Find Numbers Equal to or Greater than 1 using regex -


i'm not developer or scripter. can't contribute forum use guidance staff of developers. that's disclaimer because last time on site, reamed me out asking questions , not contributing. this, apologize.

if willing assist, or @ least give me kick-start, how find version of file if version has #.#.###; i.e., 6.1.3890?

so, goal find number equal or greater 1 , equal or greater 389. concerned digits after first l'.' , second '.'

thanks , all.

a regex number greater 389:

(39[0-9]|[4-9][0-9][0-9]|[1-9][0-9][0-9][0-9]+)

a regex number greater 1:

([2-9]|[1-9][0-9]+)

a combined regex version above 6.1.389:

(6\.1\.(39[0-9]|[4-9][0-9][0-9]|[1-9][0-9][0-9][0-9]+)|6\.([2-9]|[1-9][0-9]+)\.[0-9]+|([7-9]|[1-9][0-9]+)\.[0-9]+\.[0-9]+)

non 0 numbers should not start 0.

if version number format limited #.#.### or possibly fewer digits last part, regex can simplified to:

(6\.1\.(39[0-9]|[4-9][0-9][0-9])|6\.[2-9]\.[0-9]+|[7-9]\.[0-9]\.[0-9]+)

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