How to use variables in linux regex pattern?

问题

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

回答1:

Use double quotes for variable expansion in sed

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

回答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

来源：https://stackoverflow.com/questions/22229449/how-to-use-variables-in-linux-regex-pattern