以图像形式返回 PHP 页面

发布于 2024-07-21 18:46:24 字数 341 浏览 7 评论 0原文

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，但显示图像...

我的 index.php 有一个像这样的图像链接：

<img src='test.php?image=1234.jpeg' />

和我的 php脚本基本上执行以下操作：

1）读取1234.jpeg 2）回显文件内容... 3）我有一种感觉，我需要用 mime 类型返回输出，但这就是我迷失的地方，

一旦我弄清楚了这一点，我将删除所有输入的文件名并用图像 ID 替换它。

如果我不清楚，或者您需要更多信息，请回复。

原文

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...

my index.php has an image link like this:

<img src='test.php?image=1234.jpeg' />

and my php script does basically this:

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

Once I figure this out, I will be removing the file name input all together and replace it with an image id.

If I am unclear, or you need more information, please reply.

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阳光的暖冬 2024-07-28 18:46:24

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;
?>

重要的一点是你必须发送一个内容- 键入标题。另外，您必须小心，不要在文件中的标记之前或之后包含任何额外的空格（例如换行符）。

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格的危险：

<?php
$name = './img/ok.png';
$fp = fopen($name, 'rb');

header("Content-Type: image/png");
header("Content-Length: " . filesize($name));

fpassthru($fp);

您仍然需要小心避免脚本顶部出现空格。一种特别棘手的空白形式是 UTF-8 BOM。为了避免这种情况，请确保将脚本保存为“ANSI”（记事本）或“ASCII”或“无签名的UTF-8”（Emacs）或类似格式。

The PHP Manual has this example:

<?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.

As suggested in the comments, you can avoid the danger of extra white space at the end of your script by omitting the ?> tag:

<?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.

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羞稚 2024-07-28 18:46:24

我觉得我们可以通过从 $image_info 获取 mime 类型来使此代码更容易一些：

$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");

}

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。

I feel like we can make this code a little bit easier by just getting the mime type from $image_info:

$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-geoengineering for your contribution.

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翻了热茶 2024-07-28 18:46:24

我在没有 Content-Length 的情况下工作。也许是远程图像文件工作的原因

// 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;

I worked without Content-Length . maybe reason work for remote image files

// 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;

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川水往事 2024-07-28 18:46:24

非常非常容易。

<?php

//could be image/jpeg or image/gif or whatever
header('Content-Type: image/png')
readfile('image.png')
?>

Very, very easy.

<?php

//could be image/jpeg or image/gif or whatever
header('Content-Type: image/png')
readfile('image.png')
?>

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只为一人 2024-07-28 18:46:24

这应该有效。它可能会更慢。

$img = imagecreatefromjpeg($filename);
header("Content-Type: image/jpg");
imagejpeg($img);
imagedestroy($img);

This should work. It may be slower.

$img = imagecreatefromjpeg($filename);
header("Content-Type: image/jpg");
imagejpeg($img);
imagedestroy($img);

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渔村楼浪 2024-07-28 18:46:24

如果您不从数据库中读取数据，另一个简单的选择（没有更好，只是不同）是使用一个函数来为您输出所有代码......
注意：如果您还希望 php 读取图像尺寸并将其提供给客户端以加快渲染速度，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>

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.

<?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>

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~没有更多了~