grep command using wildcards [0-9]

Solution 1:

It's not clear from your description whether your expression is failing to match things you want, or matching things you don't want.

If it's the latter, then it may be because . in a grep regular expression matches any single character (except the newline character - however grep is normally line-based anyway). To match a literal dot (period) you need to escape it \. or place it in a character set as you have done for the decimal digit ranges:

grep "[.][0-9][0-9][0-9][0-9][0-9][0-9][0-9]"

You also mention that the expression should match data "ending with" - it's not clear whether you mean a line ending or a word boundary - these are respectively $ and \b (or \>) ex.

grep "[.][0-9][0-9][0-9][0-9][0-9][0-9][0-9]$"

grep "[.][0-9][0-9][0-9][0-9][0-9][0-9][0-9]\b"

You can also shorten the expression using a quantifier - switching to extended regular expression (ERE) mode¹:

grep -E "[.][0-9]{7}$"

¹ In GNU grep, you can use quantifiers in basic regular expression (BRE) mode by escaping the braces grep "[.][0-9]\{7\}$"