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I have this following file:-

$ cat numbers
a1
12
12345
123456
19816282
1@$%6

I am using grep "^[0-9]\{1,6\}$" numbers which is giving me the following results:-

12
12345

I was expecting 1@$%6 also in the results. Please correct me if I am wrong?

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Try this "^[0-9].*{1,6}$" – tikend Feb 14 at 13:49
it didn't work out. I want to display all lines which have a minimum of 1 digit and a maximum of 6 digits. – Ankit Feb 14 at 13:54

2 Answers

up vote 3 down vote accepted

The result is actually

12
12345
123456

Your regex means "A line consisting of 1 to 6 digits and nothing else" so 1@$%6 doesn't match.

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+1, is this because I am using $ in my regex? – Ankit Feb 14 at 13:57
Yes, without the $ it would mean "Any line that starts with 1 to 6 digits" which includes 1@$%6. – Florian Diesch Feb 14 at 14:01
up vote 2 down vote 
> grep "[0-9]\{1,6\}" numbers.txt
1
12
12345
123456
19816282
1@$%6
share|improve this answer
+1, but $ only specified that matched pattern should end with a number. please correct if I am wrong? – Ankit Feb 14 at 14:02
1  
the ^"[0-9]\{1,6\}"$ will find pattern of 1 to 6 numbers between beginning and the end of a line. "[0-9]\{1,6\}"$ will find 1 to 6 number before end of a line. – tikend Feb 14 at 15:54

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