Creating a PHP array from space delimited text file

Question

I have a text file which is a listing of directories that I would like to turn into an array. I figured space delimiting will work but the number of spaces varies between each item and the spaces in the directory name would be a problem. I would like to parse the text into a PHP array.

The text file has a very rigid structure that looks like this:

04/17/2013  09:49 PM    <DIR>          This is directory 1 (1994)
03/11/2013  06:48 PM    <DIR>          Director 2 (1951)
04/15/2013  08:34 PM    <DIR>          This is going to be number 3 (2000)
08/17/2012  09:50 PM    <DIR>          Four (1998)
10/17/2011  05:12 PM    <DIR>          And lastly 5 (1986)

I only need to keep the folder date (not time), the complete name of the directory (as one entry) and the year in parenthesis. Thanks in advance!

chris-l · Accepted Answer · 2013-08-01 05:00:31Z

3

Sure, use preg_split:

<?php
$str = "04/17/2013  09:49 PM    <DIR>          This is directory 1 (1994)
03/11/2013  06:48 PM    <DIR>          Director 2 (1951)
04/15/2013  08:34 PM    <DIR>          This is going to be number 3 (2000)
08/17/2012  09:50 PM    <DIR>          Four (1998)
10/17/2011  05:12 PM    <DIR>          And lastly 5 (1986)";

function sp($x) {
    return preg_split("/\s\s+|\s*\((\d{4}).*\)/", $x,0,PREG_SPLIT_DELIM_CAPTURE);
}
$array = preg_split("/\n/", $str);
$processed = array_map('sp', $array);

print_r($processed);

This will create an array of arrays. Each line will become an array, containing an array for each item. For instance, $processed[0][3] will contain This is directory 1

Keep in mind this code assume that spaces working as division must be 2 or more; only 1 space is considered as part of the same field. (You'll probably need to hand hack that according to your needs)

Edit: I added the part to get the year as a separated element of the array. Now $processed[0][4] has 1994. (you don't need the (), right?)

See it working with this change here: http://codepad.org/in973ijV

edited Aug 1, 2013 at 5:00

answered Aug 1, 2013 at 1:36

chris-l

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1 Comment

Beatmstrj Over a year ago

This is great but I still need the (year) to be a separate array entry. I'm sure thats easy to modify though.

mannnq · Accepted Answer · 2013-08-01 01:51:23Z

0

Why you dont forget about this txt and use scandir?

http://php.net/manual/en/function.scandir.php

$mydir = "/home/folder/";
$scan = scandir($mydir);
$i = 2 /* bypass dot and 2dots dirs */;

while($i < count($scan)){
    echo $scan[$i];
    echo "<hr>";
    $i++;
}

answered Aug 1, 2013 at 1:51

mannnq

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1 Comment

Beatmstrj Over a year ago

Because the directory I need to scan is not on a server running php. The file is created locally then uploaded to a server which DOES have php. Otherwise that would be a rather easy solution!

Casimir et Hippolyte · Accepted Answer · 2013-08-01 02:01:47Z

0

The most simple (to read) pattern is:

$pattern = '~^(?<date>\S+).*<DIR>\s+(?<name>.*) \((?<year>\d{4})\)$~m';
preg_match_all($pattern, $subject, $matches, PREG_SET_ORDER);

foreach ($matches as $match) {
    printf("<br>date: %s, name: %s, year: %s",
           $match['date'], $match['name'], $match['year']);
}

But you can optimize a little being more explicit:

$pattern = '~^(?<date>\S++)'                         . '\s++(?:\S++\s++){3}'
         . '(?<name>(?>[^(]++|\((?!\d{4}\)\s*+$))+)' . '\s++\('
         . '(?<year>\d{4})'                          . '\)\s*+$~m';

edited Aug 1, 2013 at 2:01

answered Aug 1, 2013 at 1:53

Casimir et Hippolyte

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