php 使用 file_get_contents 显示图像
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Show image using file_get_contents
提问by Belgin Fish
how can I display an image retrieved using file_get_contents in php?
如何在 php 中显示使用 file_get_contents 检索到的图像？
Do i need to modify the headers and just echo it or something?
我是否需要修改标题并只回显它或其他什么？
Thanks!
谢谢！
采纳答案by Pekka
Do i need to modify the headers and just echo it or something?
我是否需要修改标题并只回显它或其他什么？
exactly.
确切地。
Send a header("content-type: image/your_image_type");and the data afterwards.
发送 aheader("content-type: image/your_image_type");和之后的数据。
回答by robjmills
You can use readfileand output the image headers which you can get from getimagesizelike this:
您可以使用readfile并输出可以从getimagesize获得的图像标题，如下所示：
$remoteImage = "http://www.example.com/gifs/logo.gif";
$imginfo = getimagesize($remoteImage);
header("Content-type: {$imginfo['mime']}");
readfile($remoteImage);
The reason you should use readfile here is that it outputs the file directly to the output buffer where as file_get_contentswill read the file into memory which is unnecessary in this content and potentially intensive for large files.
您应该在此处使用 readfile 的原因是它将文件直接输出到输出缓冲区，因为file_get_contents会将文件读入内存，这在此内容中是不必要的，并且对于大文件可能会很密集。
回答by Ya?ar Xavan
$image = 'http://images.itracki.com/2011/06/favicon.png';
// Read image path, convert to base64 encoding
$imageData = base64_encode(file_get_contents($image));
// Format the image SRC: data:{mime};base64,{data};
$src = 'data: '.mime_content_type($image).';base64,'.$imageData;
// Echo out a sample image
echo '<img src="' . $src . '">';
回答by Mike Caron
You can do that, or you can use the readfilefunction, which outputs it for you:
您可以这样做，也可以使用readfile为您输出它的函数：
header('Content-Type: image/x-png'); //or whatever
readfile('thefile.png');
die();
Edit: Derp, fixed obvious glaring typo.
编辑：Derp，修复了明显明显的错字。
回答by Mochammad Taufiq
you can do like this :
你可以这样做：
<?php
$file = 'your_images.jpg';
header('Content-Type: image/jpeg');
header('Content-Length: ' . filesize($file));
echo file_get_contents($file);
?>
回答by darighteous1
Small edit to @seengee answer: In order to work, you need curly braces around the variable, otherwise you'll get an error.
对@seengee 答案的小编辑：为了工作，您需要在变量周围使用花括号，否则您会收到错误消息。
header("Content-type: {$imginfo['mime']}");
header("Content-type: {$imginfo['mime']}");
