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php签到逻辑实现

2026-03-13 13:24:21PHP

PHP签到逻辑实现

数据库设计

创建用户签到表，包含字段如用户ID、签到日期、连续签到天数等。示例SQL：

php签到逻辑实现

CREATE TABLE user_check_in (
    id INT AUTO_INCREMENT PRIMARY KEY,
    user_id INT NOT NULL,
    check_in_date DATE NOT NULL,
    consecutive_days INT DEFAULT 1,
    created_at TIMESTAMP DEFAULT CURRENT_TIMESTAMP
);

基础签到功能

检查用户今日是否已签到，未签到则插入记录：

php签到逻辑实现

function checkIn($userId) {
    $today = date('Y-m-d');
    $pdo = new PDO('mysql:host=localhost;dbname=test', 'username', 'password');

    // 检查今日是否已签到
    $stmt = $pdo->prepare("SELECT id FROM user_check_in WHERE user_id = ? AND check_in_date = ?");
    $stmt->execute([$userId, $today]);

    if ($stmt->fetch()) {
        return ['status' => false, 'message' => '今日已签到'];
    }

    // 插入签到记录
    $stmt = $pdo->prepare("INSERT INTO user_check_in (user_id, check_in_date) VALUES (?, ?)");
    $stmt->execute([$userId, $today]);

    return ['status' => true, 'message' => '签到成功'];
}

连续签到计算

通过查询昨日签到记录判断是否连续签到：

function getConsecutiveDays($userId) {
    $pdo = new PDO('mysql:host=localhost;dbname=test', 'username', 'password');
    $today = date('Y-m-d');
    $yesterday = date('Y-m-d', strtotime('-1 day'));

    // 获取昨日签到记录
    $stmt = $pdo->prepare("SELECT consecutive_days FROM user_check_in WHERE user_id = ? AND check_in_date = ?");
    $stmt->execute([$userId, $yesterday]);
    $lastRecord = $stmt->fetch();

    $consecutiveDays = 1;
    if ($lastRecord) {
        $consecutiveDays = $lastRecord['consecutive_days'] + 1;
    }

    // 更新今日连续天数
    $stmt = $pdo->prepare("UPDATE user_check_in SET consecutive_days = ? WHERE user_id = ? AND check_in_date = ?");
    $stmt->execute([$consecutiveDays, $userId, $today]);

    return $consecutiveDays;
}

签到奖励系统

根据连续签到天数发放不同奖励：

function giveCheckInReward($userId, $consecutiveDays) {
    $rewards = [
        1 => ['points' => 10],
        3 => ['points' => 30],
        7 => ['points' => 100]
    ];

    if (isset($rewards[$consecutiveDays])) {
        $reward = $rewards[$consecutiveDays];
        // 发放奖励逻辑
        return ['status' => true, 'reward' => $reward];
    }

    return ['status' => false];
}

完整调用示例

$userId = 123;
$checkInResult = checkIn($userId);

if ($checkInResult['status']) {
    $consecutiveDays = getConsecutiveDays($userId);
    $rewardResult = giveCheckInReward($userId, $consecutiveDays);

    if ($rewardResult['status']) {
        echo "签到成功！连续签到{$consecutiveDays}天，获得{$rewardResult['reward']['points']}积分";
    } else {
        echo "签到成功！连续签到{$consecutiveDays}天";
    }
} else {
    echo $checkInResult['message'];
}