CODEWITHSUNDEEP
php
ignaciotr
ignaciotr

Reputation: 3

Caching the output files from a php script not working as expected

I have a php script that sends different images per day. I want that images be cached during the day, but my script is not working as expected because in every request returns 200 OK .

The code is: 

<?php

$uno = '1.jpg';
$dos = '2.jpg';
$tres = '3.jpg';
$cuatro = '4.jpg';
$cinco = '5.jpg';
$seis = '6.jpg';
$siete = '7.jpg';

$today=date(l); 

header('Content-Disposition: inline');
header('Content-Type: image/jpg');
header("Content-Transfer-Encoding: Binary");

$expire=60*60*24*1; // seconds, minutes, hours, days
header('Pragma: public');
header('Cache-Control: maxage='.$expire);
header('Expires: ' . gmdate('D, d M Y H:i:s', time()+$expire) . ' GMT');
header('Last-Modified: ' . gmdate('D, d M Y H:i:s') . ' GMT');


// Find what today is? using date function
if($today==Monday){
readfile($uno);
exit;
}

elseif($today==Tuesday){
readfile($dos);
exit;
}

elseif($today==Wednesday){
readfile($tres);
exit;
}

elseif($today==Thursday){
readfile($cuatro);
exit;
}

elseif($today==Friday){
readfile($cinco);
exit;
}

elseif($today==Saturday){
readfile($seis);
exit;
}

elseif($today==Sunday){
readfile($siete);
exit;
}

?>

What's wrong with this?

UPDATE:

I've just try refactoring the code as WebnetMobile.com says, but seems that caching problem keeps alive because it still returns 200 OK.

New code is: 

<?php

$today=date("l"); 

header('Content-Disposition: inline');
header('Content-Type: image/jpg');
header("Content-Transfer-Encoding: Binary");

$expire=60*60*24*1;// seconds, minutes, hours, days
header('Pragma: public');
header('Cache-Control: maxage='.$expire);
header('Expires: ' . gmdate('D, d M Y H:i:s', time()+$expire) . ' GMT');
header('Last-Modified: ' . gmdate('D, d M Y H:i:s') . ' GMT');


// Find what today is? using date function

$map = array( 'Monday' => '1.jpg',
              'Tuesday' => '2.jpg',
              'Wednesday' => '3.jpg',
              'Thursday' => '4.jpg',
              'Friday' => '5.jpg',
              'Saturday' => '6.jpg',
              'Sunday' => '7.jpg'
             );

readfile( $map[ $today ] );

?>

Upvotes: 0

Views: 203

Answers (4)

FrancescoMM
FrancescoMM

Reputation: 2960

I think you should address this from the PHP file generating the HTML, not a file retrieving the image. If you want to load an image on Sunday, another on Monday,.. you can use something like this on the calling page.

echo('<img src="'.date('w').'.jpeg" />');

(or use any other method to find the correct filename shown in this page, like date('l')+array)

This way you let the browser/server handle the caching of real files and display the correct one for the day.

Or

<!other HTML here...>
<img src="<? echo('.date('w').'.jpeg');?>" />
<!other HTML here...>

if file is "mostly HTML".

Caching is a bad idea if you don't want to see Monday image on Tuesday, unless that's what you want, of course.

(if you really want to cache them, then $expires should be= to number of seconds till next midnight)

Upvotes: 0

moonwave99
moonwave99

Reputation: 22820

A brief comment expanding Mark answer above - please refactor your code as:

header('Content-Disposition: inline');
header('Content-Type: image/jpg');
header("Content-Transfer-Encoding: Binary");

$expire=60*60*24*1; // seconds, minutes, hours, days
header('Pragma: public');
header('Cache-Control: maxage='.$expire);
header('Expires: ' . gmdate('D, d M Y H:i:s', time()+$expire) . ' GMT');
header('Last-Modified: ' . gmdate('D, d M Y H:i:s') . ' GMT');

readfile( date('w') . '.jpeg' );
exit;

after renaming your sunday/domingo file to 0.jpg - taking advantage of date('w').

Upvotes: 1

Marcin Orlowski
Marcin Orlowski

Reputation: 75635

You need to compare with strings, otherwise you try to compare with nonexisting constant. So it should be 

if( $today == 'Monday' ) {
   // do someting...
}

but even better, instead of using bunch of if/elseif, you should use switch/case

switch( $today ) {
   case 'Monday':
       // do someting...
       break;
}

and best, you sould do this smarter way if all you need to do is to just pick up right image:

$map = array( 'Monday' => '1.jpg',
              'Tuesday' => '2.jpg',
              ...
             );

readfile( $map[ $today ] );

Upvotes: 1

Mark Baker
Mark Baker

Reputation: 212522

Monday should be a string, otherwise PHP will treat it as a constant, discover that the constant doesn't exist, issue a warning, and then decide you meant it to be a string.

if($today=="Monday"){

Same applies to all the other weekday names

The warning message is probably messing up the output

Upvotes: 1

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