2014-10-18 70 views
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我在哪里可以获得要包含在C:\xampp\php\extras文件夹中的四个魔术文件以获取正确的MIME类型($finfo = finfo_open(FILEINFO_MIME_TYPE,'C:/xampp/php/extras'))?使用PHP获取正确的MIME类型

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为什么不能谷歌? – 2014-10-18 07:48:09

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@AxelAmthor我一直在使用它,我发现这个链接http://sourceforge.net/project/showfiles.php?group_id=23617&package_id=18878,但它重定向到其他页面.. – MRS 2014-10-18 07:59:54

回答

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最后我找到了答案......以下是我用过的代码。

list($width, $height, $image_type) = getimagesize($_FILES["photo"]["tmp_name"]); $mime_photo = image_type_to_mime_type($image_type);

说明
和getimagesize()函数可以被用来获得的宽度,高度,图像类型等 然后我用image_type_to_mime_type()函数以如下参数$ image_type.This函数返回正确的MIME类型。

如果您正在使用($ _FILES [“photo”] [“type”] ==“image/jpeg”)来比较图像类型,并假设您正在更改文件的扩展名,例如'myprofile.txt 'to'myprofile.jpg'比较将是真实的,它读作'image/jpeg',但是如果你使用MIME类型进行比较,那么它将是错误的。

比较使用内容类型:

 if (($_FILES["photo"]["type"]== "image/jpeg") || ($_FILES["photo"]["type"]== "image/jpg")) 
       { 
        echo 'The image is valid and its type is '.$_FILES["photo"]["type"]; 
       } 
       else { 
        echo 'The image is invalid and its type is '.$_FILES["photo"]["type"]; 
       } 
Comparing using MIME type: 
if (($mime_photo== "image/jpeg") || ($mime_photo== "image/jpg")) 
       { 
        echo 'The image is valid and its MIME Type is '.$mime_photo; 
       } 
       else { 
        echo 'The image uploaded is invalid and its MIME Type is '.$mime_photo; 
       }