2016-03-17 09:56


I would like to get the second number present in a string. Maybe you have better ideas than mine.

From this example:

1 PLN = 0.07 Gold

I would like to get only "0.07" from that string. I obtain it from web scraping so it returns me as a string.

The problem that I have is the following. The "second number in string" might be with ".", without it, might be composed by only one number ex. "1", or by two "12", might have decimals ex. "1.2", even position may change, because for some currency I will have "1 PLN", "1 USD" for others I will have "1 TW".

So I can't work on position, I can't extract only numbers (I have the "1" at the beginning of the string), I can't extract only INT cause I could have also decimals...

So the only constant of that string - I think (but if you have better ideas pls suggest me) - is that I need the second number I find in the string.

How could I get it? Sorry If I wasn't enough clear.

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  • dounao5856 dounao5856 5年前

    Try this:

    $string = '1 PLN = 0.07 Gold';
    $pattern = '/\d+\.\d+/';
    $matches = array();
    $r = preg_match($pattern, $string, $matches);
    var_dump($matches); //$matches is array with results


    array(1) {
      string(4) "0.07"
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  • dougouqin0763 dougouqin0763 5年前

    You can use a simple regex patter to isolate all numbers, including decimals if any after the equals sign.

    The below works if the string will have the same structure, meaning an = a space and then the number that you are after.

    = - matches the equals sign
    \s - matches the space character immediately after
    (\d*\.?\d*) - matches any number of digits followed by an optional period . and then any number of digits

    $str = '1 PLN = 0.07 Gold';
    print $matches[1];

    Will output


    This works regardless of what you have before the = sign.

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  • dsfphczao23473056 dsfphczao23473056 5年前

    If its always in this format 1 PLN = 0.07 Gold You can just

    $array = explode(" ", $string) with a Space and then get the required number with

    $number = $array[3]

    Try it out, let me know if it works

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