Php 获取上传的文件名？

我正在尝试PHP脚本，用户可以上传文件。 我使用的脚本来自。 它在我的本地主机上成功运行。但问题是，如何将上传的文件名保存到数据库中

代码-

<?php

header('Content-Type: text/plain; charset=utf-8');

try {

    // Undefined | Multiple Files | $_FILES Corruption Attack
    // If this request falls under any of them, treat it invalid.
    if (
        !isset($_FILES['upfile']['error']) ||
        is_array($_FILES['upfile']['error'])
    ) {
        throw new RuntimeException('Invalid parameters.');
    }

    // Check $_FILES['upfile']['error'] value.
    switch ($_FILES['upfile']['error']) {
        case UPLOAD_ERR_OK:
            break;
        case UPLOAD_ERR_NO_FILE:
            throw new RuntimeException('No file sent.');
        case UPLOAD_ERR_INI_SIZE:
        case UPLOAD_ERR_FORM_SIZE:
            throw new RuntimeException('Exceeded filesize limit.');
        default:
            throw new RuntimeException('Unknown errors.');
    }

    // You should also check filesize here. 
    if ($_FILES['upfile']['size'] > 1000000) {
        throw new RuntimeException('Exceeded filesize limit.');
    }

    // DO NOT TRUST $_FILES['upfile']['mime'] VALUE !!
    // Check MIME Type by yourself.
    $finfo = new finfo(FILEINFO_MIME_TYPE);
    if (false === $ext = array_search(
        $finfo->file($_FILES['upfile']['tmp_name']),
        array(
            'jpg' => 'image/jpeg',
            'png' => 'image/png',
            'gif' => 'image/gif',
        ),
        true
    )) {
        throw new RuntimeException('Invalid file format.');
    }

    // You should name it uniquely.
    // DO NOT USE $_FILES['upfile']['name'] WITHOUT ANY VALIDATION !!
    // On this example, obtain safe unique name from its binary data.
    if (!move_uploaded_file(
        $_FILES['upfile']['tmp_name'],
        sprintf('./uploads/%s.%s',
            sha1_file($_FILES['upfile']['tmp_name']),
            $ext
        )
    )) {
        throw new RuntimeException('Failed to move uploaded file.');
    }

    echo 'File is uploaded successfully.';

} catch (RuntimeException $e) {

    echo $e->getMessage();

}

?>

---

但是没有运气。这里有人可以帮我吗？

要获得上传文件的信息，请使用$U文件['upfile']['tmp_name']作为临时名称，对于实名$U文件['upfile']['name']

例如，上载的文件是ccd37b2ce541f407cabfc58be4e4af952fce7bde.jpg

将此文件移动到上载目录

$path = 'uploads/' . $real_file_name; // this will be uploads/ccd37b2ce541f407cabfc58be4e4af952fce7bde.jpg

if (!move_uploaded_file($_FILES['upfile']['tmp_name'], $path)) {
    throw new RuntimeException('Failed to move uploaded file.');
}

---

表单在哪里？要获取名称，请使用$_FILES['upfile']['name']这段代码反映了图像的真实名称。不是实际存储在filesTry中的临时名称，使用$path=$\u FILES['upfile']['name']；而是使用$_文件['upfile']['tmp_name']；它显示的是这样的->/private/var/folders/dg/9hwdw1311drgdhq4k1pv1w0000gn/T/phpdbvioral文件名$\u FILES['upfile']['name']，临时文件名是$\u FILES['upfile']['tmp\u name']嘿，我试着用这个-$tmp\u file=$\u FILES['upfile']['tmp\u name']/存储临时文件名echo$tmp_文件；但它显示了类似这样的内容/private/var/folders/dg/9hwdw6q1311drgdhq4k1pv1w0000gn/T/phpyCWSRd这是不正确的路径。正确路径为/uploads/ccd37b2ce541f407cabfc58be4e4af952fce7bde。jpg@KarunKumar是，用于存储临时文件名。实名使用$tmp_file=$_FILES['upfile']['name']；您将获得类似upload.png的内容，但我的代码将文件名保存为ccd37b2ce541f407cabfc58be4e4af952fce7bde.jpg到服务器。我怎么知道这个名字？我用$_文件['upfile']['name']获取真实姓名；但我不想。。我想要存储在服务器中的临时名称。一旦重新加载页面，临时名称就不会存储在服务器上。那么，如何获取此临时名称？

$tmp_file = $_FILES['upfile']['tmp_name']; // store temporary file name
$real_file_name = $_FILES['upfile']['name']; // store the name of the file like upload.png

$tmp_file = $_FILES['upfile']['tmp_name']; // this is a random generated temp image name like /var/www/html/php‌​yCWSRd.jpg
$real_file_name = $_FILES['upfile']['name']; // which is ccd37b2ce541f407cabfc58be4e4af952fce7bde.jpg

$path = 'uploads/' . $real_file_name; // this will be uploads/ccd37b2ce541f407cabfc58be4e4af952fce7bde.jpg

if (!move_uploaded_file($_FILES['upfile']['tmp_name'], $path)) {
    throw new RuntimeException('Failed to move uploaded file.');
}