php - json_decode, foreach 只从 json 字符串返回一行

Question

这个问题已经有答案了:

How to add elements to an empty array in PHP? (9 个回答)

已关闭 去年。

在我的“购物车”表“items”列中，我存储 json_encoded 字符串。我解码、循环并将它们作为数组抓取。只要商品具有不同的产品 ID，它就可以正常工作:

[{"id":"56","size":"Small","quantity":"1"},{"id":"53","size":"Medium","quantity":"2"}]

但是当商品具有相同的产品 ID 时，它只显示 json 字符串中的最后一个，在本例中为“大”:

[{"id":"53","size":"Small","quantity":"1"},{"id":"53","size":"Large","quantity":"2"}]

我为产品创建了一个var_dump，它说这两项都在那里，但就像我说的，它只显示输出中的最后一项。

$txn_id = sanitize((int)$_GET['txn_id']);
$txnQuery = $db->query("SELECT * FROM transactions_alternative WHERE id = '{$txn_id}'");
$txn = mysqli_fetch_assoc($txnQuery);
$cart_id = $txn['cart_id'];
$cartQ = $db->query("SELECT * FROM cart WHERE id = '{$cart_id}'");
$cart = mysqli_fetch_assoc($cartQ);
$items = json_decode($cart['items'],true);
$idArray = array();
$products = array();var_dump($items);
foreach($items as $item){
   $idArray[] = $item['id'];
}
$ids = implode(',',$idArray);
$productQ = $db->query("SELECT i.id as 'id', i.title as 'title', c.id as 'cid', c.category as 'child', p.category as 'parent'
   FROM products i
   LEFT JOIN categories c ON i.categories = c.id
   LEFT JOIN categories p ON c.parent = p.id
   WHERE i.id IN ({$ids})");
  while($p =  mysqli_fetch_assoc($productQ)){

    foreach($items as $item){
      if($item['id'] == $p['id']){
        $x = $item;
        continue;
      }
    }
    $products[] = array_merge($x,$p);var_dump($products);
   }
 ?>
<h2 class="text-center">Rendelés részletei</h2>
<div class="col-md-12">
  <h3 class="text_center">Rendelt termékek</h3>
<table class="table table-condensed table-bordered table-striped">
  <thead>
    <th>Rendelt mennyiség</th><th>Termék neve</th><th>Kategória</th><th>Opció</th>
  </thead>
  <tbody>
    <?php foreach($products as $product): ?>
    <tr>
      <td><?=$product['quantity'];?> db</td>
      <td><?=$product['title'];?></td>
      <td><?=$product['parent'].' / '.$product['child'];?></td>
      <td><?=$product['size'];?></td>
    </tr>
    <?php endforeach ?>
  </tbody>
</table>

Answer 1

您正在覆盖$x

替换

$x = $item;

与

$x[] = $item;

并在foreach循环之前初始化$x = [];；