[javascript] RegEx: How can I match all numbers greater than 49?

I'm somewhat new to regular expressions and am writing validation for a quantity field where regular expressions need to be used.

How can I match all numbers greater than or equal to 50?

I tried

[5-9][0-9]+

but that only matches 50-99. Is there a simple way to match all possible numbers greater than 49? (only integers are used)

This question is related to javascript regex

The answer is


I know there is already a good answer posted, but it won't allow leading zeros. And I don't have enough reputation to leave a comment, so... Here's my solution allowing leading zeros:

First I match the numbers 50 through 99 (with possible leading zeros):

0*[5-9]\d

Then match numbers of 100 and above (also with leading zeros):

0*[1-9]\d{2,}

Add them together with an "or" and wrap it up to match the whole sentence:

^0*([1-9]\d{2,}|[5-9]\d)$

That's it!


I know this is old, but none of these expressions worked for me (maybe it's because I'm on PHP). The following expression worked fine to validate that a number is higher than 49:

/([5-9][0-9])|([1-9]\d{3}\d*)/

Try a conditional group matching 50-99 or any string of three or more digits:

var r = /^(?:[5-9]\d|\d{3,})$/

Next matches all greater or equal to 11100:

^([1-9][1-9][1-9]\d{2}\d*|[1-9][2-9]\d{3}\d*|[2-9]\d{4}\d*|\d{6}\d*)$

For greater or equal 50:

^([5-9]\d{1}\d*|\d{3}\d*)$

See pattern and modify to any number. Also it would be great to find some recursive forward/backward operators for large numbers.


Try this regex:

[5-9]\d+|\d{3,}