accessing images already in database

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nika603
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accessing images already in database

Postby nika603 » May 10th, '09, 16:21

How do i access the image location already contained in my database? I downloaded a datafeed from a merchant and want to display the associated images that are contained in the database.

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Re: accessing images already in database

Postby administrator » May 16th, '09, 08:59

What do you meed already contained?
Can you please explain in more detail, how they contained?

nika603
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Re: accessing images already in database

Postby nika603 » May 17th, '09, 15:06

My database has a field called images which contains the location of the image files. For example "http://www.thehairstyler.com/images/celebrity/Celebrity_1753.jpg" is the data contained in the field called "images". I need to have the php datagrid refer to this field and show it as an image photo.

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Re: accessing images already in database

Postby administrator » May 22nd, '09, 14:59

Write in SELECT:

Code: Select all

CONCAT('<a href=\"', image_url,  '\"><img src=\"', image_url, '\" /></a>' ) as my_image


then use my_image as a regular field and define it as a label in View mode.


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