Question

我有一个网站，我允许用户上传照片。出于安全考虑，我只希望用户能够上传jpeg，gif或png文件类型。我将如何写这些条件？

到目前为止我所拥有的：

if ($_FILES['media']['size'] != 0 && //tests file type) {
    //the file is good, upload the file
}

谢谢！

Answer 1

检查文件类型。

$fileType = $_FILES['media']['type'];
$fileSize = $_FILES['media']['size'];

if (fileSize != 0 ) {
    if ($fileType == 'image/png' || 
        $fileType == 'image/jpg' ||
        $fileType == 'image/gif') {
        //the file is good, upload the file
    }
}

您可以在此处找到您的特定mime类型：List of MIME Types by Content Type

Answer 2

使用$_FILES['media']['type']

所以你的完整代码：

if ($_FILES['media']['size'] != 0 && $_FILES["media"]["type"] == "image/jpeg" && $_FILES["media"]["type"] == "image/gif" && $_FILES["media"]["type"] == "image/png" ) {
     //the file is good, upload the file
}

Answer 3

最好的选择是@ Siamak.A.M提供给你但不是我在数组中使用mime / types字符串：

$fileType = $_FILES['media']['type'];

// add others.. and you can retrieve this from a conf. file too
$allowed = array('image/gif','image/jpg','image/png');

if (in_array($fileType, $allowed)) {
    //the file is good, upload the file
}

Answer 4

尝试

if ( $_FILES['media']['size'] != 0 && (
($_FILES["file"]["type"] == "image/gif")
|| ($_FILES["file"]["type"] == "image/jpeg")
|| ($_FILES["file"]["type"] == "image/jpg")
|| ($_FILES["file"]["type"] == "image/pjpeg")
|| ($_FILES["file"]["type"] == "image/x-png")
|| ($_FILES["file"]["type"] == "image/png") ) )
{
    //the file is good, upload the file
}

Answer 5

$ext = substr($_FILES['media']['name'], -3)

然后检查一下。

您可以将其与允许的扩展数组进行比较。

$allowed = array('jpg', 'gif');

并使用in_array进行检查。只是一种方式，但我相信还有更多。

确保上载的文件符合要求

5 个答案: