Match regex with numeric value and decimal

Question

I need to match regex to value that is numeric with decimal. Currently I have /^-?[0-9]\d*(.\d+)/ but it does not account for .00 How would I fix that

Current Valid:

1
1.0 
1.33
.00

Current Invalid:

Alpha Character 

Answer 1

You need to handle the two possibilities (numbers without a decimal part and numbers without an integer part):

/\A-?(?:\d+(?:\.\d*)?|\.\d+)\z/
#^   ^  ^            ^^     ^---- end of the string
#|   |  |            |'---- without integer part
#|   |  |            '---- OR
#|   |  '---- with an optional decimal part
#|   '---- non-capturing group
#'---- start of the string

or make all optional and ensure there's at least one digit:

/\A-?+(?=.??\d)\d*\.?\d*\z/
#  ^  ^  ^        ^---- optional dot
#  |  |  '---- optional char (non-greedy)
#  |  '---- lookahead assertion: means "this position is followed by"
#  '---- optional "-" (possessive)

Note: I used the non-greedy quantifier ?? only because I believe that numbers with integer part are more frequent, but it can be a false assumption. In this case change it to a greedy quantifier ?. (whatever the kind of quantifier you use for the "unknow char" it doesn't matter, this will not change the result.)

Answer 2

If the first part is optional you can mark it off as such with `(?:...)?:

/\A(?:-?[0-9]\d*)?(.\d+)/

The ?: beginning means this is a non-capturing group so it won't interfere with the part you're trying to snag.

Answer 3

What about simple: /\d*\.?\d*/ ?

Answer 4

Create a regex with the variants you want to match, in this case, 3:

N
N.NN
.NN

i.e.:

(\d+\.\d+|\d+|\.\d+)

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