Photo display

rkeppert

I have a page with a table on it and need to display an image in the table...

the image's name is stored in the database... I need to call that name from the database and load the image....

the directory that the images are stored in is uploads/images/

the image name is stored in the db table under 'photo'

Here's the table code that I am using...


   echo '<tr>';
   echo '<td>','</td>';
   echo '<td colspan="3">','</td>';     #///// where the image should display /////#
   echo '<td>','</td>';
   echo '<td>','</td>';
   echo '<td>','</td>';
   echo '</tr>';

And I tried the following code, with no luck:


   echo '<tr>';
   echo '<td>','</td>';
   echo '<td colspan="3">','<img src="uploads/photos/", .$row['photo'];', '</td>';
   echo '<td>','</td>';
   echo '<td>','</td>';
   echo '<td>','</td>';
   echo '</tr>';

Any ideas?

Thanks

Rob

halojoy

hey
rkeppert

can it be as simple as:
uploads/images/ ... is not same as: uploads/photos/

otherwise,
give some more info
with no luck ... means what ...

any php errors?
what does 'View source' show?

this is what I would use:

   echo '<td colspan="3"><img src="./uploads/images/' 
            . $row['photo'] . '"></td>';

thanks
halojoy

rkeppert

That did it perfectly.

Thanks again!

Weedpacket

And for future reference:

echo '<td colspan="3">','<img src="uploads/photos/", .$row['photo'];', '</td>';

You didn't put a closing quote on

'<img src="uploads/photos/"

, a "." that shouldn't be there, a ";" instead of a ',' after $row['photo'], and an extra "'" after that.

And you didn't close the <img> src attribute's value or the tag, either.

Try and pay attention to what you're writing, because neither PHP nor HTML are going to care if you've forgotten their rules of syntax.

echo '<td colspan="3">','<img src="uploads/photos/', $row['photo'], '" /></td>';