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 · 2014-03-06 16:02:38Z

5

Use double quotes for variable expansion in sed

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

edited Mar 6, 2014 at 16:02

answered Mar 6, 2014 at 15:55

Amit

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Kevin Over a year ago

The $p is not a variable, it's EOF ($) and a command (p), so the $ needs to be escaped.

Kevin · Accepted Answer · 2014-03-06 15:56:09Z

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 6, 2014 at 15:56

Kevin

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