1st file
1 4 7
c b 8
3 6 a
2nd file
z 6 1
q g w
3 t 5
suppose i have 1000 files with 3 fields.I don't know if there is value in 4th fields out of those 1000 files. I want to print those files where there is value in the 4th fields.
---------- Post updated at 12:41 PM ---------- Previous update was at 12:18 PM ----------
Can anyone help me with the requirement
suppose that all your files are named by *.txt. And fields delimiter is space " ".
cd yourDir
grep -El ".*\s.*\s.*\s.*" *.txt
will list all the filenames with 4 columns.
can you please explain how the script works
go through all the files, lookin for if there is one line contains the 4th field. if found one, print only the file name.
i am not able to understand what
-El ".*\s.*\s.*\s.*" means
-E, --extended-regexp
Interpret PATTERN as an extended regular expression (ERE, see below). (-E is specified by POSIX.)
-l, --files-with-matches
Suppress normal output; instead print the name of each input file from which output would normally have been printed.
The scanning will stop on the first match. (-l is specified by POSIX.)
".*\s.*\s.*\s.*" is regular expression to find out the pattern with 4 fields. in your case.
if you read man page of grep command, you can get all those info.
Another possible option, assuming you look for txt files:
files=$(find . -type f -name '*.txt' | awk -F"/" '{print $NF}')
for i in $files
do
awk 'NF>3{print "Files with more than 3 columns-->",FILENAME}' $i | uniq
done
Or a faster code than my first reply:
files=$(find . -type f -name '*.txt' | awk -F"/" '{print $NF}')
for i in $files
do
awk 'NF>3{print "File with at least one line with " NR " columns-->",FILENAME;exit 1}' $i
done
Regards
Hi.
Similarly:
#!/usr/bin/env bash
# @(#) s1 Demonstrate identifying lines with more than N fields.
# Utility functions: print-as-echo, print-line-with-visual-space, debug.
pe() { for i;do printf "%s" "$i";done; printf "\n"; }
pl() { pe;pe "-----" ;pe "$*"; }
db() { ( printf " db, ";for i;do printf "%s" "$i";done; printf "\n" ) >&2 ; }
db() { : ; }
C=$HOME/bin/context && [ -f $C ] && . $C awk
pl " Data files:"
for f in data*
do
pe
cat $f
done
N=3
pl " Results, looking for files with more than $N fields:"
awk -v N="$N" '
NF > 3 { print " File", FILENAME, "has more than 3 fields.";nextfile}
' data*
exit 0
producing:
% ./s1
Environment: LC_ALL = C, LANG = C
(Versions displayed with local utility "version")
OS, ker|rel, machine: Linux, 2.6.26-2-amd64, x86_64
Distribution : Debian GNU/Linux 5.0.7 (lenny)
GNU bash 3.2.39
GNU Awk 3.1.5
-----
Data files:
1 4 7
c b 8
3 6 a b
z 6 1
q g w
3 t 5
z 6 1
q g w f
3 t 5
1 4 7
c b 8
3 6 a
-----
Results, looking for files with more than 3 fields:
File data1 has more than 3 fields.
File data3 has more than 3 fields.
Showing that awk itself can deal with a number of files as parameters ... cheers, drl