Removing part of a file name and appending into a single file

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Top Forums Shell Programming and Scripting Removing part of a file name and appending into a single file
# 8  
Originally Posted by varlax
How do i extract the date part of the existing file in the process to use it in the log flog_<date part extracted>. Also i want ot remove the original files after the merge

Use this:
ls /path/directory/ | grep abc_def* > $filemerge 
for j in `cat $filemerge` 
cat $j >> $newfile 
echo "adding $filemerge to $newfile"
rm $j
rm $filemerge

---------- Post updated at 03:50 PM ---------- Previous update was at 03:44 PM ----------

filemerge=/path/merge # this is an arbitrary file that you are using. Just make up a path and filename
newfile=/path/abc_def_xxx.txt   #This is the new file that you want everything in 
ls /path/directory/ | grep abc_def* > $filemerge     #this is doing an ls and putting all of the filenames 
                                                     #into a file that you will run through
for j in `cat $filemerge`           #This is reading the file one line at a time and doing the following: 
cat $j >> $newfile                  #adding info to newfile...note that it is appending everything to it
echo "adding $filemerge to $newfile"   #just letting you know where it is in process
rm $j                       #deleting old file
rm $filemerge     #deleting file that had list of all filenames

# 9  
Try this. $1 contains the first parameter to the script
for file in ${first}_*_*.${suffix}
  cat "$file" >> "${first}_${last}"

You can try removing the files by adding && rm "$file"
cat "$file" >> "${first}_${last}" && rm "$file"

You can write to a log by using echo statements in the loop and
done >> filemerge.log

Obviously this would need to be thoroughly tested first..


( Or, in ultra-compact form just for fun Smilie)
for f in ${1%.*}_*_*.${1##*.}; do
  cat "$f" >> "${1%.*}_${f##*_}"

Last edited by Scrutinizer; 05-04-2012 at 07:43 PM..
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