Question

PHP , function question..

Asked by: www_puertoricoautoforo_com

I would like it to return "1992"  , for some reason it return "1919"


Basically this is supposed to extract the begging year of a span. like from 1992-1995 it should return "1992"  

Also from "92-95" I would like it to return "1992"

<?php
function from_year($input)
        {
        preg_match_all("(\d{2}|\d{4}\-\d{2}|\d{4})", $input, $matches);
        $input=explode('-', $matches[0][0]);
        $input=$input[0];
        $input = strtolower($input);
        $input = preg_replace('~\s{2,}~', ' ', $input);
        $input = trim($input);
        switch ($input)
                {
                    case ($input > 100):
                    {}// it's already a 4 digit year
                    break; 
                    case ($input <=99):
                    { $input = '19'.$input; }
                    break; 
                    case ($input > 30):
                    { $input = '20'.$input; }
                    break;
                }
        return $input;
        } 
echo from_year("doweh weijwe fwej oi j 1992-1995 sdfijsdfj sdioj sdfije"); 
?>

                                  
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Asked On
2009-11-07 at 11:29:38ID24880781
Tags

php

,

function

Topic

PHP Scripting Language

Participating Experts
3
Points
500
Comments
7

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Answers

 

by: Michael701Posted on 2009-11-07 at 12:25:53ID: 25767873

change to this

        preg_match_all("(\d{4}|\d{2}\-\d{4}|\d{2})", $input, $matches);

                                              
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by: Michael701Posted on 2009-11-07 at 12:30:41ID: 25767903

that was fun. btw, here's what happened

with the old way it returned

Array ( [0] => Array ( [0] => 19 [1] => 92 [2] => 19 [3] => 95 ) )

----------------

btw, you have another issue (to be found later)

echo from_year("doweh weijwe fwej oi j 07-09 sdfijsdfj sdioj sdfije");

returns 1907

---

one last thing case = 100 what should that return?

 

by: felixjetPosted on 2009-11-07 at 12:43:21ID: 25767955

it should be:

case (($input > 30) ($input <= 99)):
 { $input = '19'.$input; }
 break; 
case (($input >= 0) && ($input <= 30)):
 { $input = '20'.$input; }
 break;
31 up to 99 will be 1931 to 1999
00 up to 30 will be 2000 to 2030

                                              
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by: felixjetPosted on 2009-11-07 at 12:44:26ID: 25767957

forgot one thing, line 1 is this:

case (($input > 30) && ($input <= 99)):

:P

 

by: Michael701Posted on 2009-11-07 at 12:51:51ID: 25767981

hey I wanted him to find that one. lol :)

and think about it, by the time you have that case statement you might as well just have if statments. it reduces down to this

if ($input <= 30)
  $input += 2000;
if ($input <= 99)
  $input += 1900;

 

by: silver_surfer_44Posted on 2009-11-07 at 13:01:50ID: 25768023

Your regexp pattern should be changed as pointed by Michael, Otherwise it check for the matches in first expression in OR and not check for other regexp patterns.
So change regexp to
preg_match_all("(\d{4}|\d{2}\-\d{4}|\d{2})", $input, $matches);

 

by: www_puertoricoautoforo_comPosted on 2009-11-07 at 15:25:47ID: 31651440

woo hoo it works now!!!

20120131-EE-VQP-002

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