Tables, Images, Drop-Downs

Mykl

I've just finished codecademy's PHP tutorial so I've covered loops, arrays, functions, objects and stuff like that but I still can't do what i need to do. I've wasted quite a few hours searching for this online and now I give up and am asking you guys.

I need to build a site like a VERY tiny, VERY simplified version of Amazon: a dropdown menu populates a table with the relevant information, including an image.

I know I'll need PHP to make the page and MySQL to store the data. I also know the images will be stored in a folder just like with a regular HTML/CSS site. I just cannot figure out how to display the images through PHP and MySQL commands. Can someone PLEASE point me in the right direction for tutorials or something? It's driving me nuts!

sid

Assuming the images are stored already in a certain directory, you either can name the files, so they can be implicitely mapped to the corresponding product in your database (f.ex. filename: PRODUCT_ID.jpg, making it 24.jpg for the product with the id 24). Alternatively (and more flexible), you can store the filename with the products in the database and retrieve them like this. There are numerous ways on how to lay out your database, even for such a "simple" (I don't think there is such a thing as a "simplified version of Amazon") application. You even could save the binary image data right within the database, however, I wouldn't advise to do so, if you're a novice.

Hope this helps!

Mykl

Thanks, but no, it doesn't help! The images will be stored in a folder in a directory, and I can make them appear in a page with php. The trouble is that I don't know how to store the info in the database so that the image will appear in the table. I can't find any tutorials about this. Perhaps I don't know what keywords to search for.

sid

just take the filename and add a field to your products-table that holds this string.
Then, just as you display the product title on the result page, also display an img-tag with the dynamically created src attribute.

// $field is an array of fields, fetched from the db

$src = 'path/to/file/' . $field['imagefile'];

print '<img src="' . $src . '" alt="image"/>';

Mykl

Thanks for the suggestions, but it turns out the easiest thing is to put html img code in the database.

johanafm

However that also means storing a mix of data and presentational information. While this may seem simpler at the moment, it inevitably leads to more problems in the long run.

For example, what happens when you want to either remove an image or move images from one place to another, such as from your webserver.example.com to your mediaserver.example.com? I recommend you follow the original advice. I also prefer keeping local image file path info (or ftp server and path info) in environment variables rather than in the code.

dalecosp

+1. I've just added "image support" to a system we have for remote inventory. We used to harvest URL's for images and keep them in the DB. Now we harvest images and keep them in a special directory on a special server with the filename keyed to the product's ID number. Simple, fast, effective, awesome. 😉