将 PHP 页面作为图像返回
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Return a PHP page as an image
提问by MichaelICE
I am trying to read a image file (.jpeg to be exact), and 'echo' it back to the page output, but have is display an image...
我正在尝试读取图像文件（确切地说是 .jpeg），并将其“回显”回页面输出，但是显示图像...
my index.php has an image link like this:
我的 index.php 有一个像这样的图片链接：
<img src='test.php?image=1234.jpeg' />
and my php script does basically this:
我的 php 脚本基本上是这样的：
1) read 1234.jpeg 2) echo file contents... 3) I have a feeling I need to return the output back with a mime-type, but this is where I get lost
1) 读取 1234.jpeg 2) 回显文件内容... 3) 我觉得我需要用 mime 类型返回输出，但这是我迷路的地方
Once I figure this out, I will be removing the file name input all together and replace it with an image id.
一旦我弄清楚这一点，我将一起删除输入的文件名并将其替换为图像ID。
If I am unclear, or you need more information, please reply.
如果我不清楚，或者您需要更多信息，请回复。
回答by Martin Geisler
The PHP Manual has this example:
PHP 手册有这个例子：
<?php
// open the file in a binary mode
$name = './img/ok.png';
$fp = fopen($name, 'rb');
// send the right headers
header("Content-Type: image/png");
header("Content-Length: " . filesize($name));
// dump the picture and stop the script
fpassthru($fp);
exit;
?>
The important points is that you must send a Content-Type header. Also, you must be careful not include any extra white space (like newlines) in your file before or after the <?php ... ?>tags.
重要的一点是您必须发送一个 Content-Type 标头。此外，您必须注意不要在<?php ... ?>标签之前或之后的文件中包含任何额外的空格（如换行符）。
As suggested in the comments, you can avoid the danger of extra white space at the end of your script by omitting the ?>tag:
正如评论中所建议的，您可以通过省略?>标记来避免脚本末尾出现额外空白的危险：
<?php
$name = './img/ok.png';
$fp = fopen($name, 'rb');
header("Content-Type: image/png");
header("Content-Length: " . filesize($name));
fpassthru($fp);
You still need to carefully avoid white space at the top of the script. One particularly tricky form of white space is a UTF-8 BOM. To avoid that, make sure to save your script as "ANSI" (Notepad) or "ASCII" or "UTF-8 without signature" (Emacs) or similar.
您仍然需要小心避免脚本顶部的空白。一种特别棘手的空白形式是UTF-8 BOM。为避免这种情况，请确保将您的脚本另存为“ANSI”（记事本）或“ASCII”或“无签名的 UTF-8”（Emacs）或类似格式。
回答by ban-geoengineering
I feel like we can make this code a little bit easier by just getting the mime type from $image_info:
我觉得我们可以通过从 $image_info 获取 mime 类型来使这段代码更容易一些：
$file_out = "myDirectory/myImage.gif"; // The image to return
if (file_exists($file_out)) {
$image_info = getimagesize($file_out);
//Set the content-type header as appropriate
header('Content-Type: ' . $image_info['mime']);
//Set the content-length header
header('Content-Length: ' . filesize($file_out));
//Write the image bytes to the client
readfile($file_out);
}
else { // Image file not found
header($_SERVER["SERVER_PROTOCOL"] . " 404 Not Found");
}
With this solution any type of image can be processed but it is just another option. Thanks ban-geoengineeringfor your contribution.
使用此解决方案可以处理任何类型的图像，但这只是另一种选择。感谢ban-geoengineering的贡献。
回答by Drew LeSueur
This should work. It may be slower.
这应该有效。它可能会更慢。
$img = imagecreatefromjpeg($filename);
header("Content-Type: image/jpg");
imagejpeg($img);
imagedestroy($img);
回答by Berk
I worked without Content-Length . maybe reason work for remote image files
我在没有 Content-Length 的情况下工作。也许是远程图像文件的原因
// open the file in a binary mode
$name = 'https://www.example.com/image_file.jpg';
$fp = fopen($name, 'rb');
// send the right headers
header('Cache-Control: no-cache, no-store, max-age=0, must-revalidate');
header('Expires: January 01, 2013'); // Date in the past
header('Pragma: no-cache');
header("Content-Type: image/jpg");
/* header("Content-Length: " . filesize($name)); */
// dump the picture and stop the script
fpassthru($fp);
exit;
回答by Joe Bubna
Another easy Option (not any better, just different) if you aren't reading from a database is to just use a function to output all the code for you... Note: If you also wanted php to read the image dimensions and give that to the client for faster rendering, you could easily do that too with this method.
如果您不从数据库中读取数据，另一个简单的选项（没有更好，只是不同）是仅使用函数为您输出所有代码... 注意：如果您还希望 php 读取图像尺寸并给出为了更快地呈现给客户端，您也可以使用此方法轻松做到这一点。
<?php
Function insertImage( $fileName ) {
echo '<img src="path/to/your/images/',$fileName,'">';
}
?>
<html>
<body>
This is my awesome website.<br>
<?php insertImage( '1234.jpg' ); ?><br>
Like my nice picture above?
</body>
</html>
