CODEWITHSUNDEEP
ubuntuawkscriptinggrep
Babbaranish
Babbaranish

Reputation: 69

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

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 :)

Upvotes: 4

Views: 258

Answers (6)

Faisal
Faisal

Reputation: 53

awk -F: '{if (substr($5,1)~"ore")print $0}' emp.txt

This works for exact col match

grep -Ri "ore" emp.txt

This works for whole doc match

Upvotes: 0

tomc
tomc

Reputation: 1207

fgrep "ore:" employee.txt

f(ast)grep is exact string match no regex (same as grep -F)

It would not know anything about which column it was matching, so that filter would have to happen before or after.

Upvotes: 0

RavinderSingh13
RavinderSingh13

Reputation: 133610

We could go with this simple awk solution here. Without regex approach as per OP's requirement. Simple explanation would be: check if 5th field last 3 characters are ore then print that line.

awk 'BEGIN{FS=OFS=":"} substr($5,length($5)-2)=="ore"' Input_file

Generic answer: As per Ed sir's nice suggestion adding here more generic solution. Where one could set value of tail as per string needs to be looked upon.

awk 'BEGIN{FS=OFS=":"; tail="ore"} substr($5,length($5)-length(tail)+1)==tail' Input_file

Upvotes: 5

Darkman
Darkman

Reputation: 2981

Just in case you would change your mind to use regex.

Using awk:

$ awk -F: -v ends="ore" '$5~".*"ends' file.txt
1002:Johnson:TE::Bangalore:18-Jun-2000:50000
1006:LilySE:IVS::Bangalore:17-Dec-2015:40000

Using grep:

$ grep 'ore:' file.txt
1002:Johnson:TE::Bangalore:18-Jun-2000:50000
1006:LilySE:IVS::Bangalore:17-Dec-2015:40000

Or this:

$ grep -E '(.*:){4}.*ore:' file.txt
1002:Johnson:TE::Bangalore:18-Jun-2000:50000
1006:LilySE:IVS::Bangalore:17-Dec-2015:40000

Upvotes: 4

anubhava
anubhava

Reputation: 785541

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

Upvotes: 5

Shawn
Shawn

Reputation: 52529

Using just grep (With a regular expression; the only way you can avoid them in grep is using grep -F, which does literal string matching):

$ grep -E '^([^:]*:){4}[^:]*ore:' input.txt
1002:Johnson:TE::Bangalore:18-Jun-2000:50000
1006:LilySE:IVS::Bangalore:17-Dec-2015:40000

Explanation:

Using Extended instead of Basic Regular Expression syntax for readability:

Starting at the beginning of the line, matches four fields (0 or more non-: characters followed by a :), and then a fifth field that ends in ore (Again, 0 or more non-: characters, then o, r, e, and finally the : at the end of the field).

Upvotes: 4

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