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Try this:

awk '{sum+=$1} END {print sum}' file

Try this:
cat file | awk '{ sum+=$1} END {print sum}'

thats equivalent to print $0, means prints all the contents.

# awk 'BEGIN{FS=OFS=","}
$1 !~ /^0/{$1="0"$1}
$2 !~ /^0/{$2="1234"$2} {print $0}' file
...

awk 'BEGIN{FS=OFS=","}
$1 !~ /^0/{$1="0"$1}
$2 !~ /^0/{$2="1234"$2}1' file

ls -m > output.txt should do the trick

The below will find all files in current directory that were created/last modified on given date and then will search in those files for any word.


ls -l | grep "Dec 12" | awk '{print $NF}' |...

find /path/to/dir -type f -name "filename" -mtime -1 2>/dev/null | xargs grep -iw "word"

filename could be *.txt , *abc*txt etc.
-mtime -1 means modification(or creation time) less than 1 day