使用file_get_contents显示图像

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回显它或什么？

Thanks!

谢谢！

6 个解决方案

#1

Do i need to modify the headers and just echo it or something?

我是否需要修改标题并回显它或什么？

exactly.

究竟。

Send a header("content-type: image/your_image_type"); and the data afterwards.

发送标题（“content-type：image / your_image_type”）;以及之后的数据。

#2

You can use readfile and output the image headers which you can get from getimagesize like this:

你可以使用readfile并输出你可以从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_contents will read the file into memory which is unnecessary in this content and potentially intensive for large files.

你应该在这里使用readfile的原因是它将文件直接输出到输出缓冲区，其中file_get_contents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是密集的。

#3

$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 . '">';

#4

You can do that, or you can use the readfile function, which outputs it for you:

你可以这样做，或者你可以使用readfile函数，它为你输出：

header('Content-Type: image/x-png'); //or whatever
readfile('thefile.png');
die();

Edit: Derp, fixed obvious glaring typo.

编辑：Derp，修复了明显的明显错字。

#5

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);
?>

#6

Small edit to @seengee answer: In order to work, you need curly braces around the variable, otherwise you'll get an error.

小编辑@seengee回答：为了工作，你需要围绕变量的花括号，否则你会得到一个错误。

header("Content-type: {$imginfo['mime']}");

header（“Content-type：{$ imginfo ['mime']}”）;

#1