正则表达式以匹配带点号的数字

Question

I have a string:

Product, Q.ty: 1, Price: 120.00

I want to select everything after the first comma up to the last two decimal digits (.00) - or, in other words, select the Product , which will be variable though; what is not variable is , Qt and it is also known that the last two characters in the string will be two digits preceded by a dot . - However only the last one will be always 0 , the one preceding it could be anything 0-9 , but always a digit.

I've used this to match the string:

preg_replace('/' . preg_quote(', Q.t') . '.*?' . preg_quote('.00') . '/', '', $data );

the problem is that it fails when the last two digits are not 00 but something else like 50, 40, 30 etc. If I use the same regex with a single digit '0', it won't work either because it will catch the first 0 in a string like in my earlier example and will leave out the remaining 0.

How to adjust this expression to catch a group of digits preceded by a '.' dot?

*one further note: this preg_replace is inside a foreach loop; some data won't match at all the pattern I'm trying to pass; which is ok, so in those cases I can print the strings the way they are; but for the cases in the foreach where there's a match, I want to replace part of the string with nothing*

Thank you

Answer 1

Why not just

/(\d+\.\d{2})$/

which would capture any trailing "numbers" with a decimal place?

Answer 2

 /([^,]*), Q\.ty: (\d*), Price: (\d*\.\d{2})/

By using ([^,]*), it will use the comma in the string as the first delimiter. This will capture the beginning of the string up to the first comma, the second match will be quantity and the last match will be the price.

So your provided string:

 Product, Q.ty: 1, Price: 120.00

will return

$1 = Product
$2 = 1
$3 = 120.00

on a side note I don't know if that period after Q in Q.ty is intentional in your example or just a typo.

Answer 3

You can try

(.+?), (Q\.ty: \d+, .+?\.\d{2})

This should capture everything from the first comma to the last two decimal digits into $2, with the product label being kept in $1

Answer 4

I figured someone (there always is) would say "You can get the pieces with str_replace() and explode() ." However it's not faster.

<?php

$string = "Product, Q.ty: 1, Price: 120.00";
$removals = array(",",":");


$stime = microtime();
    $nstring = str_replace($removals,'',$string);
    $parts = explode(" ",$nstring);

echo microtime()-$stime."secs\n";
    print_r($parts);

$pattern = "!^([A-Za-z]+),\s([A-Za-z.]+)\:\s([0-9]+),\s([A-Za-z]+):\s([0-9.]+)$!";

$ptime = microtime();
    $m = preg_match($pattern,$string,$matches);
echo microtime()-$ptime."secs\n";

print_r($matches);

?>

Output

4.0999999999958E-5secs
Array
(
    [0] => Product
    [1] => Q.ty
    [2] => 1
    [3] => Price
    [4] => 120.00
)
3.5000000000007E-5secs
Array
(
    [0] => Product, Q.ty: 1, Price: 120.00
    [1] => Product
    [2] => Q.ty
    [3] => 1
    [4] => Price
    [5] => 120.00
)

Using a more literal approach ,providing the $string doesn't deviate, does not improve performance of the preg_match function.

$pattern = "!^(Product), (Q\.ty): ([0-9]+), (Price): ([0-9.]+)$!";

Answer 5

如果需要文字点，则应对其进行换行：\\。