如何从文件的grep中拆分行？

Question

Given Employee.txt,

PD:198801221:Jason:Jason@gmail.com
PU:19941029:Tan:tan@gmail.com
PS:19960601:May:may@gmail.com

and a form,

echo -n "Enter BirthDate:";read bdate
grep "$bdate" Employee.txt
echo "Employee name (auto display):"
echo "Employee email (auto display):"

which takes an employee birth date and displays the name and email based on the input file.

For example, if I input in the BirthDate as 19941029 , the result I get from the file is PU:19941029:Tan:tan@gmail.com

The above script displays the entire line.

How can I separate the line into the auto display name and email fields?

Answer 1

You can use IFS to read the input into 4 variables

echo -n "Enter BirthDate:";read bdate

match=$(grep "$bdate" Employee.txt)
IFS=':' read -r first second third fourth <<<$match


echo "Employee name (auto display):" $third
echo "Employee email (auto display):" $fourth

References

https://bash.cyberciti.biz/guide/$IFS

https://unix.stackexchange.com/questions/209123/understanding-ifs-read-r-line

Answer 2

You can pipe the grep output through cut and pick up the fields you want.

#!/usr/bin/env bash
set -eu

echo -n "Enter BirthDate: "
read -r bdate

found=$(grep "$bdate" Employee.txt)
name=$(echo "$found" | cut -d: -f 3)
email=$(echo "$found" | cut -d: -f 4)

echo "Employee name (auto display): $name"
echo "Employee email (auto display): $email"

This is fairly simplistic and breaks if your fields contain separator characters (colons in this case). In that case you should get a CSV reader.