regex and commas

Question

Hi this is working now but I am confused at to why.

I am learning regex and need to pull the numbers out of strings like

'Discount 7.5%' should get 7.5
'Discount 15%' should get 15
'Discount 10%' should get 10
'Discount 5%' should get 5

etc.

/\d,?.\d?/ //works

/\d,?.\d,?/ //doesn't works

/\d?.\d?/ //doesn't works

I thought one of the second two would work could someone explain this.

**1.** You should escape the dots `.` in all three regex by preceding it with \, **2.** Use `\d+` to match more than one digits **3.** No need of `,` in any regex — Tushar, Nov 05 '15 at 04:36
Possible duplicate of [regular expression for DOT](http://stackoverflow.com/questions/3862479/regular-expression-for-dot) — Basilevs, Nov 05 '15 at 04:39
IMO, use [`(\d+(\.\d+)?)`](https://regex101.com/r/iL3dR1/1) with first capture group — Tushar, Nov 05 '15 at 04:42

Derek 朕會功夫 · Accepted Answer · 2015-11-05T04:50:55.057

2

Quick and dirty with easy to understand regex.

//Let the system to optimize the number parsing for efficiency
parseFloat(text.replace(/^\D+/, ""));

Demo: http://jsfiddle.net/DerekL/4bnp8381/

edited Nov 05 '15 at 04:50

answered Nov 05 '15 at 04:47

Derek 朕會功夫

92,235
44
185
247

1

Will replace `.` decimal too – Tushar Nov 05 '15 at 04:48
2

@Tushar Nope It won't. – Derek 朕會功夫 Nov 05 '15 at 04:48
@Tushar True, but it's there just to make sure it only removes the non-digits in the front. – Derek 朕會功夫 Nov 05 '15 at 04:53
@Derek朕會功夫 , this would match `Fee 20%` too while it is not needed – Mi-Creativity Nov 05 '15 at 06:19
I like this and you can add parseFloat(t.replace(/^Discount /, "")) if you want to have only Strings that start with Discount good way to get rid of the comma it even works with 'Discount 10% extra' and i thought that would break it https://jsfiddle.net/4bnp8381/2/ – Thomas K Nov 05 '15 at 21:37
@ThomasK If you know `"Discount "` is always at the beginning of the string, you can even do `parseFloat("Discount 99% off".slice(9))` to completely get rid of the regex. – Derek 朕會功夫 Nov 05 '15 at 23:11

score 0 · Answer 2 · answered Nov 05 '15 at 04:40

0

Try this. Make the second part with dot . optional with ?

(\d)*(\.\d*)?

answered Nov 05 '15 at 04:40

Tuan Anh Tran

6,807
6
37
54

score 0 · Answer 3 · edited May 23 '17 at 12:15

0

IF you check this Regex, this is what you need so it will return the float number only if the word before it is Discount by making use of the lookbehind operator:

(?<=Discount\s)\d+(\.\d+)?

But the problem is that works with PHP(prce) I can't get it working with javascript.

EDIT: As mentioned in here regex and commas Javascript does not support lookbehind regex so here is what I did to work around it JS Fiddle

var div = document.getElementById('test');
var text = div.innerHTML;
div.innerHTML = text.replace(/Discount (.*)%/ig,'$1');

<div id="test">
    Discount 7.5%<br>
    Discount 15%<br>
    discount 10%<br>
    Discount 5.433%<br>
    Fee 11%<br>
    Discount 22.7%
</div>

As you see it does match it only if it was followedd by the word Discount, the word Fee 11% does not match

edited May 23 '17 at 12:15

Community

1
1

answered Nov 05 '15 at 05:26

Mi-Creativity

9,554
10
38
47

@TimBiegeleisen, I'm sorry but it is not the same – Mi-Creativity Nov 05 '15 at 05:29
My inputs are pretty standard but I do like that this checks for the word Discount. I tried starting using replace but in something like this i could not get it to work. var re = /\d,?.\d?/; var vipextract = re.exec(cusvipdis); var skuextract = re.exec(skutext); – Thomas K Nov 05 '15 at 21:16

Tim Biegeleisen · Answer 4 · 2015-11-05T04:58:27.827

-1

You can use the following regex, which will match one or more digits, optionally followed by a decimal point and any number of digits:

^Discount\s(\d+(\.\d+)?)%$

Regex101

edited Nov 05 '15 at 04:58

answered Nov 05 '15 at 04:39

Tim Biegeleisen

502,043
27
286
360

@Tushar Thanks for catching this. I did not test thoroughly enough. – Tim Biegeleisen Nov 05 '15 at 04:41
`(\d+(\.\d+)?)` better? – Tushar Nov 05 '15 at 04:43
1

@Tushar I would avoid the `+` operator since this is a percentage. We don't want to allow for, say 4 digits in the beginning. – Tim Biegeleisen Nov 05 '15 at 04:46
how about `\d+(\.\d+)?` – Mi-Creativity Nov 05 '15 at 04:46
@Mi-Creativity It's the same, I just added brackets to get captured group – Tushar Nov 05 '15 at 04:47
That what I wonder about, why do we need an outer group here? – Mi-Creativity Nov 05 '15 at 04:48
@Mi-Creativity The other group is not used for capturing the result, it is just used for grouping two things `.` and `\d+`. – Tushar Nov 05 '15 at 05:03

regex and commas

4 Answers4