使用 PHP echo 从数据库中获取图像
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Getting image out of database with PHP echo
提问by Awais Umar
I have a database in which there is a row containing the images. Now in my code, I want to get images out of the database, but somehow I am unable to do it. Following is my code – kindly help me out –:
我有一个数据库，其中有一行包含图像。现在在我的代码中，我想从数据库中获取图像，但不知何故我无法做到。以下是我的代码——请帮助我——：
<?php
$qry = mysql_query("SELECT * FROM products ORDER BY products.id DESC LIMIT 0, 1", $con);
if (!$qry)
{
die("Query Failed: ". mysql_error());
}
while ($row = mysql_fetch_array($qry))
{
echo "<h2>".$row['title']."</h2>";
echo "<img src=".'Image/'.$row['image']." />";
echo "<p>".substr($row['body'],0,200)."<a href=articles.php?id=".$row['id']." > Read more</a></p>";
echo "<p>".$row['price']."</p>";
}
?>
If I don't use PHP and just use simply the <img>tag, then the path must be src="Image/passbook.jpg"and it works fine but it's not working with PHP. I am creating an admin panel so that the client can delete or update the images as he want, so I must not use simple <img>tag.
如果我不使用 PHP 而只使用<img>标记，那么路径必须是 src="Image/passbook.jpg"并且它可以正常工作但它不适用于 PHP。我正在创建一个管理面板，以便客户可以根据需要删除或更新图像，所以我不能使用简单的<img>标签。
采纳答案by Rikesh
Try changing,
尝试改变，
echo "<img src=".'Image/'.$row['image']." />";
to
到
echo "<img src='Image/".$row['image']."' />";
回答by Dipesh Parmar
You have error in image declaration code. Correct it as below,
您在图像声明代码中有错误。改正如下，
while( $row = mysql_fetch_array($qry) )
{
echo "<h2>".$row['title']."</h2>";
echo "<img src = 'Image/".$row['image']."' alt = ""/>";
echo "<p>".substr($row['body'],0,200)."<a href=articles.php?id=".$row['id']." > Read more</a></p>";
echo "<p>".$row['price']."</p>";
}
回答by Nirmal Ram
Try this
尝试这个
echo "<img src='Image/".$row['image']."' />";
回答by tattvamasi
try this
尝试这个
while($row=mysql_fetch_array($qry))
{
$title = $row['title'];
$src = $row['image'];
$whatever = $row['body'];
echo "$title <br/><img src="$src" alt="my fancy photo" height="" width=""/><br/>$watever";
}
回答by Ravinder Singh
Do this :
做这个 ：
while($row = mysql_fetch_array($qry))
{
echo "<h2>".$row['title']."</h2>";
echo "<img src = 'Image/".$row['image']."' alt = ""/>";
echo "<p>".substr($row['body'],0,200)."<a href=articles.php?id=".$row['id']." > Read more</a></p>";
echo "<p>".$row['price']."</p>";
}
回答by Tapas Pal
Just use this
就用这个
echo "<img src=Image/".$row['image']." />";
