php 如何在php中显示数据库中的图像？
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how to display images from database in php?
提问by Jaydipsinh
i write bellow code to display images from database
我编写波纹管代码来显示数据库中的图像
<?php
mysql_connect("localhost","root","");
mysql_select_db("test");
$qry = "select id from image";
$res = mysql_query($qry) or die(mysql_error());
while($row = mysql_fetch_array($res))
{
echo "<img src=image.php?id='$row[0]'>";
}
?>
Image.php
图片.php
<?php
$query = mysql_query("SELECT img FROM images WHERE id = ".$_GET['id']);
$row = mysql_fetch_array($query);
if($type=="pjpeg")
$type="jpeg";
header("Content-type:$type");
echo $row['img'];
?>
but this will not work. it display blank image icon.
但这行不通。它显示空白图像图标。
回答by nkamm
You also might use base64 encodingto build in the image. Like
您也可以使用 base64 编码来构建图像。喜欢
<img src="data:image/jpeg;base64,/9j/4AAQSkZJRgABAQEAWgBaAAD/4gxYSUNDX1BST0ZJTEUAAQEAAAxITGlubwIQAAB..." />
UPDATE, base64 encoding example
更新，base64 编码示例
You can do that easily:
你可以很容易地做到这一点：
<?php
$imageId = intval($_GET["id"]);
$query = mysql_query("SELECT img FROM images WHERE id = ". $imageId);
$row = mysql_fetch_array($query);
$mime = null;
// place $type init. here
if ($type=="pjpeg") // <<< where do you get $type btw?
$mime = "image/jpeg";
$b64Src = "data:".$mime.";base64," . base64_encode($row["img"]);
echo '<img src="'.$b64Src.'" alt="" />';
?>
回答by Jeroen
jpegis not a valid Content-Type, it should be image/jpeg
jpeg不是有效的内容类型，它应该是 image/jpeg
回答by Tim
As above
+ be sure to pass the replace = trueparameter to the header function.
如上 + 确保将replace = true参数传递给标头函数。
header( 'Content-Type: image/jpeg', true );
header( 'Content-Type: image/jpeg', true );
