How to use variables in linux regex pattern?

Question

I need to execute a sed command with pattern /^03.06.2014/ in a .sh script & command line. The date is a variable not a constant. How could i implement this? When I use a variable inside a regex pattern, the command breaks. Do I need to escape something here? Any help is appreciated. Thanks!

date=$(date +%m.%d.%Y)
sed -n '/^$date/,$p' filename

Welcome to SO. You've made a good first impression. Please read this next: http://stackoverflow.com/help/someone-answers — glenn jackman, Mar 06 '14 at 16:01

Amit · Accepted Answer · 2014-03-06T16:02:38.347

3

Use double quotes for variable expansion in sed

sed -n "/^$date/,\$p" filename

edited Mar 06 '14 at 16:02

answered Mar 06 '14 at 15:55

Amit

18,038
6
44
52

The `$p` is not a variable, it's EOF (`$`) and a command (`p`), so the `$` needs to be escaped. – Kevin Mar 06 '14 at 15:57
@Kevin Thanks, Updated. – Amit Mar 06 '14 at 16:03

score 2 · Answer 2 · answered Mar 06 '14 at 15:56

2

You need to use double quotes to allow for shell expansion. You'll need to escape the $ meaning EOF.

sed -n "/^$date/,\$p" filename

answered Mar 06 '14 at 15:56

Kevin

51,293
15
96
128

How to use variables in linux regex pattern?

2 Answers2

Linked