PHP- 通过 URL 获取文件类型

PHP- Get file type by URL

提问by Ashish Srivastava

I want to get the file type (eg. image/gif) by URL using PHP.I had tried

我想使用 PHP.I 尝试通过 URL 获取文件类型（例如图像/gif）

<?php
$image_path="http://fc04.deviantart.net/fs71/f/2010/227/4/6/PNG_Test_by_Destron23.png";
exif_imagetype($image_path);
?>

The above code gave me a blank page and the following code returned "3":

上面的代码给了我一个空白页，下面的代码返回“3”：

<?php
$image_path="http://fc04.deviantart.net/fs71/f/2010/227/4/6/PNG_Test_by_Destron23.png";
echo exif_imagetype($image_path);
?>

Where am I going wrong? Solved: using Fileinfo to fetch content type

我哪里错了？已解决：使用 Fileinfo 获取内容类型

采纳答案by Avis

<?php
 $image_path="http://fc04.deviantart.net/fs71/f/2010/227/4/6/PNG_Test_by_Destron23.png";
 echo exif_imagetype($image_path);
?>

It returned 3because pngresponse type as maciej said.

它返回3是因为pngmaciej 所说的响应类型。

Try this to get like this image/png:

试试这个得到这样的image/png：

echo mime_content_type($image_path);

Try this:

尝试这个：

$finfo = finfo_open(FILEINFO_MIME_TYPE); // return mime type ala mimetype extension   
echo finfo_file($finfo, $image_path) . "\n";
finfo_close($finfo);

回答by pgee70

Here is a PHP function I came up with:

这是我想出的一个 PHP 函数：

/**
 * @param $image_path
 * @return bool|mixed
 */
function get_image_mime_type($image_path)
{
    $mimes  = array(
        IMAGETYPE_GIF => "image/gif",
        IMAGETYPE_JPEG => "image/jpg",
        IMAGETYPE_PNG => "image/png",
        IMAGETYPE_SWF => "image/swf",
        IMAGETYPE_PSD => "image/psd",
        IMAGETYPE_BMP => "image/bmp",
        IMAGETYPE_TIFF_II => "image/tiff",
        IMAGETYPE_TIFF_MM => "image/tiff",
        IMAGETYPE_JPC => "image/jpc",
        IMAGETYPE_JP2 => "image/jp2",
        IMAGETYPE_JPX => "image/jpx",
        IMAGETYPE_JB2 => "image/jb2",
        IMAGETYPE_SWC => "image/swc",
        IMAGETYPE_IFF => "image/iff",
        IMAGETYPE_WBMP => "image/wbmp",
        IMAGETYPE_XBM => "image/xbm",
        IMAGETYPE_ICO => "image/ico");

    if (($image_type = exif_imagetype($image_path))
        && (array_key_exists($image_type ,$mimes)))
    {
        return $mimes[$image_type];
    }
    else
    {
        return FALSE;
    }
}

回答by Gentle153

You are not going wrong anywhere. exif_imagetypereturns the value of one of the image type constants: http://php.net/manual/en/image.constants.php

你在任何地方都不会出错。exif_imagetype返回图像类型常量之一的值：http: //php.net/manual/en/image.constants.php

If you would like to convert this to an extension string, you could use a switch statement:

如果要将其转换为扩展字符串，可以使用 switch 语句：

$typeString = null;
$typeInt = exif_imagetype($image_path);
switch($typeInt) {
  case IMG_GIF:
    $typeString = 'image/gif';
    break;
  case IMG_JPG:
    $typeString = 'image/jpg';
    break;
  case IMG_JPEG:
    $typeString = 'image/jpeg';
    break;
  case IMG_PNG:
    $typeString = 'image/png';
    break;
  case IMG_WBMP:
    $typeString = 'image/wbmp';
    break;
  case IMG_XPM:
    $typeString = 'image/xpm';
    break;
  default: 
    $typeString = 'unknown';
}

You may want to change the order to most to least frequently expected for better performance.

您可能希望将顺序更改为最频繁到最不常见，以获得更好的性能。

回答by Alex Howansky

In the first example, you're getting a blank page because you're not doing anything with the return value from the function call. In the second example, you're getting a valid response. See the manual page for exif_imagetype()for a list of what the values mean.

在第一个例子中，你得到一个空白页，因为你没有对函数调用的返回值做任何事情。在第二个示例中，您将获得有效响应。有关值的含义列表，请参阅exif_imagetype() 的手册页。

回答by Grice

exif_imagetypereturns the image type. The response, 3, indicates it is IMAGETYPE_PNG, the correct response.

exif_imagetype返回图像类型。响应 3 表示它是 IMAGETYPE_PNG，即正确的响应。

回答by Maciej Asembler

3 is image type response for PNG image. See: http://php.net/manual/en/function.exif-imagetype.php

3 是 PNG 图像的图像类型响应。请参阅：http: //php.net/manual/en/function.exif-imagetype.php