php 如何从 MySQL 数据库中检索图像并显示在 html 标签中
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How to retrieve images from MySQL database and display in an html tag
提问by exxcellent
I created a MySQL database with a table using phpmyadmin. I created this table with a BLOB column to hold a jpeg file.
我使用 phpmyadmin 创建了一个带有表的 MySQL 数据库。我用一个 BLOB 列创建了这个表来保存一个 jpeg 文件。
I have issues with regards to the php variable $resulthere.
我有关于 php 变量的$result问题。
My code so far: (catalog.php):
到目前为止我的代码：（catalog.php）：
<body>
<?php
$link = mysql_connect("localhost", "root", "");
mysql_select_db("dvddb");
$sql = "SELECT dvdimage FROM dvd WHERE id=1";
$result = mysql_query("$sql");
mysql_close($link);
?>
<img src="" width="175" height="200" />
</body>
How can I get the variable $result from PHP into the HTML so I can display it in the <img>tag?
如何将变量 $result 从 PHP 获取到 HTML 中，以便我可以在<img>标签中显示它？
回答by daiscog
You can't. You need to create another php script to return the image data, e.g. getImage.php. Change catalog.php to:
你不能。您需要创建另一个 php 脚本来返回图像数据，例如 getImage.php。将 catalog.php 更改为：
<body>
<img src="getImage.php?id=1" width="175" height="200" />
</body>
Then getImage.php is
然后 getImage.php 是
<?php
$id = $_GET['id'];
// do some validation here to ensure id is safe
$link = mysql_connect("localhost", "root", "");
mysql_select_db("dvddb");
$sql = "SELECT dvdimage FROM dvd WHERE id=$id";
$result = mysql_query("$sql");
$row = mysql_fetch_assoc($result);
mysql_close($link);
header("Content-type: image/jpeg");
echo $row['dvdimage'];
?>
回答by Ilmari Karonen
Technically, you cantoo put image data in an img tag, using data URIs.
从技术上讲，您也可以使用数据 URI将图像数据放入 img 标签中。
<img src="data:image/jpeg;base64,<?php echo base64_encode( $image_data ); ?>" />
There are some special circumstances where this could even be useful, although in most cases you're better off serving the image through a separate script like daiscog suggests.
在某些特殊情况下，这甚至可能很有用，尽管在大多数情况下，您最好通过像 daiscog 建议的单独脚本来提供图像。
回答by James Williams
You need to retrieve and disect the information into what you need.
您需要检索信息并将其分解为您需要的信息。
while($row = mysql_fetch_array($result)) {
echo "img src='",$row['filename'],"' width='175' height='200' />";
}
回答by Marcus
First off you need to fetch the resulting row from the resultset of the query. For that you can use mysql_fetch_row. Now that you have the fetched row you can access the retrieved value and echo it into the src.
首先，您需要从查询的结果集中获取结果行。为此，您可以使用mysql_fetch_row. 现在您有了获取的行，您可以访问检索到的值并将其回显到 src 中。
For example:
例如：
$sql = "SELECT dvdimage FROM dvd WHERE id=1";
$result = mysql_query($sql);
$row = mysql_fetch_row($result);
?>
<img src="<?=$row[0]?>" width="175" height="200" />
<?
回答by Punit
add $row = mysql_fetch_object($result);after your mysql_query();
$row = mysql_fetch_object($result);在您的 mysql_query() 之后添加；
your html <img src="<?php echo $row->dvdimage; ?>" width="175" height="200" />
你的html <img src="<?php echo $row->dvdimage; ?>" width="175" height="200" />
回答by Ganesh
I have added slashes before inserting into database so on the time of fetching i removed slashes again stripslashes()and it works for me. I am sharing the code which works for me.
我在插入数据库之前添加了斜杠，因此在获取时我再次删除了斜杠stripslashes()，它对我有用。我正在分享对我有用的代码。
How i inserted into mysql db (blob type)
我如何插入 mysql db（blob 类型）
$db = mysqli_connect("localhost","root","","dName");
$image = addslashes(file_get_contents($_FILES['images']['tmp_name']));
$query = "INSERT INTO student_img (id,image) VALUES('','$image')";
$query = mysqli_query($db, $query);
Now to access the image
现在访问图像
$sqlQuery = "SELECT * FROM student_img WHERE id = $stid";
$rs = $db->query($sqlQuery);
$result=mysqli_fetch_array($rs);
echo '<img src="data:image/jpeg;base64,'.base64_encode( stripslashes($result['image']) ).'"/>';
Hope it will help someone
希望它会帮助某人
Thanks.
谢谢。
