我想在shell脚本的sed -f命令中使用变量

Question

I am using this command to copy certain line from one file to another.Its working fine.No issue with it.

sed -f <(sed -e '1,10d; 12,$d; x; s/.*/10a\\/;p; x' ../log/file2.txt ) ../log/file4.txt > ../log/file5.txt

The problem is instead of 10, I want to use variable VAR1 (where var1=10). The $VAR1 is not working.

I tried this command.

sed -f <(sed -e '1,$VAR1d; 12,$d; x; s/.*/10a\\/;p; x' ../log/file2.txt ) ../log/file4.txt > ../log/file5.txt

Please help me.

Answer 1

At first, use double quotes. It will enable BASH to process string.

Next, escape backslashes (even those that supposed to be escape symbols for sed) - because bash will escape them too.

I suppose it should be

sed -f <(sed -e "1,${VAR1}d; 12,\$d; x; s/.*/10a\\\\/;p; x" ../log/file2.txt ) ../log/file4.txt > ../log/file5.txt

Answer 2

尝试双qoutes和花括号：

 sed -f <(sed -e "1,${VAR1}d; 12,\$d; x; s/.*/10a\\\\/;p; x" ../log/file2.txt ) ../log/file4.txt > ../log/file5.txt

Answer 3

I prefer mix of single & double quotes:

1. Single quotes around the text that must not be expanded. (Removes need to escape special characters.)
2. Double quotes around the text that needs to be expanded.

eg Your case would look like:

sed -f <(sed -e '1,'"$VAR1"'d; 12,$d; x; s/.*/10a\\/;p; x' ../log/file2.txt ) ../log/file4.txt > ../log/file5.txt