php - 去年这个日期的MySQL语句

我对这些php mysql语句感到困惑。

我需要从表中的同一列（金额）中获得四个不同的结果。

这些函数为以下项找到（金额）的总和：
今年迄今为止
去年给时间
去年捐款总额
最近30天内捐赠

它们似乎都是相同的变化。我不知道如何获取LastYearNow以显示正确的结果。如何显示从2016年初到今天的日期2016的结果？

public function getSum() {
        $query = $this->db->query("SELECT SUM(amount) as amount_sum FROM " . DB_PREFIX . "donate Where date_added >= (CURDATE() - INTERVAL 12 MONTH) ");
       return $query->row;
}
    public function lastThirty() {
        $query = $this->db->query("SELECT SUM(amount) as amount_sum FROM " . DB_PREFIX . "donate Where date_added >= (CURDATE() - INTERVAL 1 MONTH)" );
       return $query->row;
}
public function lastYearNow() {
        $query = $this->db->query("SELECT SUM(amount) as amount_sum FROM " . DB_PREFIX . "donate Where date_added >= (CURDATE() - INTERVAL 12 MONTH) ");
       return $query->row;
}
    public function lastYearTotal() {
        $query = $this->db->query("SELECT SUM(amount) as amount_sum FROM " . DB_PREFIX . "donate Where YEAR(date_added) = YEAR(DATE_SUB(CURDATE(), INTERVAL 1 YEAR))" );
       return $query->row;
}

最佳答案

查询lastYearNow当前正在计算总数的最后12个月。相反，如果要计算从一年的第一天到当前日期的去年总计，则需要使用BETWEEN子句：

SELECT SUM(amount) as amount_sum
FROM donate
WHERE date_added BETWEEN
MAKEDATE(YEAR(DATE_SUB(CURDATE(), INTERVAL 1 YEAR)), 1, 1)
AND
(CURDATE() - INTERVAL 12 MONTH)

警告：我前面没有安装MySQL，因此语法可能不太正确。