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POBuilding a photo array from database tables
 

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  1. POBuilding a photo array from database tables
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    <p>So I have been fighting this for months now and am hoping to find a solution here once and for all. I am using <a href="http://www.bradpineau.com/PHPPwinty/" rel="nofollow">phppwinty</a> as a php library for <a href="http://pwinty.com" rel="nofollow">pwinty.com</a>. </p> <p>In the default example provided by the author, the images array is created like the example below with all the variables hard coded as an example. </p> <pre><code>$photo1 = $pwinty-&gt;addPhoto($order, "4x6", "http://www.picisto.com/photos/picisto-20120608100135-925870.jpg", "1", "ShrinkToFit"); $photo2 = $pwinty-&gt;addPhoto($order, "6x6", "http://www.picisto.com/photos/picisto-20120601010631-895061.jpg", "4", "ShrinkToFit"); $photo3 = $pwinty-&gt;addPhoto($order, "5x7", "http://www.picisto.com/photos/picisto-20120626210525-763773.jpg", "3", "ShrinkToFit"); </code></pre> <p><em>(note: in training, I have found out we don't necessarily need to add the integer after photo. instead of: <code>photo1</code> it can just be <code>photo</code>.)</em></p> <p>Now the $order variable is fine to stay. Next is the size, image source, and quantity. These are the only 3 variables I am worried about at the moment. With the shopping cart I have these unfortunately are broken up into 3 different tables.</p> <pre><code>-- order_options -- The table for the size variable --option_value -- The column for the size variable --order_id -- Relationship column -- order_products -- The table for the product id's and qty in the order -- product_id -- The column for the product id variable -- product_quantity -- The column for the quantity variable -- order_id -- Relationship column -- products -- The table for the image source variable -- product_image -- The column for the image source variable -- product_id -- Relationship column </code></pre> <p>I get the order_id into the script with a get variable: <code>$order_id = $_GET['id'];</code></p> <p>I have attempted to <code>JOIN</code> all the tables together with relationships and have got nothing but errors, Though I don't beleive the problem is in the <code>JOIN</code> but the way I have the <code>foreach</code> set up.</p> <p>My <code>JOIN</code> and <code>foreach</code> looks like this right now: </p> <pre><code>foreach ($db-&gt;query("SELECT orderproducts.product_id, orderoptions.option_value, products.product_image FROM order_products as orderproducts INNER JOIN order_options as orderoptions ON orderproducts.order_id = orderoptions.order_id INNER JOIN products as products ON orderproducts.product_id = products.product_id WHERE orderproducts.order_id = $order_id AND orderproducts.product_id = $product_id") as $row) $order_variables[] = $row; if (count($order_variables) &gt; 0): foreach ($order_variables as $row): $product_id = $row['product_id']; $product_quantity = $row['product_quantity']; $size = $row['option_value']; $product_source = "https://www.alphahq.org/uploads/images/".$row['product_image'].""; $photo = $pwinty-&gt;addPhoto($order, $size, $product_source, $product_quantity, "ShrinkToFit"); endforeach; endif; </code></pre> <p>The reason I say I believe the problem is in how the foreach is set up is because the 1 error I get in my browser when I run the script is as follows: </p> <pre><code>Warning: Invalid argument supplied for foreach() in /home/www/alphahq.org/libraries/phppwinty/print.php on line 35 </code></pre> <p>According to my code editor line 35 is the foreach statement from above. </p> <pre><code> foreach ($db-&gt;query("SELECT orderproducts.product_id, orderoptions.option_value, products.product_image FROM order_products as orderproducts INNER JOIN order_options as orderoptions ON orderproducts.order_id = orderoptions.order_id INNER JOIN products as products ON orderproducts.product_id = products.product_id WHERE orderproducts.order_id = $order_id AND orderproducts.product_id = $product_id") as $row) $order_variables[] = $row; </code></pre> <p>I hope I have been descriptive enough to finally get a solution to this darn problem once in for all. Thank you for your help in advance and hope we can work out a solution together. Happy coding.</p> <p><strong>If I can provide any more information that you believe would be helpful, please simply comment kindly below requesting the information.</strong> </p>
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