ferdri3
-
Posts
5 -
Joined
-
Last visited
Never
Posts posted by ferdri3
-
-
The PHP will be gone because it is processed before it hits your browser.
Anyway, in your code you are assigning the image path to a variable but arent doing anything with it
switch (basename($_SERVER['PHP_SELF'])) { case 'test1.php': $image = 'images/image1.png'; break; case 'test2.php': $image = 'images/image2.png'; break; } echo '<img src="', $image , '">';The echo made it work!
HUGE thanks to all of you!
-
images/image1.png
images/image2.png
tried:
../images/image1.png
../images/image2.png
no difference
When I view source in chrome, all php code is gone (even the "<?php include" part).. I don't know if that's normal.
-
Thanks for the fast replies!
I've set up a test server:
www.meesterferdi.net/test1.php
www.meesterferdi.net/test2.php
This is what I have now:
test1.php
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta content="text/html; charset=utf-8" http-equiv="Content-Type" /> <!-- START title & keywords & description --> <title>Test Homepage</title> <meta content="keywords" name="keywords" /> <meta content="testing page" name="description" /> <!-- !END! title & keywords & description --> <!-- START CSS stylesheetlinks --> <link href="css/test.css" rel="stylesheet" type="text/css" /> <!-- !END! CSS stylesheetlinks --> </head> <body style="margin: 0"> <?php include("menu.php"); ?> </body> </html>test2.php looks the same.
menu.php
<div id="menu"> <ul> <li> <a href="menu1.php" title="menu1">menu1</a> </li> <li> <a href="menu2.php" title="menu2">menu2</a> </li> <li> <a href="menu3.php" title="menu3">menu3</a> </li> <li> <a href="menu4.php" title="menu4">menu4</a> </li> <li> <a href="test1.php" title="test1">test1</a> </li> <li> <a href="test2.php" title="test2">test2</a> </li> </ul> </div> <div id="title_img"> <?php switch (basename($_SERVER['PHP_SELF'])) { case 'test1.php': $image = 'images/image1.png'; break; case 'test2.php': $image = 'images/image2.png'; break; } ?> </div>test.css
/* START menu style */ ul { list-style: none; margin: 0; padding: 0; } img { border: none; } #menu { position: relative; float: left; width: 200px; margin: 10px; } #menu li a { height: 32px; voice-family: "\"}\""; voice-family: inherit; height: 24px; text-decoration: none; } #menu li a:link, #menu li a:visited { font-family:Arial, Helvetica, sans-serif; color: #000; display: block; background: url(../images/include/menu.gif); padding: 8px 0 0 30px; } #menu li a:hover { color: #2f5bb7; background: url(../images/include/menu.gif) 0 -32px; padding: 8px 0 0 30px; } /* !END! menu style */ #title_img { position: relative; width: 670px; height: 192px; float: left; }The menu displays just fine, but the images dont appear.
-
Hello person who reads this now
I work a lot with <?php include("..."); ?> and I never thought beyond the html I put in there.
In 1 of the include files I have the set-up of a simple vertical html/css menu and on the right next to it
I have an image. However, not every page on my website needs the same image and I probably will change
a lot of them frequently.
Now I was wondering if (and how) I could include a piece of PHP code which does the following:
If the web-url is "page1.php" show "image1.png"
If the web-url is "page2.php" show "image2.png"
If the web-url is "page3.php" show "image3.png"
etc.
This way I only have to change the "imageX.png" in 1 file.
I hope you can help me.
Best regards and thanks in advance!
Ferdri3
<?php include(" Question
in PHP Coding Help
Posted
for if anyone wants to know, here is the text version:
<?php switch (basename($_SERVER['PHP_SELF'])) { case 'index.php': $txt = "Line of Text 1"; break; case 'anotherpage.php': $txt = "Line of Text 2"; break; } echo '<h2>', $txt ,'</h2>'; ?>If you dont want the text to be a header,
replace:
with: