Listing files with full path

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# 1  
Old 08-02-2007
Question Listing files with full path


I need to store all the files in a directory to a text file with its full path.
The example below can explain:


The ls -R1 command won't give the result as I desired. Please help.

# 2  
Old 08-02-2007
find . -type f | sed "s#^.#$(pwd)#"

# 3  
Old 08-02-2007
Dear vgersh99,

Thanks a lot. It is working fine.

Is it possible to get the files ordered by Last modified date in the same pattern?

Thanks again for all.

# 4  
Old 08-02-2007
Originally Posted by r_sethu

Is it possible to get the files ordered by Last modified date in the same pattern?

If you can use
ls -ltR yourpathandpipes

instead of `find` that might make you happy. You can then awk print the filename and append the sed pwd. Or run from another location onto the catalogue where files are placed, then ls should return fullpath.

Do you need hardcode examples ?
# 5  
Old 08-02-2007
Hi Congo,

Thank you. It will be very useful for me if you could give one hardcoded example please.

# 6  
Old 06-15-2009
Hi Guys,
I did something good today:

My situation:

I had 2 servers say server A and server B. Server A had got millions of files and directories. I had to keep the same ownership and group ownership in server B as in Server A. For this I did below:

1. Get to know the file ownerships in server A and put it in a file

find . -type f | sed "s#^.#$(pwd)#" | xargs ls -l | tr -s " " " " | cut -f3,4,9 -d\ > perm.conf

find . -type d | sed "s#^.#$(pwd)#" | xargs ls -ld | tr -s " " " " | cut -f3,4,9 -d\ >> perm.conf

2. copy the perm.conf on the server B under the same hierachy as in server A.

3. Then I wrote a script:


k=`wc -l perm.conf | cut -f1 -d\ `
while [ $k -ne $i ]
RECORD=`head -$i perm.conf | tail -1`
OWNER=`echo $RECORD | cut -f1 -d\ `
GROUP=`echo $RECORD | cut -f2 -d\ `
FILE=`echo $RECORD | cut -f3 -d\ `
chown $OWNER:$GROUP $FILE 2>> logfile
i=`expr $i + 1`

and I was done...

Hope this help someone...

Thanks and regards,
Pravin Goyal
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