[英]Can you give a help understanding how to properly solve “Image not found or type unkown.”?
[英]Image unkown with myphpadmin
嘗試使用數據庫來引用和圖像位置時,它顯示為空白
<td><img width="100" height="100" src=<?php echo $item[image]; ?></td>
表中的其他字段工作正常,如名稱和價格,圖像返回圖像未知錯誤。 我將如何 go 關於在 myphpadmin 中引用圖像位置?
echo count($_SESSION['cart']); ?> items.</p>
<p><a href="?cart">View your cart</a></p>
<table border="1">
<thead>
<tr>
<th>Item Description</th>
<th>Price</th>
<th>image</th>
</tr>
</thead>
<tbody>
<?php foreach ($items as $item): ?>
<tr>
<td><?php htmlout($item['desc']); ?></td>
<td>
$<?php echo number_format($item['price'], 2); ?>
</td>
<td>
<img width="100" height="100" src=<?php echo $item[image]; ?>
</td>
<form action="" method="post">
<div>
<input type="hidden" name="id" value="<?php
htmlout($item['id']); ?>">
<input type="submit" name="action" value="Buy">
</div>
</form>
</td>
</tr>
<?php endforeach; ?>
</tbody>
</table>
以下是項目包含的內容:
$items = array();
$query = "SELECT * FROM catalog";
$result = $conn->query($query);
while($row = $result->fetch_assoc())
{
$item = array('id' => $row['Item_id'], 'desc' => $row['description'], 'price' => $row['price'], 'image' => $row[image]);
$items[] = $item;
}
它應該是:
src="<?php echo $item['image']?>"
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.