在上传文件之前重命名为随机名称，然后使用文件名

Question

我有这些代码并上传图片很好，但在一个部分检查＆＃39;如果文件存在＆＃39;不要upload.i想删除这部分并设置一个随机名称到文件。并使用新文件名。 代码：

    let syncConc = DispatchQueue(label:"con",attributes:.concurrent)


    DispatchQueue.global(qos: .utility).async {
        syncConc.sync{
            for _ in 0...10{
                print("XYZ - \(Thread.current)")
            }
            for _ in 0...10{
                print("ABC - \(Thread.current)")
            }
        }
    }


    DispatchQueue.global(qos: .userInitiated).async {
        syncConc.sync{
            for _ in 0...10{
                print("HHH - \(Thread.current)")
            }

            for _ in 0...10{
                print("XXX - \(Thread.current)")
            }
        }
    }

感谢

答案：因为它已被标记为重复，我也可以编辑在这里发布答案。

$target_dir = "pics/";
$target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]);
$uploadOk = 1;
$imageFileType = pathinfo($target_file,PATHINFO_EXTENSION);
// Check if image file is a actual image or fake image
if(isset($_POST["submit"])) {
    $check = getimagesize($_FILES["fileToUpload"]["tmp_name"]);
    if($check !== false) {
        echo "File is an image - " . $check["mime"] . ".";
        $uploadOk = 1;
    } else {
        echo "File is not an image.";
        $uploadOk = 0;
    }
}
// Check if file already exists
if (file_exists($target_file)) {
    echo "Sorry, file already exists.";
    $uploadOk = 0;
}
// Check file size
if ($_FILES["fileToUpload"]["size"] > 500000) {
    echo "Sorry, your file is too large.";
    $uploadOk = 0;
}
// Allow certain file formats
if($imageFileType != "jpg" && $imageFileType != "png" && $imageFileType != "jpeg"
&& $imageFileType != "gif" ) {
    echo "Sorry, only JPG, JPEG, PNG & GIF files are allowed.";
    $uploadOk = 0;
}
// Check if $uploadOk is set to 0 by an error
if ($uploadOk == 0) {
    echo "Sorry, your file was not uploaded.";
// if everything is ok, try to upload file
} else {
    if (move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $target_file)) {
        echo "The file ". basename( $_FILES["fileToUpload"]["name"]). " has been uploaded.";
    } else {
        echo "Sorry, there was an error uploading your file.";
    }
}