Grep search for any number in a range

grepsearch

So I have a file with a number of records that look like the following:

Peugeot:206:2000:Red:1

And I'm trying to grep search for any that contain a year 1995-1999. I have tried

grep '[1995-1999]' file.txt

to no avail. Any ideas? Thanks.

Best Answer

This one is easy:

grep -E "199[5-9]"

does the job. It is easy, because the intended number range matches a character code range. For a more complicated example, e.g., 1998-2003, you will have to split the range appropriately:

grep -E "199[8-9]|200[0-3]"

Explanation

^[^0-9]*1234: find the first match of 1234, which may be preceded by non-digit characters.
([^0-9]+1234)*: there may be further iterations of 1234, but these must be separated from the first 1234 (and other 1234) by non-digit characters (hence use +).
[^0-9]*$: match the entire line (with $). There may be non-digit characters after the final 1234 (but not necessarily, hence *).

EDIT

If 1234 must be delimited by spaces (or be at the beginning or end of the line), then use

grep -E '^([^0-9]* )?1234(( [^0-9]*)? 1234)*( [^0-9]*)?$'

Explanation

^([^0-9]* )?: there may be non-digit characters to start with, as long as they end with a space.
1234: find the first (required) match of 1234.
(( [^0-9]*)? 1234)*: I'll work through the parentheses backwards. There may be (zero or more) further copies of 1234, but these must be preceded by a space, i.e. 1234. Before this space, there may be non-digit characters, which is fine as long as these are separated from the preceding copy of 1243 by another space, i.e.( [^0-9]*)?.
( [^0-9]*)?$: there may be non-digit characters to end with, as long as they are preceded by a space.

Best Answer

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