Arguments inside sed loop

问题

Im trying to understand sed command and loops. I need to take part of the text (20 lines) and append it to the csv with the file name. here is my code

for i in ~/mspl/*.doc
do 
    catdoc "$i" > auto.txt
    sed -e '1,20!d' auto.txt > auto1.txt
    sed -e '1s/^/"$i" ;/' auto1.txt > auto2.txt
    sed -e '20s/$/~/' auto2.txt > auto3.txt
    cat auto3.txt >> lines.csv
done

the problem is that my second "i" argument doesn't convert to the file name in csv.

in the line

sed -e '1s/^/"$i" ;/' auto1.txt > auto2.txt

Please tell what is my mistake here?

回答1:

The issue is that variables are not expanded within single quotes '. Use double quotes instead ":

# this is assuming you did want to add explicit " around your $i
sed -e "1s/^/\"$i\" ;/" auto1.txt > auto2.txt

If your double-quoted string needs to contain explicit double quotes, you have to escape them (\"). If using double quotes is not an option (e.g. if you have a very complex command with multiple levels of quoting already) you can always exit the quoting around your variable:

sed -e '1s/^/"'$i'" ;/' auto1.txt > auto2.txt

回答2:

You might find it useful to run this script with the -x setting, to see what's actually going on. (Just do set -x.)

But just to be blunt, what's going on is that your $i is inside single quotes, so it doesn't get expanded.

来源：https://stackoverflow.com/questions/14402121/arguments-inside-sed-loop