What is the regex for these cases?

Question

What is the regex for these cases:

29000.12345678900, expected result 29000.123456789

29000.000, expected result 29000

29000.00003400, expected result 29000.000034

In short, I want to eliminate the 0 point if there is no 1-9 found again behind decimal and I also want to eliminate the dot (.) if actually the number can be considered as integer.

I use this regex

(?:.0*$|0*$)

but it gives me this result:

29123.6 from 29123.6400, 4 is gone from there.

When I tested the regex separately, it works perfectly,

.0*$ gives me 29123 from 29123.0000

0*$ gives me 29123.6423 from 29123.642300

Am I missing something with the combined regex?

Answer 1

I don't know whether Objective-C supports something like the following construct, but in Python you can do it completely without regular expressions using str.rstrip():

In [1]: def shorten_number(number):
   ...:     return number.rstrip('0').rstrip('.')

In [2]: shorten_number('29000.12345678900')
Out[2]: '29000.123456789'

In [3]: shorten_number('29000.000')
Out[3]: '29000' 

Answer 2

You simply want this:

^\d*(\.?\d*[1-9])?

1. ^\d* that means one or more digit before the first group.
2. In the () that describes matching group.
3. \.? means single DOT(.) can be there but optional. eg. (.)
4. \d* there can be one or more digits. eg. (1234)
5. \.?\d* there can be one DOT and one or more digit eg. (.123)
6. [1-9] this includes only digit from 1 to 9 only excluding 0. eg. (2344)

Regex

Answer 3

You can use this regex

^\d+(\.\d*[1-9])?
-   -------------
|        |->this would match only if the digits after . end with [1-9] 
|
|->^ depicts the start of the string..it is necessary to match the pattern

that solves your problem

try it here

Answer 4

If you think regex is the best way of doing it, you can just use something like this:

\.?0+$

It works for both cases:

> '12300000.000001130000000'.replace(/\.?0+$/g, '')
"12300000.00000113"
> '12300000.000000000000'.replace(/\.?0+$/g, '')
"12300000"