在 PHP 中输出图像
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Output an Image in PHP
提问by steven
I have an image $file( eg ../image.jpg)
我有一个图像$file（例如../image.jpg）
which has a mime type $type
它有一个 mime 类型 $type
How can I output it to the browser?
如何将其输出到浏览器？
回答by Emre Yazici
$file = '../image.jpg';
$type = 'image/jpeg';
header('Content-Type:'.$type);
header('Content-Length: ' . filesize($file));
readfile($file);
回答by Benjamin Wohlwend
If you have the liberty to configure your webserver yourself, tools like mod_xsendfile(for Apache) are considerably better than reading and printing the file in PHP. Your PHP code would look like this:
如果您可以自由配置自己的网络服务器，像mod_xsendfile（用于 Apache）这样的工具比在 PHP 中读取和打印文件要好得多。您的 PHP 代码如下所示：
header("Content-type: $type");
header("X-Sendfile: $file"); # make sure $file is the full path, not relative
exit();
mod_xsendfile picks up the X-Sendfile header and sends the file to the browser itself. This can make a real difference in performance, especially for big files. Most of the proposed solutions read the whole file into memory and then print it out. That's OK for a 20kbyte image file, but if you have a 200 MByte TIFF file, you're bound to get problems.
mod_xsendfile 获取 X-Sendfile 标头并将文件发送到浏览器本身。这会对性能产生真正的影响，尤其是对于大文件。大多数建议的解决方案将整个文件读入内存，然后将其打印出来。对于 20kbyte 的图像文件来说这没问题，但是如果你有一个 200MB 的 TIFF 文件，你肯定会遇到问题。
回答by Mike
$file = '../image.jpg';
if (file_exists($file))
{
$size = getimagesize($file);
$fp = fopen($file, 'rb');
if ($size and $fp)
{
// Optional never cache
// header('Cache-Control: no-cache, no-store, max-age=0, must-revalidate');
// header('Expires: Mon, 26 Jul 1997 05:00:00 GMT'); // Date in the past
// header('Pragma: no-cache');
// Optional cache if not changed
// header('Last-Modified: '.gmdate('D, d M Y H:i:s', filemtime($file)).' GMT');
// Optional send not modified
// if (isset($_SERVER['HTTP_IF_MODIFIED_SINCE']) and
// filemtime($file) == strtotime($_SERVER['HTTP_IF_MODIFIED_SINCE']))
// {
// header('HTTP/1.1 304 Not Modified');
// }
header('Content-Type: '.$size['mime']);
header('Content-Length: '.filesize($file));
fpassthru($fp);
exit;
}
}
回答by Carlos Lima
Try this:
尝试这个：
<?php
header("Content-type: image/jpeg");
readfile("/path/to/image.jpg");
exit(0);
?>
回答by code_burgar
header('Content-type: image/jpeg');
readfile($image);
回答by Hans
For the next guy or gal hitting this problem, here's what worked for me:
对于下一个遇到这个问题的人或女孩，这对我有用：
ob_start();
header('Content-Type: '.$mimetype);
ob_end_clean();
$fp = fopen($fullyQualifiedFilepath, 'rb');
fpassthru($fp);
exit;
You need all of that, and only that. If your mimetype varies, have a look at PHP's mime_content_type($filepath)
您需要所有这些，而且仅此而已。如果您的 mimetype 有所不同，请查看 PHP 的 mime_content_type($filepath)
回答by Tomo Hadi Sutomo
$file = '../image.jpg';
$type = 'image/jpeg';
header('Content-Type:'.$type);
header('Content-Length: ' . filesize($file));
$img = file_get_contents($file);
echo $img;
This is works for me! I have test it on code igniter. if i use readfile, the image won't display. Sometimes only display jpg, sometimes only big file. But after i changed it to "file_get_contents" , I get the flavour, and works!! this is the screenshoot: Screenshot of "secure image" from database
这对我有用！我已经在代码点火器上对其进行了测试。如果我使用 readfile，图像将不会显示。有时只显示jpg，有时只显示大文件。但是在我将其更改为 "file_get_contents" 之后，我得到了味道，并且有效！！这是屏幕截图：来自数据库的“安全图像”的屏幕 截图
回答by Fabien Sa
You can use finfo(PHP 5.3+) to get the right MIME type.
您可以使用finfo(PHP 5.3+) 来获取正确的 MIME 类型。
$filePath = 'YOUR_FILE.XYZ';
$finfo = finfo_open(FILEINFO_MIME_TYPE);
$contentType = finfo_file($finfo, $filePath);
finfo_close($finfo);
header('Content-Type: ' . $contentType);
readfile($filePath);
PS: You don't have to specify Content-Length, Apache will do it for you.
PS：您不必指定Content-Length，Apache 会为您指定。
回答by Pascal MARTIN
You can use headerto send the right Content-type :
您可以使用header发送正确的 Content-type ：
header('Content-Type: ' . $type);
And readfileto output the content of the image :
并readfile输出图像的内容：
readfile($file);
And maybe (probably not necessary, but, just in case)you'll have to send the Content-Length header too :
也许（可能没有必要，但是，以防万一）您也必须发送 Content-Length 标头：
header('Content-Length: ' . filesize($file));
Note : make sure you don't output anything else than your image data (no white space, for instance), or it will no longer be a valid image.
注意：确保除了图像数据之外不输出任何其他内容（例如，没有空格），否则它将不再是有效图像。
回答by Matthew Scharley
<?php
header("Content-Type: $type");
readfile($file);
That's the short version. There's a few extra little things you can do to make things nicer, but that'll work for you.
那是简短的版本。你可以做一些额外的小事情来让事情变得更好，但这对你有用。
