finding a string

file1="your file"
cat $file | while read line
do file_date=`echo $line|cut -d"|" -f4`
echo "Here is your date : "$file_date
done

It will read all line in your file and the variable file_date will have your date info for you.

Try this

script.sh
--- begin ---
#/bin/sh
cat $1 | awk -F "|" '$4 ~ /1999/ {print $0}'
--- end ---

U can use the script by running script.sh [filename]

This will match the fourth column for year 1999 and return the whole line that match the relevant results. Modify the value 1999 to your needs and if you want to show specific colum, modify the $0 to something else like $1 to show first column, or {print $1, $5} to show column 1 and 5

Observation:

- You have a file which has the "Pipe" symbol <|> as
the "Field Separator".
- Field 4 and 5 are the date in the format : DD/MM/YYYY

- You didnot specify that you want to find the year 1999 in field 4 , or 5, or either, or both .

Answer:

Case 1: Look for "/1999" in either field 4 or 5:

A simple grep can do the job:

cat file_name | grep '/1999'

Explanation: the date has the forward slash in it.

Case 2: Look for "/1999" in field 4 only.

cat file_name | awk -F"|" '( $4 ~ "/1999" ) {print $0}'

Case 3: Look for "/1999" in field 5 only.

cat file_name | awk -F"|" '( $5 ~ "/1999" ) {print $0}'

Case 4: Look for "/1999" in BOTH field 4 and field 5:

cat file_name | | awk -F"|" '( $4 ~ "/1999" ) && ( $5 ~ "/1999" ) {print $0}'

Note:
{print $0} == print the whole line, replace $0 with other number if you want to print certain field only, SEPARATE them by comas.
Ex.: {print $1, $3, $5}

Good luck,

This question was closed by the author