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I've been following multiple tutorials on uploading and retrieving images using MySQL and PHP, and they all seem to be using the same technique; upload the image as normal (no problem there) and then create another document and use a PHP script to select the image from the database then change the 'content-type' to image/jpeg, then display said document in an iframe.

That's fine and dandy; if I wanted to display one single image, and I was fine with it being in an iframe. However, I'm developing a CMS(like) system where the end user will be changing many styled images on a particular page and that technique doesn't quite work, as it appears to only work for single image.

I'm not looking to upload the files to a folder, then store the path in the database either. I'd like to store the image itself in the database.

I know this is a broad question and I definitely don't expect anyone to give me a step by step tutorial. But if you could direct me to one that would be great.

Thanks in advance.

VantaRabbit
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    what exactly is your question? maybe you should add some code, that we see what you are trying to achieve. and why do you want to store the data in your database instead of a path or guid? – Raphael Müller Jun 01 '16 at 07:55
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    Why don't you want to upload the files to a folder & store the path in the database? Here's a thread discussing the best practices for storing image in the database. http://stackoverflow.com/questions/3748/storing-images-in-db-yea-or-nay – Alan Tan Jun 01 '16 at 07:56
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    Why doesn't this work with more than one image? What do iframes have to do with this? I have the impression that you are trying to load several pictures inside a single `` tag—that doesn't work with static files either. – Álvaro González Jun 01 '16 at 08:02
  • Thank you to the above. I didn't want to use the path in the database as the rest of the page's editable content is also stored in the database. And would like to have all the content in one place. "Why?" I guess 'just because', I knew that if it really came down to it I would have no other option. I'm guessing the reason the tutorial said to use an iframe was because if you include the document into the page the header "image/jpeg" will remove the rest of the content on the page and just display the image. – VantaRabbit Jun 01 '16 at 08:44
  • And it was my fault for not thinking of using image source. thank you to Martin for clarifying that. – VantaRabbit Jun 01 '16 at 08:46

1 Answers1

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The page used to display the image in the way you outline the tutorials show, with retrieval and then a header output is a way that sets the PHP image-calling file to act as an image file .

So you can have 

<img src='phpimagefile.php?id=imgreference' alt'my image'>

for example.

If you want to display multiple images like this then you can simply roll the above HTML code into a foreach or similar loop on another page. 

foreach ($imagesSet as $image){
    print "<img src='phpimagefile.php?id=".$image['id']."' alt'my image'>\n";
}

Edit:

You can get really smart and with some mod_rewrite you can actually store images with filepath names such as :

<img src='/images/45757/goldenShower.jpg' alt='My Image'>

And the mod_rewrite will reorganise this to be able to find the image of:

<img src='/images/phpimagefile.php?id=45757' alt'my image'>

.

Storing the images directly in the database is cumbersome and does not solve your (apparent) problem of retrieval.

If you can clarify why being in the database is better for the image than a image output file as I describe above, this will probably show more people (judging from the comments) exactly what your current issue is.

Some Notes:

  • Don't use iframes. Use the HTML img tag as I example above.
  • We can't direct you to any tutorial until we actually understand what you're trying to gain.
  • Even if you upload the files to a database table, you are still going to need to store a reference to that table/row in order to reference the correct image, so imageId = 'files/image/goldenShower.jpg' is just another reference in the Database like imageId = 4365 would be.
Martin
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