Displaying image from database


Posted on 16th Feb 2014 07:03 pm by admin

Hi,

I've got a site where that's got a database behind it. Currently it has loads of items in rows that all have different pictures. There is a field called "Image" that has the name of the image ie. example.jpg.

Did you know?Explore Trending and Topic pages for more stories like this.
Then I have a folder on my webserver called image where the pictures are stored with the same file names and extensions.

Now, I want to display the correct image for the item that the page is showing. I have this code but it's just showing the usual box with a red X in the top left.

I should also point out that the 'Item' line is working correctly and displaying the name of the item.

Quote<table border='1' align="left" width="100%">
<?php
while($row = mysql_fetch_array( $result )) {
$dir = '/image';
$image = $row['Image'];
echo('<b><h1>'.$row['Item'].'</h1></b><br />');

echo "<tr>";
echo "<td width=15%>"."<img width=100% src='$dir/$image /'></td>";
echo "<td width=85% align=center>Description</td>";
echo "</tr>";
}
?>
</table>
Any ideas why this isn't working?
No comments posted yet

Your Answer:

Login to answer
104 Like 35 Dislike
Previous forums Next forums
Other forums

Java API in PHP?
I have an application that we use internally here at the office.

The software company provide

FAGL_FC_TRANSLATION FAS52 New GL ECC 6.0
Hello,

The new program for Translating GL Balances (FAS 52) gives the option to use diffe

Do While statement
hi guys,

This may sound trivial but im new to php and as part of an assignmenti have to const

mysql query with single quotes in a variable
$sitedetails = "INSERT INTO vars (address, sitename, description, ownername, theme) VALUES ('$u

fwrite error
Hi All,
Does anyone know what is causing the error in this code?

Code: <?
$error

Solution to the FindControl problem
I have seen may posts about having problems with the FindControl method. Most seem to come about bec

Custom list order
Hi there,

I have checked this tutorial and it's great till the point where I want to display

php calculate
this code echoes correctly the sum but the inserted result is 0??
Code: <?php
$TotalNum

Why is my row count 0?
Here's the MySQL query i'm running. It basically pulls data from 2 tables based on some data passed.

I have a parse error in this query help..
Code: $query1="INSERT INTO `rating` (`item_name`, `rating`, `ip_address`, `date_rated`) VALUES

Sign up to write
Sign up now if you have flare of writing..
Login   |   Register
Follow Us
Indyaspeak @ Facebook Indyaspeak @ Twitter Indyaspeak @ Pinterest RSS



Play Free Quiz and Win Cash