Displaying image from database

[Confuzzled]

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.

 

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.

 

<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?

 

Many thanks for your time!

[taquitosensei]

try viewing the source and see if it shows the correct value for the image, then try to browse to the location the source shows to see if it comes up.

[Calver]

I can get this to work if I change this line...

 

echo "<td width=15%>"."<img width=100% src='$dir/$image /'></td>";

 

... to this...

 

echo "<td width=15%>"."<img width=100% src='$dir/$image' /></td>";

[mrMarcus]

echo '<td width="15%"><img width="100%" src="'.$dir/$image.'" /></td>';

 

that'll work .. you have the closing / in with the image name .. make sure you use quotations in your tags or it will never validate.

[Confuzzled]

Ok so, the example given by mrMarcus pulls me an error saying "Division by 0 in (filepath)"

 

The example given by Calver doesn't appear to affect the page as it still remains the same without the image showing.

 

If you both say they work, is it something to do with my browser or something?

 

Many thanks for the oh so quick replies  :D

 

Quick EDIT:  Both examples show the path as correct in the source as suggested by the first reply so i'm even more confused 

[mrMarcus]

my bad, this should do it:

 

echo '<td width="15%"><img width="100%" src="'.$dir.'/'.$image.'" /></td>';

 

my previous code was dividing $dir by $image, which is your error.

[Confuzzled]

my bad, this should do it:

 

echo '<td width="15%"><img width="100%" src="'.$dir.'/'.$image.'" /></td>';

 

my previous code was dividing $dir by $image, which is your error.

 

I'm sure it was, however, placing that code you just supplied in leaves the page with no image shown again

[Calver]

Ok so, the example given by mrMarcus pulls me an error saying "Division by 0 in (filepath)"

 

The example given by Calver doesn't appear to affect the page as it still remains the same without the image showing.

 

If you both say they work, is it something to do with my browser or something?

 

Many thanks for the oh so quick replies  :D

 

Quick EDIT:  Both examples show the path as correct in the source as suggested by the first reply so i'm even more confused 

 

Sorry, I also took the '/' out of this line...

$dir = 'image';

 

I didn't use a database for testing, just a local image.

 

[mrMarcus]

Ok so, the example given by mrMarcus pulls me an error saying "Division by 0 in (filepath)"

 

The example given by Calver doesn't appear to affect the page as it still remains the same without the image showing.

 

If you both say they work, is it something to do with my browser or something?

 

Many thanks for the oh so quick replies  :D

 

Quick EDIT:  Both examples show the path as correct in the source as suggested by the first reply so i'm even more confused 

 

Sorry, I also took the '/' out of this line...

$dir = 'image';

 

I didn't use a database for testing, just a local image.

 

that code will show an image if there is an image to show .. check your source code and see what's being displayed in the <img> tag, and match it to what it should be.

 

also, turn error_reporting on:

error_reporting(1);

at the top of your script.

[Confuzzled]

Ok so, the example given by mrMarcus pulls me an error saying "Division by 0 in (filepath)"

 

The example given by Calver doesn't appear to affect the page as it still remains the same without the image showing.

 

If you both say they work, is it something to do with my browser or something?

 

Many thanks for the oh so quick replies  :D

 

Quick EDIT:  Both examples show the path as correct in the source as suggested by the first reply so i'm even more confused 

 

Sorry, I also took the '/' out of this line...

$dir = 'image';

 

I didn't use a database for testing, just a local image.

 

that code will show an image if there is an image to show .. check your source code and see what's being displayed in the <img> tag, and match it to what it should be.

 

also, turn error_reporting on:

error_reporting(1);

at the top of your script.

 

This worked! Thankyou very much!

 

I will also turn on error reporting as you suggest. Cheers guys. What a great forum! I love it when people reply so quickly!