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Answer 1

+4 A:

you could use the date() function to check for the current month/day/year and some simple if/else constructs to show different images.

ZeissS 2010-02-07 16:57:44

Thanks for your suggestion. If/else wasn't my preferred way of doing it initially, but in the end seemed the simplest way to accomplish it. Probably the way I've done it isn't the most efficient code-wise, but it works for now. Thanks again.

NeonBlue Bliss 2010-02-12 19:30:04

Answer 2

+1 A:

The best way to do it would probably be just to use GD to display it.

<?php

// Create a 75*15 image
$im = imagecreate(75, 15);

// White background and black text
$bg = imagecolorallocate($im, 255, 255, 255);
$textcolor = imagecolorallocate($im, 0, 0, 0);

// Write the date at the top left, offset by 2px to the right
imagestring($im, 5, 2, 0, date("m/d/y"), $textcolor);

// Output the image
header('Content-type: image/png');

imagepng($im);
imagedestroy($im);

?>

Xorlev 2010-02-07 17:01:48

Thanks - that's really useful. I've never really looked into GD and what it can do, but that's a nice example

NeonBlue Bliss 2010-02-07 18:27:10

You can also use actual fonts...just upload arial.ttf and such. ImageMagick can do even more advanced compositing.

Xorlev 2010-02-07 21:32:02

Answer 3

+1 A:

switch(date('Y-m-d')) {
  // multiple dates with same image
  case '2010-02-15':
  case '2010-02-07':
    print '<img src="/path/to/image.jpg" alt="" />';
    break;
  // really long ranges don't work all that well in this solution
  case '2010-03-01':
  case '2010-03-02':
  case '2010-03-03':
  case '2010-03-04':
  case '2010-03-05':
  case '2010-03-06':
  case '2010-03-07':
  case '2010-03-08':
  case '2010-03-09':
  case '2010-03-10':
    print '<img src="/path/to/image.jpg" alt="" />';
    break;
  // specific date image
  case '2010-12-25':
    print '<img src="/path/to/christmas.jpg" alt="" />';
    break;
  // fallback image
  default:
    print '<img src="/path/to/default.jpg" alt="" />';
    break;
}

If you're doing a lot of largeish date ranges, this'll break down, but it works well for a few smallish ranges and specific dates.

ceejayoz 2010-02-07 17:10:07

Thanks - that looks very similar to what I remember of the script I had. The date ranges will be a month or maybe two months on odd occasions at most - the rest will either be weeks or days. There's just a couple of potential problems I'm wondering about. The first is whether it'd worth without the year (i.e. with just a month and day), and also what would happen if there was a day say with an image that was within a date range that was specified with a different image. I want to use it to display awareness ribbons for different cancers, and there may also be a day within a date range.

NeonBlue Bliss 2010-02-07 21:28:26

Yep, you could skip the year entirely if that suits your usage case.In the case of a date matching multiple conditions, the first one specified will be the one that shows up (due to the `break` statement). Put the most specific dates first and the more general ranges towards the bottom and you'll be in good shape.

ceejayoz 2010-02-08 14:23:31

NeonBlue Bliss 2010-02-08 22:59:25

Really long ranges don't work that well in this solution - it's more designed for short ranges or individual dates. Just updated with an example. If you're dealing with a lot of ranges, you may be better off with a database to store your criteria - they handle selecting this sort of data much better.

ceejayoz 2010-02-09 03:47:06

Answer 4

+1 A:

It sounds like you were copy/pasting your code before, but this is really a good snippet to learn on if you're up for it. The basic construct you're looking for is:

$today = getdate();
$day = $today['wday'];
if ($day == 1 OR $day == 2 OR $day == 3){
    echo "<img src='whatever image you want'>";
}
elseif ($day == 4 OR $day ==5){
    echo "<img src='another image' />";
}
else { echo "<img src='default image' />";}

Basically you find the date, and using IF/ELSE loops, determine if today is in one range or another. IF it is, echo out the image you want to show.

In this example, the variable $day is set to a number 1-7. 1==Monday, 2==Tuesday, 3==Wednesday... If it's Monday Tuesday or Wednesday, it shows one image, (the if ($day == 1 OR $day == 2 OR $day == 3) line), Thursday/Friday shows another, and otherwise it shows the "default" image, which shows on the weekend. Obviously this isn't the EXACT case of days/dates you want, but something like this construct will help you out.

Check out the PHP Date reference for help with picking just what dates you want to compare.

Alex Mcp 2010-02-07 17:10:34

Thanks Alex - that's a really useful snippet to know.

NeonBlue Bliss 2010-02-07 18:23:36

Answer 5

+1 A:

This is my solution, I guest maybe you do not care about which year, so this solution just consider month and date:

function image_of_date($default, $options) {
    $today = date('md');
    foreach ($options as $item) {
        $src = $item[0];
        $begin = $item[1];
        $end = (3===count($item))? $item[2] : $begin;
        $begin = date('md', strtotime($begin));
        $end = date('md', strtotime($end));
        if ($today >= $begin && $today <= $end) {
            return $src;
        }
    }
    return $default;
}


echo image_of_date('default.png', array(
    array('jan-01.png', 'Jan 1'),
    array('feb.png', 'Feb 1', 'Feb 29')
));

XUE Can 2010-02-07 19:28:03

Thanks Xue - that solves the problem of just using the month and day, so I can just specify the month and day though. As in my comment below though, I'm wondering what would happen if there was a day say with an image that was within a date range that was specified with a different image. I want to use it to display awareness ribbons for different cancers, and there may also be a day within a date range - i.e. there may be more than one image for a particular date

NeonBlue Bliss 2010-02-07 23:44:46

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Display image based on date in PHP

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