How to extract integer and decimal numbers in PHP string correctly?

Question

I have a string in PHP where I want to extract only integer and decimal numbers, this is an example:

<?php 

$string = 'María vive en un pueblo de 25 957 habitantes y cobra 1859 euros al mes. OJO: no sé si os habéis fijado, pero los números del último ejemplo 
no llevan un punto o una coma separando los millares (25.957 o 1.859). Sé que resulta extraño, pero la nueva normativa de la R.A.E. 
dice que los números de cuatro cifras NO llevarán separación (1859) y los números de cinco cifras o más NO llevarán ni puntos ni comas, 
sino una separación (25 957 o 1 343 392). El 94% de los niños ha traído los deberes hechos. He pagado $50,95 dólares.';

?>

In this case I try to get numbers using this line:

$numbers = preg_replace('/[^0-9\.,]/', ' ', $string);

But the problem is the result is:

25 957 1859 . , 25.957 1.859 . , . . . 1859 , 25 957 1 343 392 . 94 . 50,95 .

And I want to get this result:

25 957 1859 25.957 1.859 1859 25 957 1 343 392 94 50,95

How can I fix it?

I'd like your help.

If it's possible I want to consider those details

NekoLopez
– NekoLopez

2019-08-16 23:21:31 +00:00
Commented Aug 16, 2019 at 23:21 — NekoLopez
– NekoLopez, Commented Aug 16, 2019 at 23:21

rickdenhaan · Accepted Answer · 2019-08-16 23:25:24Z

4

You're using a . for thousands and a , for decimals, and it looks like your text is intended for humans so there will always be a number immediately before a . and ,.

So what you want is 1 or more numbers, followed by 0 or more thousand-groups (a . followed by 3 more numbers), optionally followed by exactly one , and one or more numbers.

One or more numbers: [0-9]+
0 or more thousand-groups: (\.[0-9]{3})*
Optionally a `,` and one or more numbers: (,[0-9]+)?

Combined: [0-9]+(\.[0-9]{3})*(,[0-9]+)?

$string = 'María vive en un pueblo de 25 957 habitantes y cobra 1859 euros al mes. OJO: no sé si os habéis fijado, pero los números del último ejemplo 
no llevan un punto o una coma separando los millares (25.957 o 1.859). Sé que resulta extraño, pero la nueva normativa de la R.A.E. 
dice que los números de cuatro cifras NO llevarán separación (1859) y los números de cinco cifras o más NO llevarán ni puntos ni comas, 
sino una separación (25 957 o 1 343 392). El 94% de los niños ha traído los deberes hechos. He pagado $50,95 dólares.';

$matches = [];
preg_match_all('/[0-9]+(\.[0-9]{3})*(,[0-9]+)?/', $string, $matches);

print_r($matches[0]);

/*
Array
(
    [0] => 25
    [1] => 957
    [2] => 1859
    [3] => 25.957
    [4] => 1.859
    [5] => 1859
    [6] => 25
    [7] => 957
    [8] => 1
    [9] => 343
    [10] => 392
    [11] => 94
    [12] => 50,95
)
*/

answered Aug 16, 2019 at 23:25

rickdenhaan

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2 Comments

rickdenhaan Over a year ago

@user3783243 yes, my expression is only using a , as a decimal separator. It is using a . for a thousand separator. The given text does not have any numbers with a , as a thousand separator (although it does have numbers with a space as a thousand separator, but that wasn't part of what OP wanted).

NekoLopez Over a year ago

Thanks for your answer, this works great, I just needed to add an implode() function, thanks again!

tcj · Accepted Answer · 2019-08-16 23:53:11Z

2

Probably not the most elegant way to do it but you could use two preg_replace functions based on your first REGEX and remove the decimals without any numbers wrapped around them like so :

$numbers = preg_replace('/[^0-9\.,]+/', ' ', $string);
$decimals = preg_replace('/([^0-9][\.,])/', '', $numbers);
var_dump($decimals); // string(56) " 25 957 1859 25.957 1.859 1859 25 957 1 343 392 94 50,95"

Live example here : https://3v4l.org/9aUCt

edited Aug 16, 2019 at 23:53

answered Aug 16, 2019 at 23:47

tcj

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Comments

Brad · Accepted Answer · 2019-08-16 23:35:04Z

1

Use this regular expression:

(\d*\.)?\d+

NOTE
This exp will match 25 957 like 2 matches (for the space between numbers).

edited Aug 16, 2019 at 23:35

Brad

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answered Aug 16, 2019 at 23:15

Giacomo M

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2 Comments

Giacomo M Over a year ago

what you mean? I did not understand

Giacomo M Over a year ago

according to the first post it is exactly what he wanted to do

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