如何轻松调整大小和重命名上载的图像？ （php）

Question

我将使用什么代码：

1. 将所有上传的图片调整为600px的宽度，同时保持高宽比
2. 重命名要上传到当前时间戳的文件

以及如何精确地将其添加到或编辑现有代码（最好不使用类）：

$target_dir2 = "creature_pics/";
$target_file2 = $target_dir2 . basename($_FILES["u_c2pic"]["name"]);
$uploadOk2 = 1;
$imageFileType2 = pathinfo($target_file2,PATHINFO_EXTENSION);

if(isset($_POST["submit"])) {
    $check2 = getimagesize($_FILES["u_c2pic"]["tmp_name"]);
    if($check2 !== false) {
        $uploadOk2 = 1;
    } else {
        $uploadOk2 = 0;
    }
}

if (file_exists($target_file2)) {
    $uploadOk2 = 0;
}

if ($_FILES["u_c2pic"]["size"] > 5000000) {
    $uploadOk2 = 0;
}

if($imageFileType2 != "jpg" && $imageFileType2 != "png" && $imageFileType2 != "jpeg"
&& $imageFileType2 != "gif" ) {
    $uploadOk2 = 0;
}

if ($uploadOk2 == 0) {
    $ercode2 = "Sorry, your file was not uploaded.";
} else {
    if (move_uploaded_file($_FILES["u_c2pic"]["tmp_name"], $target_file2)) {
    } else {
        $ercode2 = "Sorry, there was an error uploading your file.";
    }
}

我是php编码的新手，并且希望有最简单的解决方案。

Answer 1

如果可以在服务器上使用Imagick，则以下代码应为您提供帮助。

首先该功能进行大小调整。

if(!defined('APPLICATION_PATH')) define('APPLICATION_PATH',  dirname(__FILE__));

function popupImage($width, $height, $code, $name) {
    $file = APPLICATION_PATH.'/creature_pics/'.$name.'.jpg';

    $im = new Imagick($file);

    $imageprops = $im->getImageGeometry();
    $r_width = $imageprops['width'];
    $r_height = $imageprops['height'];
    if($width > $height){
        $newHeight = $height;
        $newWidth = ($width / $r_height) * $r_width;
    }else{
        $newWidth = $width;
        $newHeight = ($height / $r_width) * $r_height;
    }
    $im->resizeImage($newWidth,$newHeight, imagick::FILTER_LANCZOS, 0.9, true);
    $im->cropImage ($width,$height,0,0);    
    $im->writeImage(APPLICATION_PATH.'/creature_pics/'.$code.'.jpg');
    $im->destroy();

    $newFile = APPLICATION_PATH.'/creature_pics/'.$code.'.jpg';
}

然后在您的代码中只需添加

if (move_uploaded_file($_FILES["u_c2pic"]["tmp_name"], $target_file2)) {
    //This is the new line calling the function
    $timestamp = microtime(true);
    $newFile = popupImage(600,1200,$timestamp,$_FILES["u_c2pic"]["tmp_name"]);
} else {
    $ercode2 = "Sorry, there was an error uploading your file.";
}

那应该在/creature_pics/创建一个新文件，其名称将类似于142023023.9777.jpg ，这是作为代码传递给函数的时间戳。

编辑

错过了时间戳元素，因此现在添加了