php – 我想显示图像而不是下载它(图像在数据库blob中)

发布时间：2020-12-13 17:04:57 所属栏目：PHP教程来源：网络整理

导读：我想要显示图像而不是下载它. 我的数据库表中有图像,列为BLOB. 此代码段会下载图片,但我想显示它： $query = "SELECT * FROM upload";$result = mysql_query($query);$row = mysql_fetch_array($result);$content = $row['content'];$size = $row['size'];$t

我想要显示图像而不是下载它.

我的数据库表中有图像,列为BLOB.

此代码段会下载图片,但我想显示它：

$query = "SELECT * FROM upload";
$result  = mysql_query($query);
$row = mysql_fetch_array($result);
$content =  $row['content'];
$size =  $row['size'];
$type =  $row['type'];
header("Content-length: $size");
header("Content-type: $type");

// The following headers make the image download,but I don't want it to
// download,I want to show the image. What should I do?
// header("Content-Disposition: attachment; filename=$name");

echo $content;

解决方法

如果你想使用它更动态地制作原始代码的脚本并像这样调用它：

<img src="image.php?imageid=$myImageID" />

你的脚本是：

$myImageID = $_GET["myImageID"];
$query = "SELECT * FROM upload where id='"+$myImageID+"'";
$result  = mysql_query($query);
$row = mysql_fetch_array($result);
$content =  $row['content'];
$size =  $row['size'];
$type =  $row['type'];
header("Content-length: $size");
header("Content-type: $type");
//header("Content-Disposition: attachment; filename=$name");---> this headers make system to download,but i dont want to download,i want to show image,what  should i do,echo $content; ?>"

（编辑：李大同）

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