say, the file 'listall' contains the directory list and I want to list up all the files in those directories including the directory path. I wrote a bash script below.
#!/bin/bash
\rm listZ
echo ' ' > listZ
for i in `cat listall`; ls $i | sed -e '1,$s/^/$i\//g' >> listZ; done
But in the sed command, the $i is not replaced with the directory name.
I tried using double quotes for sed command but in no avail. This is in cygwin.
How can I do it?
The single-quoted string is immune to variable substitution. However, it's a simple matter to move the variable to outside the single-quoted string, such as with:
pax> one=1 ; echo '
...> 1
...> 101
...> 3.14159' | sed 's/'$one'/x/g'
x
x0x
3.x4x59
In your case, it would be:
for i in `cat listall`; do ls $i | sed -e '1,$s/^/'$i'\//g' >> listZ; done
# ^^^^
Keep in mind that, as per one of your comments, if your file contains characters special to sed, they'll cause you grief. For example, if they contain / characters, you'll need to stop that from being interpreted by the sed s command.
Perhaps the easiest way to do this is to use a different separator that's not likely to be in the input file:
for i in `cat listall`; do ls $i | sed -e '1,$s?^?'$i'/?g' ; done
DontParseLs and DontReadLinesWithFor.
The problem you are asking about could be fixed by using double quotes instead of single quotes, but then, you need to escape any shell metacharacters which you do not intend for the shell to interpret (in other words, you'd have to backslash-escape the dollar sign in the address range 1,$, or use single quotes next to double quotes, and probably change to a different delimiter to cope with slashes in the value of the variable $i; '1,$'"s:^:$i:").
But your entire task can be performed as a single command.
IFS=$'\n' find $(cat listall) -mindepth 1 -maxdepth 2 -print >listZ
The IFS is only strictly necessary if listall contains file names with whitespace in them. You would still in principle need to escape any other shell metacharacters (asterisks, square brackets, question marks). A more robust workaround would be
while read -r directory; do
printf '%s\n' "$directory"/* "$directory"/*/*
done <listall >listZ
(I misunderstood your question at first, and assumed listall contained file names, not directory names. I'm keeping my original answer below.)
sed "s:^:$i/:" listall >listZ
provided you don't actually need an empty space on the first line (maybe add '1i\ ' to the sed script then). The address range 1,$ is the default in sed, so specifying it is redundant. Because there can only be one substitution per line, the /g flag is also redundant (it means, substitute all occurrences on one line, instead of just the first). If there could be colons in $i, use a different delimiter.
As noted by paxdiablo(and thanks to tripleee), my problem was that the shell variable contained several '/' characters, which messed up the sed substitute command. So we can avoid the problem using another delimiter like ';' instead of '/' for sed. So the correct way is
#!/bin/bash
\rm listZ
echo ' ' > listZ
for i in `cat listall`; ls $i | sed 's;^;$i' >> listZ; done
Note, in my case the file listall looks like
/usr/bin/
/usr/local/bin/