从字符串中提取数据

Question

I have a string and want to extract data from it.

$str = "Online (UVD) - 154,842 - Last Updated: Nov 23 2015 02:24 PM";

I want this 154,842 extract and this 2015 I've successfully extracted the first part. with this method

trim(str_replace("Online (UVD) - ", "", str_replace(",", "", substr_replace($str, "", strpos($str, " - Last Updated"))), $str))

Now, I'm unsure how to extract the other one. Data can vary for instance,

$str = "Online (UVD) - 1123123 - Last Updated: Nov 23 2015 02:24 PM";
$str = "Online (UVD) - 12 - Last Updated: Nov 23 2015 02:24 PM";
$str = "Online (UVD) - 1546546 - Last Updated: Nov 23 2015 02:24 PM";
$str = "Online (UVD) - 3525252525 - Last Updated: Nov 23 2015 02:24 PM";

Is there a better method to extract?/

Answer 1

If the strings will always have the same number of values perhaps explode and then using specific array positions would work for you.

$str = "Online (UVD) - 154,842 - Last Updated: Nov 23 2015 02:24 PM";
$pieces = explode(' ',$str);
echo 'Value is ' . $pieces[3] . ' and the year is ' . $pieces[9];

Answer 2

You can do it without using regex if all the words in the string are in same order that you provided. Let's try with explode() -

<?php
$str = "Online (UVD) - 1123123 - Last Updated: Nov 23 2015 02:24 PM";
$str = "Online (UVD) - 12 - Last Updated: Nov 23 2015 02:24 PM";
$str = "Online (UVD) - 1546546 - Last Updated: Nov 23 2015 02:24 PM";
$str = "Online (UVD) - 3525252525 - Last Updated: Nov 23 2015 02:24 PM";

$digit = explode(' ',$str);
echo trim($digit[3]); // returns digits
echo trim($digit[9]); // returns date
?>

DEMO: https://3v4l.org/ttBDG

Answer 3

I know this is answered but I think on also providing a regex solution for this:

To extract your 1st group , you can use bellow regex:

preg_match('/.-.(\d+).-/', $str, $numExtracted);

if (!empty($numExtracted)) {
    echo $numExtracted[1].PHP_EOL;   
}

To extract your Year :

preg_match('/(\w\w\w).(\d\d).(\d\d\d\d)/', $str, $year, PREG_OFFSET_CAPTURE);
$year = $year[3][0];
echo $year.PHP_EOL;

This worked on all of the below trials:

Online (UVD) - 1123123 - Last Updated: Nov 23 2015 02:24 PM
Online (UVD) - 12 - Last Updated: Nov 23 2015 02:24 PM
Online (UVD) oi oi    -            1546546 - Last Updated: Nov 23 2015 02:24 PM
Online -sdtgstg346fg - (UVD) - 3525252525 - Last Updated:             Nov 23 2015 02:24 PM

You can check the working code here

As per you comment question, you can enhance your regex to consider such cases:

.-.(\d+)?[\,\#\!\?\$\£\;\:]*(\d+)?.-

It will match all of the above plus this cases:

Online (UVD) - 1123,123 - Last Updated: Nov 23 2015 02:24 PM
Online (UVD) - 1123#!,123 - Last Updated: Nov 23 2015 02:24 PM

But I think there is a time you need to consider if you want to have a hold on the information you received or just consider it corrupt.

You can even introduce cycles to parse to every single case scenario but if I am expecting a number and suddenly the regex that triggers a match is for something like 1A2B3C4G5D8D2F I will discard it as it goes far from what I initially expected. But it all depends from where you receive your information, how likely is it to change, etc :)

Still, I think regex will make you happier and assert far more possibilities

PS: For the special cases introduced, because the number is interrupted by special chars (or even words if you consider them) it now interprets and 2 numbers.