通过 PHP 加载图片
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Loading an image through PHP
提问by Adam
I'm trying to load an image through PHP, but I don't know how.
The filename is stored in the database, such as image.jpg
我正在尝试通过 PHP 加载图像，但我不知道如何。文件名存储在数据库中，如image.jpg
if($_GET['image']){
// Client requesting image, so retrieve it from DB
$id = mysql_real_escape_string($_GET['image']);
$sql = "SELECT * FROM $tbl_name WHERE id = '$id' LIMIT 1";
}
The client needs to request an image like so
客户端需要像这样请求图像
http://example.com/index.php?image=1
And then it should return the image, so it can be embedded like this:
然后它应该返回图像，所以它可以像这样嵌入：
<img src="http://example.com/index.php?image=1" alt="" />
Is this possible?
这可能吗？
回答by joschua011
$img = 'path/to/image.jpg';
header('Content-Type: image/jpeg');
readfile($img);
just tested it
刚刚测试过
回答by mishu
You can use the GD library for that. You start by creating a resource using a function like http://php.net/imagecreatefromjpeg. You will need to provide the path as a parameter.
您可以为此使用 GD 库。您首先使用像http://php.net/imagecreatefromjpeg. 您需要提供路径作为参数。
After that, you just output the resource using a function like http://php.net/imagejpeg.
之后，您只需使用类似的函数输出资源http://php.net/imagejpeg。
Don't forget to send the content type header, and also to use imagedestroyon the resource.
不要忘记发送内容类型标头，并imagedestroy在资源上使用。
Update:
更新：
Consider this sample:
考虑这个示例：
$im = imagecreatefromjpeg('path/to/image.jpg');
header('Content-Type: image/jpeg');
imagejpeg($img);
imagedestroy($img);
回答by safarov
Here's my solution:
这是我的解决方案：
$mime = 'image/jpg';
$out_image = 'default.jpg'; //default no photo image
$user_img = 'image.jpg'; //it can be gif, png, jpg
//Check if file exist in directory
if(file_exist('/path/to/'.$user_img)) {
$s = imagesize($user_img);
$mime = $s['mime'];
$out_image = $img_data;
}
header('content-type: '.$mime);
readfile($img_data);
回答by David
I suggest you first make a file called image.php. So you will call image.php?id=1
我建议你先创建一个名为 image.php 的文件。所以你会调用 image.php?id=1
Have image.php header be the image type. header('Content-Type: image/jpeg');
将 image.php 标头作为图像类型。header('Content-Type: image/jpeg');
Then you can use the GDImage library in PHP to load the image, and output it. Or you can read the file and output it. The header() is key.
然后就可以使用PHP中的GDImage库来加载图片，并输出了。或者您可以读取文件并输出它。header() 是关键。
