如何从PHP下载清单中的项目

$projects = [
    [
        'name' => 'Project 1',
        'download_link' => 'http://example.com/project1.zip'
    ],
    [
        'name' => 'Project 2',
        'download_link' => 'http://example.com/project2.zip'
    ],
    // 添加更多项目...
];

<?php
$requestedProject = $_GET['project']; // 从URL参数中获取项目名称

// 在项目清单中查找请求的项目
$project = null;
foreach ($projects as $p) {
    if ($p['name'] === $requestedProject) {
        $project = $p;
        break;
    }
}

// 如果找到了项目，则发送文件给用户
if ($project) {
    $file = file_get_contents($project['download_link']);
    header('Content-Type: application/zip');
    header('Content-Disposition: attachment; filename="' . $project['name'] . '.zip"');
    echo $file;
} else {
    echo '项目不存在';
}

http://yourdomain.com/download.php?project=Project%201