我正在制作一个用户上传文件的页面.如果文件类型是其他 jpg、gif 和 pdf,我想要一个 if 语句来创建一个 $error 变量.
I'm making a page where the user upload a file. I want an if statement to create an $error variable if the file type is anything other jpg, gif, and pdf.
这是我的代码:
$file_type = $_FILES['foreign_character_upload']['type']; //returns the mimetype
if(/*$file_type is anything other than jpg, gif, or pdf*/) {
$error_message = 'Only jpg, gif, and pdf files are allowed.';
$error = 'yes';
}
我在构建 if 语句时遇到了困难.怎么说呢?
I'm having difficulty structuring the if statement. How would I say that?
将允许的类型放入一个数组中并使用 <代码>in_array().
Put the allowed types in an array and use in_array()
.
$file_type = $_FILES['foreign_character_upload']['type']; //returns the mimetype
$allowed = array("image/jpeg", "image/gif", "application/pdf");
if(!in_array($file_type, $allowed)) {
$error_message = 'Only jpg, gif, and pdf files are allowed.';
$error = 'yes';
}
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