Question

我现在已经尝试了几次，这是我认为我能让它正常工作的最接近的。我在其他地方有类似的代码并且它正常工作但是当我执行这个时，mime_content_type不会返回任何内容。我试图以很多不同的方式让它工作，如果你看到我忽视的东西，请告诉我。

for($i = 0; $i < 5; ++ $i) {

                $mime = false;

                if (preg_match ( '/(jpeg|png|gif|jpg|jpe)/i', $_FILES ['listing'] ['type'] ['images'] [$i] )) {

                    $new_image = new image_handler ( $_FILES ['listing'] ['tmp_name'] ['images'] [$i] );

                    $m = mime_content_type ( $new_image );

                    if ($m == 'image/png' || $m == 'image/jpeg' || $m == 'image/gif') {
                        $mime = true;
                    }

                    if ($mime) {
                        $new_images [$i] ['name'] = date ( 'ymdgis' ) . $_FILES ['listing'] ['name'] ['images'] [$i];
                        $new_images [$i] ['default'] = ($_POST ['listing'] ['default_image'] == $i) ? true : false;

                        $new_image->save ( IMAGE_SIZE, IMAGE_SIZE, REAL_PATH . 'uploads/listings/' . $new_images [$i] ['name'] );
                        $new_image->save ( THUMB_SIZE, THUMB_SIZE, REAL_PATH . 'uploads/listings/thumbnails/' . $new_images [$i] ['name'] );
                    }
                } elseif ((! preg_match ( '/(jpeg|png|gif|jpg|jpe)/i', $_FILES ['listing'] ['type'] ['images'] [$i] )) && ($_FILES ['listing'] ['name'] ['images'] [$i] != '')) {
                    $pass_message .= '<p>The File ' . $_FILES ['listing'] ['name'] ['images'] [$i] . ' was not uploaded due to its filetype.</p>';
                }
                if (! $mime && ($_FILES ['listing'] ['name'] ['images'] [$i] != '')) {
                    $pass_message .= '<p>The File ' . /*$_FILES ['uploads'] ['name'] ['image']*/ $m . ' was not uploaded due to its mime type.</p>';
                }
            }

Answer 1

根据文档，mime_content_type将文件名作为输入参数。

在您的示例中，您将实例化传递给mime_content_type（）函数的新image_handler（）对象。

我相信你的班上应该有一个方法来获取文件路径。

这样的事情：

$new_image = new image_handler ( $_FILES ['listing'] ['tmp_name'] ['images'] [$i] );
$filename = $new_image->get_filename_method();

$m = mime_content_type ( $filename );

PHP mime_content_type没有返回任何内容

1 个答案: