match a string with an array of values and return the key

Question

so i have an array with months:

$arr_month = array(
    1=>'january',
    2=>'february',
    3=>'march',
    4=>'april',
    5=>'may',
    6=>'june',
    7=>'july',
    8=>'august',
    9=>'september',
    10=>'october',
    11=>'november',
    12=>'december'
);

i also have strings (i know, they look weird...) that look like:

23 de january del 2003
12 de may de 1976
6 de february de 1987

What i want is to find and match the month in the string with the array and return the array key.

so:

23 de january del 2003 returns 1
12 de may de 1976 returns 5
6 de february de 1987 returns 2

and so on...

Any idea how?

Jeffory J. Beckers · Accepted Answer · 2012-09-26 00:03:56Z

1

This should work for you.

$dateString = "23 de january del 2003"; // as an example
$exploded = explode (" ", $dateString); // separate the string into an array of words
$month = $exploded[2]; // we want the 3rd word in the array;
$key = array_search ($month, $arr_month); // lets find the needle in the haystack
print $key;

yields 1.

See http://php.net/manual/en/function.array-search.php for more.

answered Sep 26, 2012 at 0:03

Jeffory J. Beckers

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RichardTheKiwi · Accepted Answer · 2012-09-25 23:56:21Z

1

$index = -1;
foreach ($arr_month as $k => $v) {
   if (strstr($mystring, $v)) {
      $index = $k;
      break;
   }
}
// exists with $index set

answered Sep 25, 2012 at 23:56

RichardTheKiwi

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Comments

Esben Tind · Accepted Answer · 2012-09-26 00:15:20Z

You should be able to do like this:

<?php

function getMonth($time){
    $matches = array();
    preg_match("/[0-9] de ([a-z]+) de/", $time, $matches);
    return date('n',strtotime($matches[1]));
}
?>

Then you don't even need your month array :D

EDIT if you for some reason want the array in there:

<?php
function getMonth($time){
    $matches = array();
    $pattern = "/[0-9] de ([a-z]+) de/";
    preg_match($pattern, $time, $matches);
    $month = $matches[1];
    $arr_month (
        1=>'january',
        2=>'february',
        3=>'march',
        4=>'april',
        5=>'may',
        6=>'june',
        7=>'july',
        8=>'august',
        9=>'september',
        10=>'october',
        11=>'november',
        12=>'december'
    );
    return in_array($month, $arr_month) ? array_search($month, $arr_month) : false;
}
?>

AlexP · Accepted Answer · 2012-09-26 00:17:39Z

1

You can do this with the optional search param with array keys:

$matchedKeys = array_keys($arr_month, "november");
var_dump($matchedKeys);
// array(1) { [0]=> int(11) }

As you can see this will return an array of all matching keys.

answered Sep 26, 2012 at 0:17

AlexP

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Comments

John Carter · Accepted Answer · 2012-09-26 00:32:31Z

If you're trying to parse a date, bear in mind that PHP can do that for you (though it depends on how arbitrary your input data is - this is obviously more reliable if you know the date format in advance).

eg:

print_r(date_parse("6 de february de 1987"));

gives:

Array
(
    [year] => 1987
    [month] => 2
    [day] => 
    [hour] => 
    [minute] => 
    [second] => 
    [fraction] => 
    [warning_count] => 2
    [warnings] => Array
        (
            [14] => Double timezone specification
            [22] => The parsed date was invalid
        )

    [error_count] => 2
    [errors] => Array
        (
            [0] => Unexpected character
            [2] => The timezone could not be found in the database
        )

    [is_localtime] => 1
    [zone_type] => 0
)

So it's given up on the day, but it did correctly identify the month and year.

Using the more modern DateTime api fails on your input (is that a mixture of French and English?)

But it does work for the following:

$d = new DateTime("6th february 1987", new DateTimeZone("UTC"));
print $d->format("Y-m-d H:i:s") . "\n";'

Gives:

1987-02-06 00:00:00

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