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Whats wrong with this code
Posted: Thu May 01, 2008 9:51 pm
by definewebsites
Hi there,
I am developing a content management system, and I have some details stored in a mysql database. I am trying to retrieve an image whose path is stored in the database.
I obtained the path using the following:
Code: Select all
$uploadDir = $_SERVER['DOCUMENT_ROOT'] . "/images/portfolio/";
and after retrieving the details from the database, I am trying to display it using the below as part of a table:
Code: Select all
echo "<td rowspan=\"3\"><img src=\" ".$content . "\"/></td>";
but the image is not showing. I would appreciate it if someone could point me in the right direction as I have been staring at this for the past 30 minutes without a clue..
Thanks
Re: Whats wrong with this code
Posted: Fri May 02, 2008 12:43 am
by Zoxive
Code: Select all
echo "<td rowspan=\"3\"><img src=\" " . $uploadDir . $content . "\"/></td>";
A bit difficult trying to understand your problem. Your current output of this would help, and some more information as well.
Re: Whats wrong with this code
Posted: Fri May 02, 2008 6:31 am
by aceconcepts
From your second segment of code you would simply be outputting $content.
What is the value of $content? If $content is simply the file name then Zoxive has helped you out already.
Re: Whats wrong with this code
Posted: Fri May 02, 2008 7:01 am
by definewebsites
Hi guys,
Thanks for the responses. Well, the "$content" contains the full path to the image, which looks like this:
/home/content/k/a/b/user/html/images/portfolio/myImage.jpg
Any suggestions?
Thanks
Re: Whats wrong with this code
Posted: Fri May 02, 2008 7:43 am
by highjo
Hey! don't know on what u are developing but what i do to reference my root is this
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$uploaddir = "./images/portfolio";
the dot means the root and if you change your server name or something has changed then there would not be much problem.
Somebody correct me if i'm wrong
Re: Whats wrong with this code
Posted: Fri May 02, 2008 8:05 am
by definewebsites
Hi,
I tried your method, but just got an error stating that no such directory exists, as the dot does not seem to take it to the document root of the server.
I should think that since I have the image path stored as the complete path on the server it should show the image when I use it as a source, but do not understand why it is not working.
Thanks for all the help..
Re: Whats wrong with this code
Posted: Fri May 02, 2008 8:07 am
by aceconcepts
Have you tried using the literal absolute path in the img src to see if your syntax is right?
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echo'<img src="/home/content/k/a/b/user/html/images/portfolio/myImage.jpg" />';
Re: Whats wrong with this code
Posted: Fri May 02, 2008 10:56 am
by remyx187
All in all this should work (ideally):
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$uploadDir = "/home/content/k/a/b/user/html/images/portfolio/";
echo "<td rowspan=\"3\"><img src=\"" . $uploadDir . $content . "\"/></td>";
a fusion of previous posts.
if its not that, double check your paths, make sure the images exist, and the $content variable acquires the right image names.
Re: Whats wrong with this code
Posted: Fri May 02, 2008 6:45 pm
by definewebsites
I'd like to thank all of you for your assistance.
Well, I did try using the absolute URL to load the image, but that did not work so I take it that the setup of the server my website is not compatible with loading content via that path, so I developed a workaround.
I simply just extracted a substring of the full path that contains the end bit: /images/portfolio/myImage.jpg , and appended the starting URL to it. Its now working perfectly.
I do appreciate all the assistance.
Cheerio
Re: Whats wrong with this code
Posted: Sat May 03, 2008 2:20 am
by Mordred
Err, people, that's not how HTML works. Back to first grade all of you, except the OP who got it right in the end
