Question

我要求您提供有关如何通过表单上传照片的帮助，发布后，然后以某种方式修改照片并立即在浏览器上下载

- 现在我知道关于html表单的部分。有没有办法可以在没有首先在服务器中保存文件的情况下使用PHP来完成？

让我在这里给我的代码请帮助:(

    // Loop $_FILES to execute all files
    foreach ($_FILES['photos']['name'] as $f => $name) {     
        if ($_FILES['photos']['error'][$f] == 4) {
            continue; // Skip file if any error found
        }          
        if ($_FILES['photos']['error'][$f] == 0) {             
            if ($_FILES['photos']['size'][$f] > $max_file_size) {
                $message[] = "$name is too large!.";
                continue; // Skip large files
            }
            elseif( ! in_array(pathinfo($name, PATHINFO_EXTENSION), $valid_formats) ){
                $message[] = "$name is not a valid format";
                continue; // Skip invalid file formats
            }
            else{
                // No error found! Move uploaded files 
                //if(move_uploaded_file($_FILES["photos"]["tmp_name"][$f], $path.$name))
                $count++; // Number of successfully uploaded file
                $img = $_FILES["photos"]["tmp_name"][$f];
                header('Content-Description: File Transfer');
                header('Content-Type: image/png');
                header('Content-Disposition: attachment; filename=Image.png');
                header('Content-Transfer-Encoding: binary');
                header('Expires: 0');
                header('Cache-Control: must-revalidate, post-check=0, pre-check=0');
                header('Pragma: public');
                $image = imagecreatefromjpeg($img);
                imagejpeg($image);
                imagedestroy($image);
                //echo $name;
                exit;
                //$zip->addFromString($_FILES["photos"]["name"][$f], $result['contents']);
            }
        }
    }

Answer 1

<?php

if(!$_FILES || is_uploaded_file($_FILES['html-input-name'])){
    //someone is trying to hack your upload
}else{
    $img = $_FILES['html-input-name'];
    header(...);//set a header


    ...// work your magic with the image

    echo $img;//finally echo the result
}

以此为例，访问http://php.net/并尝试解决问题，然后在遇到问题时再回来。

html和php表单上传图片，然后在处理后下载相同的图片

1 个答案: