Shell Programming and Scripting

I have a requirement where I need to add a variable value based on a pattern match. When I try this it works:

Code:

sed '/IS_ETL_DEV/a\
ABCD' File1.txt > File1.tmp

Now I want to replace IS _ETL_DEV with a variable like $Pattern and replace the values ABCD with a variable $Var the value "$$XYZ=1". So, I replaced the code like this:

Code:

sed '/${Pattern}/a\
${Var}' File1.txt > File1.tmp

So, when the pattern is found, it will add a line $$XYZ=1 after it. I am using Solaris 10.

Hi guy,
the $'s special mean would not be effect in single quote, so you may try this.

Code:

sed '/'"${Pattern}"'/a\
'"${Var}"'' File1.txt > File1.tmp

and let Var='$$XYZ=1'

Or only use double quotes and escape special characters:

Code:

sed "/${Pattern}/a\\
${Var}" file

Thank you. Both the solutions work. But I have another question. Why do we have to give the substitution part in the 2nd line. Cant I give the whole command in one line? I tried and it did not work. Appreciate your inputs.

For the append command you need two lines with regular sed. You can try with GNU sed (gsed):

Code:

sed "/${Pattern}/a${Var}" file

Code:

sed "/${Pattern}/s/\$/\\n${Var}/" file