Search for a precise number with grep

Question

I am trying to look for lines in a file that have one or more instances of 1234, but no other numbers (non-digit characters are allowed). Any other number should cause the line not to match.

Valid examples:

• 1234
• 1234 xxx
• xxx 1234
• 1234 1234
• 1234 xxx 1234

Invalid examples:

• 12341234
• 12345
• 1234xxx345
• 1234 345
• 1234xxx
• xxx1234
• 1234xxx1234

This is what I have used:

grep -E '^([^0-9]*1234)+[^0-9]*$' file.txt

But this command also outputs 12341234 as valid, how do I prevent that?

Answer 1

awk might be handier than grep at this. Set the field separator to runs of non-digit characters, then loop through fields and only print line if every non-empty field is exactly equal to 1234

awk -F'[^[:digit:]]+' '{
  for(i=1; i<=NF; ++i) 
  if (($i) && ($i != 1234)) next
  }; {print}' file.txt

Answer 2

grep -E '^[^0-9]*1234([^0-9]+1234)*[^0-9]*$' file.txt

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.

Answer 3

Search for the number either at the beginning of the (^) line or preceded by a non-digit ([^0-9]), and likewise at the end ($ or [^0-9]).

grep -E '(^|[^0-9])1234($|[^0-9])' file.txt

If you want to allow leading zeros, add 0* before the number.

grep -E '(^|[^0-9])0*1234($|[^0-9])' file.txt

If you also want to reject lines containing any sequence of one or more digits that isn't part of an occurrence of this particular number, then you're looking for lines consisting of sequences alternating one or more nondigits and that particular number, which is basically ([^0-9]+1234)*. The line may start and end with either the number or a nondigit, whereas internal sequences of nondigits may not be empty. Furthermore the line must contain the number at least once. Putting it all together:

grep -xE '[^0-9]*1234([^0-9]+1234)*[^0-9]*' file.txt