i need regex for grep(posix) to print only(return true) if every floating point separate by blank space in single line is valid like:
3.14 12 18.9
and not print anything (return false) if one or more invalid floating point appear in that line between blank space like:
8.12 1.1.1 78
or
1..28 1.09 46
the floating point can appears as much as possible in that single line, as long is valid floating point separate by blank space it will return true/print by grep(posix).
currently i have regex for grep:
grep -E "^[[:blank:][:digit:]]*.+[[:digit:]]+$" FILE
it work for pattern like:
1.13 1 1.2.3 1. 1
but i don't wont pattern like 1.2.3 and 1. 1 to be match, just i need is 1.13 or 1 between blank space.
note: i just need that work single line only.
You can use
grep -E '^[0-9]+(\.[0-9]+)?([[:blank:]][0-9]+(\.[0-9]+)?)*$'
Details:
^ - start of string[0-9]+ - one or more digits(\.[0-9]+)? - an optional occurrence of a . and one or more digits([[:blank:]][0-9]+(\.[0-9]+)?)* - zero or more occurrences of
[[:blank:]] - a tab or space[0-9]+(\.[0-9]+)? - one or more digits followed with an optional occurrence of a . and one or more digits$ - end of string.