How to print and insert a variable in the beginning of the file in Unix?

Question

Length=$(cat Final.txt|wc -l)
echo $Length
sed -i '1 i\<records>$(echo $Length)</records>' Final.txt

I need to the add the line XXX where XXX should be the variable which represents the no.of lines.I have tried the above code and other combinations.None of them seems to work.

single quotes don't expand anything inside them, double quotes do. Compare echo '$Length' and echo "$Length". (Besides, the command substitution and echo in the sed command are unnecessary). shellcheck.net would actually tell you both of those. — ilkkachu, Mar 08 '18 at 09:23
references: What's a good mnemonic for shell double vs. single quotes? and http://mywiki.wooledge.org/Quotes — ilkkachu, Mar 08 '18 at 09:25

score 0 · Answer 1 · answered Mar 08 '18 at 09:33

0

Length=$(wc -l < Final.txt)
echo $Length
sed -i "1 i\<records>$Length</records>" Final.txt

answered Mar 08 '18 at 09:33

Kamaraj

4,365

user1404316 · Answer 2 · 2018-03-08T16:10:23.403

0

sed -i "1 i\<records>$(wc -l Final.txt < Final.txt)</records>" Final.txt

You don't need an intermediate variable.
Edit: Per user kamaraj & per user DonChrissti's commened, removed claim that you need the cut command, or some other way to parse out the file name which wc outputs along with the number of lines. A simple I/O redirection hides that from wc, so without knowing what file is being processed, it has nothing to print.

edited Mar 08 '18 at 16:10

answered Mar 08 '18 at 10:09

user1404316

3,078

$(wc -l Final.txt | cut -d" " -f1) This is my output in the file. It is not working. – M.Nehru Mar 08 '18 at 10:52

How to print and insert a variable in the beginning of the file in Unix?

2 Answers2