How to match some words from file and list all the rows of that matching word? (without regex)

Question

Display the Employee details of all those working in a location which ends with ore in its location name

EmpID:Name:Designation:UnitName:Location:DateofJoining:Salary
1001:Thomson:SE:IVS:Mumbai:10-Feb-1999:60000
1002:Johnson:TE::Bangalore:18-Jun-2000:50000
1003:Jackson:DM:IMS:Hyderabad:23-Apr-1985:90000
1004:BobGL::ETA:Mumbai:05-Jan-2004:55000
1005:Alice:PA:::26-Aug-2014:25000
1006:LilySE:IVS::Bangalore:17-Dec-2015:40000
1007:Kirsten:PM:IMS:Mumbai:26-Aug-2014:45000
1004:BobGL::ETA:Mumbai:05-Jan-2021:55000

Expected output:

1002:Johnson:TE::Bangalore:18-Jun-2000:50000
1006:LilySE:IVS::Bangalore:17-Dec-2015:40000

Here's the code I tried, it's only showing the location but I want full details

cut -d ":" -f4 employee.txt | grep 'ore\>' 

EDIT: SOLVED

grep "`cut -d ":" -f5 employee.txt | grep 'ore\>'`$" employee.txt

got output:

1002:Johnson:TE::Bangalore:18-Jun-2000:50000
1006:LilySE:IVS::Bangalore:17-Dec-2015:40000

Thanks everyone for help :)

Answer 1

Here a non-regex approach using awk:

awk -F: -v s="ore" '(n=index($5,s)) && (n + length(s)-1) == length($5)' file

1002:Johnson:TE::Bangalore:18-Jun-2000:50000
1006:LilySE:IVS::Bangalore:17-Dec-2015:40000

Details:

• index($5,s) function finds position of input string ore in fifth column i.e $5 of each line
• (index($5,s) + length(s)-1) == length($5) check is to ensure that ore is the ending substring of $5

A regex approach would be simpler:

awk -F: -v s="ore" '$5 ~ s "$"' file