Really simple ??

TheBeginner

Hi all of you,

Are doing my first website in php and have someproblems.

My structure in the root folder is following:

index.php

header.php

work\

family\

music\

film\

were work, family, music and film are FOLDERS.

In my header.php a I have a image. The things is that I want to change the image depending of which file I'm showing.
For example:
- If the file is from the work folder I want the picture to show
my office,
- If the file is from the family folder I want the picture to show
my family....
............ and so on.....

So I have so have some code like this:

php:

if (something)
{
print("<img src=\"work.gif\" width=\"15\" height=\"15\">");
}
elseif (something)
{
print("<img src=\"music.gif\" width=\"15\" height=\"15\">");
}
elseif (something)
{
print("<img src=\"film.gif\" width=\"15\" height=\"15\">");
}

The question is: WHAT should I put in "something" ???
It has to recognize in which folder the file is.....how to do that?

/TheBeginner

drawmack

actually what I would do is something like this

in header.php

<?php
//make sure header_image conatins something
if(!isset($header_image)) {
    $header_image = 'default.gif';
} //end if

//display the header_image
echo "<img src=\"" . $header_image . "\" width=\"15\" height=\"15\">\n";
?>

in other files

<?php
$header_image = 'thisimage.gif';
include('header.php');
?>

groovything

you can feed the variable through the querystring like:

the link to the page would be:

<a href="index.php?img=work">referring link</a>

then in the body where you want the image to display you just type:

<img src="<? echo $img ?>.gif" width="15" height="15">

(you could print the entire line with php, but I prefer to plug in the <? echo?> because it makes the page layout viewable in dreamweaver in page layout view.)

also, I would get rid of width="15' and height="15" in the image tag and make sure that the images are the right size. That way if you ever need to change the images, you don't have to go back and redefine dimensions. that can be a huge pain when your page has a lot of images.

TheBeginner

Thanks for the answer, but your solution is not that good.

If I refer to some files under my WORK folder, I then have to add to all my links "myfile.php?img=work" in all the links. That I knew already!

It can be a hell of a work to add that to all links in files, and what happen if I forget - no picture.....

I want a solution that php-engine recognize that a file that is running is comming from the WORK folder and put in the right image......or it recodnize it from the URL........

TheBeginner

Originally posted by drawmack
actually what I would do is something like this

in header.php

<?php
//make sure header_image conatins something
if(!isset($header_image)) {
    $header_image = 'default.gif';
} //end if

//display the header_image
echo "<img src=\"" . $header_image . "\" width=\"15\" height=\"15\">\n";
?>

in other files

<?php
$header_image = 'thisimage.gif';
include('header.php');
?>

[/B]

It is not a optimal solution but it is a pretty good one!
Thanks.

groovything

Originally posted by TheBeginner
Thanks for the answer, but your solution is not that good.

If I refer to some files under my WORK folder, I then have to add to all my links "myfile.php?img=work" in all the links. That I knew already!

It can be a hell of a work to add that to all links in files, and what happen if I forget - no picture.....

I want a solution that php-engine recognize that a file that is running is comming from the WORK folder and put in the right image......or it recodnize it from the URL........

sorry... I was assuming that you were not hard coding your links, but rather generating your content using a db call like:

<? $db = mysql_connect("localhost","userName","password");

mysql_select_db (dbname); 
$result = mysql_query ("select * from table");

while ($row = mysql_fetch_array($result))
{
$STRfile = $row["filename"];
$STRdir = $row["category"];
?>
<a href="<? echo $STRfile ?>?img=<? echo $STRdir ?>">
<? } ?>

TheBeginner

Yes....I'm hard coding some of the links and some not, but still if I'm coding let's say 10 and also have some FORM in my code than I have to add this variable to every FORM in the code.
That's why I put a post here, since I couldn't find a better solution myself.

But thanks anyway. Every attempt to communication gives our brains a kick of new ideas - that's good!

groovything

Originally posted by TheBeginner
But thanks anyway. Every attempt to communication gives our brains a kick of new ideas - that's good!

yep I'm still trying to refine my code all the time. I look at stuff I wrote a couple months ago and think "what was I thinking!!!?"

webblackes

You can try this code :


if (strpos($_SERVER["SCRIPT_FILENAME"],'work')) 
{ 
print("<img src=\"work.gif\" width=\"15\" height=\"15\">"); 
} 
elseif (strpos($_SERVER["SCRIPT_FILENAME"],'music')) 
{ 
print("<img src=\"music.gif\" width=\"15\" height=\"15\">"); 
} 
elseif (strpos($_SERVER["SCRIPT_FILENAME"],'film')) 
{ 
print("<img src=\"film.gif\" width=\"15\" height=\"15\">"); 
}

😃