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

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

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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

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

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How to use variables in linux regex pattern?

2 Answers2

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