这是我的数据库连接代码。
我想将blob图像显示为td
<?php while ($news = mysql_fetch_assoc($content))
{
echo "<tr>";
echo "<td>" .$news['<img src = "data:image/jpg;base64,'.base64_encode($img).'"/>']. "</td>";
echo "</td>";
echo "<tr>";
echo "<td>" .$news['title']."</td>";
echo "</td>";
echo "<tr>";
echo "<td>" .$news['content']."</td>";
echo "</td>";
echo "<tr>";
echo "<td>" .$news['date']."</td>";
echo "</td>";
echo "<tr>";
echo "<td>" .$news['author']."</td>";
echo "</td>";
}
?>
我得到一个错误说未定义索引
答案 0 :(得分:0)
您的代码中有拼写错误。改变你的行:
echo "<td><img src='data:image/jpg;base64,".base64_encode($news['img'])."'/></td>";
通过
{{1}}