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how to display layered jpegs using php??


dazzathedrummer

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Hi,

 

I have an events page on my site (gigs) and we use an iPhone style graphic depicting a tear off calendar to illustrate the dates: -

 

http://www.the-guards.org.uk/gigs.html

 

I'm wondering if I can use a script to select and layer three images to build the final calendar image - ie, a background image, with a month name image and a number image for the day layered on top. At the moment we hard code the whole list in html and create a new image for each date that we have a booking for.

I can then store 1 image for the background, 12 month images and 31 day number images and the date can be determined by using date functions on the date field in my database (which I can do - its just the graphic layering I'm unsure about).

 

?

 

 

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Thanks for that link, that works perfectly.

 

My next question is, what am I doing wrong here: -

 

I'm trying to set up an if statement to change the graphic image based on the results of a form.

So, for example, if the form posts 'Month = 1' and 'Day = 1' then show the graphics relevent to January 1st.

 

Im pretty sure i'll be able to figure out the file selection - I just cant seem to get any kind of if statement to work.

 

This is what I created to try and echo it out: -

 

www.the-guards.org.uk/private/calendartest does this: -

 

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"

"http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> 
<html xmlns="http://www.w3.org/1999/xhtml" lang="en"> 
<head> 
<title>Calendar test</title> 

<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> 
<style type="text/css"> 

#background {
position: absolute;
left: 100px;
top: 100px;
z-index: 1;
}

#layer_1 {
position: absolute;
left: 100px;
top: 100px;
z-index: 2;
}

#layer_2 {
position: absolute;
left: 100px;
top: 100px;
z-index: 3;
}
#selectform {
position: absolute;
left: 100px;
top: 200px;

}
</style>
</head> 
<body> 


<h1>Show Calendar pic test</h1> 

<div id="background"> 
<img src="calpics/background.png" width="79" height="79" alt="Background"> 
</div> 

<div id="layer_1"> 
<img src="calpics/25.png"width="79" height="79" alt="layer1"> 
</div> 

<div id="layer_2"> 
<img src="calpics/mar.png"width="79" height="79" alt="layer2"> 
</div> 


<div id="selectform">
<form action="calposttest.php" method="post"> 
<select name="month" id="month">
      <option selected="selected">1</option>
      <option>2</option>
      <option>3</option>
      <option>4</option>
      <option>5</option>
      <option>6</option>
      <option>7</option>
      <option>8</option>
      <option>9</option>
      <option>10</option>
      <option>11</option>
      <option>12</option>
        </select>Month
	<br>
<select name="day" id="day">
      <option selected="selected">1</option>
      <option>2</option>
      <option>3</option>
      <option>4</option>
      <option>5</option>
      <option>6</option>
      <option>7</option>
      <option>8</option>
      <option>9</option>
      <option>10</option>
      <option>11</option>
      <option>12</option>
      <option>13</option>
      <option>14</option>
      <option>15</option>
      <option>16</option>
      <option>17</option>
      <option>18</option>
      <option>19</option>
      <option>20</option>
      <option>21</option>
      <option>22</option>
      <option>23</option>
      <option>24</option>
      <option>25</option>
      <option>26</option>
      <option>27</option>
      <option>28</option>
      <option>29</option>
      <option>30</option>
      <option>31</option>
    </select>Day<br>
<input type="submit" name="submit" value="submit">	
</form>
</div> 
</body> 
</html>

 

and calposttest.php does this: -

<?php
if ($_POST["month"]=="1")
{
echo $_POST["month"]}
else echo "something";


?>

 

All i'm trying to do at the moment is get the if statement to work so I can adapt it to echo the filenames of the different calendar images - once I get that going i'll then conect it to a db so the image is generated from a date field.

 

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Your code is producing a fatal syntax error due to a missing semi-colon -

Parse error: syntax error, unexpected '}', expecting ',' or ';' in ... ...\calposttest.php on line 4

 

You should be developing and debugging php code on a system with error_reporting set to E_ALL and display_errors set to ON in your master php.ini so that php will help you. You will save a TON of time.

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Your code is producing a fatal syntax error due to a missing semi-colon -

Parse error: syntax error, unexpected '}', expecting ',' or ';' in ... ...\calposttest.php on line 4

 

You should be developing and debugging php code on a system with error_reporting set to E_ALL and display_errors set to ON in your master php.ini so that php will help you. You will save a TON of time.

 

ah well thats highlighted a problem - my hosting provider upgraded their servers over the weekend - looks like I'm not getting error messages any more - i'll get in touch with them as I dont think I can access the php.ini myself.

 

thanks,

 

Darren

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You should be developing php code on a local development system. Constantly uploading code changes to a live server just to test what it does wastes a huge amount of development time. Code that is not fully tested often has security holes that would allow a hacker to exploit your live server. Also, a live server running with display_errors ON allows hackers to gain information about your account and server when they deliberately trigger errors.

 

If necessary, you can set the error_reporting/display_errors settings in a local php.ini (when php is running as a CGI application) or in a .htaccess file (when php is running as an Apache Module.)

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