Question

I’m trying to delete a part of a file using sed in Linux (Ubuntu). Specifically, I want to delete the first lines of a log file until the first occurrence of the current system date (using the pattern ’10 Jan 13′).

So, I store the date in a variable

root@server:/# VAR_DATE=`date -R | cut -c6-11`

And after that, I use sed

root@server:/# cat log_file.txt | sed -n -e '/$VAR_DATE/,$p'

But it doesn’t work. I’ve tried a lot of combinations with the same result:

root@server:/# cat log_file.txt | sed -n -e '/"$VAR_DATE"/,$p'
root@server:/# cat log_file.txt | sed -n -e '/"${VAR_DATE}"/,$p'
root@server:/# cat log_file.txt | sed -n -e "/$VAR_DATE/,$p"

What I’m doing wrong?

Answer 1

Use double quotes so the variable $vardate gets expanded by the shell and escape the last $ so it’s not expanded by the shell sed -n "/$vardate/,\$p" file:

$ cat file
6 Jan 13
7 Jan 13
8 Jan 13
9 Jan 13
10 Jan 13
11 Jan 13
12 Jan 13
13 Jan 13

$ vardate="10 Jan 13"

$ sed -n "/$vardate/,\$p" file
10 Jan 13
11 Jan 13
12 Jan 13
13 Jan 13