php 调用未定义的函数 exif_imagetype()

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时间:2020-08-25 10:39:30  来源:igfitidea点击:

Call to undefined function exif_imagetype()

phpimagefile-uploaduploadmime-types

提问by Subedi Kishor

I am trying to get Mime-Typefor image-typesas follow:

我试图让Mime-Typeimage-types如下:

if(!empty($_FILES['uploadfile']['name']) && $_FILES['uploadfile']['error'] == 0){    

    $file = $_FILES['uploadfile']['tmp_name'];
    $file_type = image_type_to_mime_type(exif_imagetype($file));

    switch($file_type){

        // Codes Here

    }

}

But it always gives the error Call to undefined function exif_imagetype(). What am I doing wrong here?

但它总是给出错误Call to undefined function exif_imagetype()。我在这里做错了什么?

回答by samayo

Enable the following extensions in php.iniand restart your server.

启用以下扩展php.ini并重新启动服务器。

extension=php_mbstring.dll
extension=php_exif.dll

扩展名=php_mbstring.dll
扩展名=php_exif.dll

Then check phpinfo()to see if it is set to on/off

然后检查phpinfo()它是否设置为开/关

回答by aesede

I think the problem is PHP config and/or version, for example, in my case:

我认为问题在于 PHP 配置和/或版本,例如,就我而言:

We know exif_imagetype()takes a file path or resource and returns a constant like IMAGETYPE_GIF and image_type_to_mime_type()takes that constant value and returns a string 'image/gif', 'image/jpeg', etc. This didn't work (missing function exif_imagetype), so I've found that image_type_to_mime_type()can also take an integer 1, 2, 3, 17, etc. as input, so solved the problem using getimagesize, which returns an integer value as mime type:

我们知道exif_imagetype()需要一个文件路径或资源,并返回像IMAGETYPE_GIF一个常数,image_type_to_mime_type()采用的是恒定值,并返回一个字符串'image/gif''image/jpeg'等等。这并没有工作(缺少功能exif_imagetype),所以我发现, image_type_to_mime_type()还可以采取整数1 , 2, 3, 17 等作为输入,所以使用 getimagesize 解决了这个问题,它返回一个整数值作为 mime 类型:

function get_image_type ( $filename ) {
    $img = getimagesize( $filename );
    if ( !empty( $img[2] ) )
        return image_type_to_mime_type( $img[2] );
return false;
}

echo get_image_type( 'my_ugly_file.bmp' );
// returns image/x-ms-bmp
echo get_image_type( 'path/pics/boobs.jpg' );
// returns image/jpeg

回答by Imane Fateh

Add this to your code so as we could know which version of php you do have because this function is only supported by (PHP version 4 >= 4.3.0, PHP 5).

将此添加到您的代码中,以便我们知道您使用的是哪个版本的 php,因为此功能仅受 (PHP version 4 >= 4.3.0, PHP 5) 支持。

<?php 
    phpinfo(); 
?> 

It may be not installed, you can add this part of code to make sure it is :

它可能没有安装,您可以添加这部分代码以确保它是:

<?php
if (function_exists('exif_imagetype')) {
    echo "This function is installed";
} else {
    echo "It is not";
}
?>