Shell Programming and Scripting

Hi All,

I have about 10000 files having names of the same pattern

I am using awk to match the pattern.

Inside the awk block how can i delete all the 10000 files..


Or is there any other way to delete files in bulk...


"rm" command fails with the message that argument list is too large

You can use the find

Ex:

Code:

find dir_name -type f -exec rm -f {} \;

Awk doesn't haev any mechanism to delete file/files. However, you can make use of system("command").

will this be able to delete 10000 files....

---------- Post updated at 05:40 AM ---------- Previous update was at 05:37 AM ----------

where is filename coming in the command that you have written

in awk itself, you can use system(), assuming no spaces in files

Code:

awk '{cmd="rm " FILENAME; print cmd; system(cmd) }' file*

#!/sbin/sh
###################################################
# Write This Shell Script and run it. Edit As u Wish.
###################################################

# Give ur dirctory location
DIRLOC=
export DIRLOC

# Give ur file Pattern
# This is one eg.

FILEEXT=DB.*.log;
export FILEEXT

find $DIRLOC -name "$FILEEXT" -exec /bin/rm {} \;

echo "Deletion Completed"

#
#
##################################################

This is a very dangerous script because it makes no attempt to limit the find to the target directory, hence it will cherrfully delete it's way down all subdirectories.Another problem is that when using -exec it will launch rm as a new process every time it deletes a file. Better to write code that passes a list of file names to rm vi xargs.Better still is to write a function that explicitly tests - ownership, presence, writability and so on. You could also pass it parameters such as recursive = yes/no etc

Thanks Dude, Actually I am new in shell scripting and in UNIX environment.

If u possible plz give some solution, to overcome this problem.

Thanks,

---------- Post updated at 11:17 AM ---------- Previous update was at 11:11 AM ----------

#Suppose I have two different directory ...
#with some pattern logs as FILEEXT
#eg. as below

DIRLOC=/opt/logs/db/logs;
export DIRLOC
FILEEXT=DB.*.log;
export FILEEXT

DIR1LOC=/opt/logs/csv/logs;
export DIR1LOC
FILE1EXT=csv.*.log;
export FILE1EXT


find $DIRLOC -name "$FILEEXT" -exec /bin/rm {} \;
find $DIR1LOC -name "$FILE1EXT" -exec /bin/rm {} \;


# so how can i write code which passes a list of file names to rm vi xargs.

# Plz Help !!! :-)