Question

我在一个数据库表中有图像产品ID，在另一个表中有图像Source。这些都是订单。一个订单中可以有多个图像。我想把每张照片发给Pwinty。

因此，对于我需要制作的每张图片：

$photo = $pwinty->addPhoto($order, "$size", "$source", "$qty", "ShrinkToFit");
// I have not got to the size and qty variable yet

// Because of that I have this instead
$photo = $pwinty->addPhoto($order, "4x6", "$source", "1", "ShrinkToFit");

现在，因为它们在多个表格中，所以我有以下代码:(根本不会返回上面的照片数组。

// Get the customers pictures for this order 
foreach ($db->query("SELECT * FROM order_products WHERE order_id=$order_id") as $row)
$picture_info[] = $row;

if (count($picture_info) > 0):
    foreach ($picture_info as $row):

        $product_id = $row['product_id'];

        echo $product_id;

        foreach ($db->query("SELECT * FROM products WHERE product_id=$product_id") as $row)
        $picture_source[] = $row;

            foreach ($picture_source as $row):

                $source = $row['product_image'];

                echo $source;

                // add some photos
                $photo = $pwinty->addPhoto($order, "4x6", "$source", "1", "ShrinkToFit");


            endforeach; 

    endforeach;

endif;

它返回的是：

14138646283376c471632817da60f95964cb2d57dc46.png
Array
(
[id] => 7776
[address1] => 2002 E Blain
[address2] => 
[postalOrZipCode] => 68460
[country] => United States
[addressTownOrCity] => Belvidear
[recipientName] => Joe Dohn
[textOnReverse] => Photos by AlphaHQ
[stateOrCounty] => Kansas
[status] => NotYetSubmitted
[payment] => 
[paymentUrl] => 
[photos] => Array
    (
    )

[documents] => Array
    (
    )

[stickers] => Array
    (
    )

    )

查看图像数组是如何为空的？我需要修理什么？

Answer 1

优化注意：最好从select语句列出你想要的内容，而不是说SELECT *。 SQL将执行得更快

您能告诉我们您从所有回声中获得的输出吗？

此外，当你将事物作为测试回应时，最好输入一些描述你所回应的内容的文本，这样即使它被设置为NULL，你也会看到文本，现在该变量是空的。

现在看来

foreach ($db->query("SELECT * FROM products WHERE product_id=$product_id") as $row)

实际上并没有获得任何行。

您可以做的简单测试是

echo "My row is " + $row + "\n";

因此，您可以看到实际获得的行数。

从foreaches中的两个表中收集信息

1 个答案: