

Having ^ when grepping ensures that the words NAME and VERSION appear at the beginning of the line, so that you won't grep other lines such as PRETTY_NAME . The = ensures that you won't grep lines such as VERSION_ID .

If you just want to print the values of NAME and VERSION , however, you can do the following:

$ grep "^NAME=\|^VERSION=" /etc/*-release | grep -E -o ".[a-z]\w+"
Ubuntu
Trusty 
Tahr

Answer 2

Suppose your text file name is test.txt. Then run following command for name and version lines only

cat test.txt | grep "NAME\|VERSION"

For complete value as you informaed in comments

cat text.txt | grep "NAME=\"Ubuntu\"\|VERSION=\"14.04.3 LTS, Trusty Tahr\""