用php解决了这个问题

// Get the dates m-d for the next 45 days
for ($i = 0; $i <= 45; $i++) {
    $days = '+' . $i . ' days';
    $date = date('m-d', strtotime($days));
    if ($i != 45) {
        $DateQuery .= "'$date',";
    } else {
        $DateQuery .= "'$date'";
    }
}

并将我需要的日期直接插入到查询中

SELECT
       p.*,
       c.company 
    FROM
       products p 
       LEFT JOIN
          customers c 
          ON c.id = p.id 
    WHERE
       DATE_FORMAT(p.service_date, '%m-%d') IN 
       (
           $DateQuery
       )
       AND c.email_service = 1 
    ORDER BY
       p.service_date ASC

解决这个问题的最好办法是使用 DAYOFYEAR . 困难的部分是在年底45天内处理案件，在这种情况下，你必须拆分支票，看看日期是现在到年底还是年初到现在+45天；否则，您只需检查一年中的某一天是否介于现在和现在+45天之间。尝试用以下内容替换您的状况：

CASE WHEN DAYOFYEAR('2018-12-31')-DAYOFYEAR(NOW()) < 45 THEN
         DAYOFYEAR(p.service_date) BETWEEN DAYOFYEAR(NOW()) AND DAYOFYEAR('2018-12-31') OR
         DAYOFYEAR(p.service_date) BETWEEN 1 AND DAYOFYEAR(NOW() + INTERVAL 45 DAY)
     ELSE
         DAYOFYEAR(p.service_date) BETWEEN DAYOFYEAR(NOW()) AND DAYOFYEAR(NOW() + INTERVAL 45 DAY)
     END

您的完整查询应如下所示：

SELECT
   p.*,
   c.company 
FROM
   products p 
   LEFT JOIN
      customers c 
      ON c.id = p.id 
WHERE
    (CASE WHEN DAYOFYEAR('2018-12-31')-DAYOFYEAR(NOW()) < 45 THEN
              DAYOFYEAR(p.service_date) BETWEEN DAYOFYEAR(NOW()) AND DAYOFYEAR('2018-12-31') OR
              DAYOFYEAR(p.service_date) BETWEEN 1 AND DAYOFYEAR(NOW() + INTERVAL 45 DAY)
          ELSE
              DAYOFYEAR(p.service_date) BETWEEN DAYOFYEAR(NOW()) AND DAYOFYEAR(NOW() + INTERVAL 45 DAY)
          END)
    AND c.email_service = 1 
ORDER BY
    p.service_date ASC

如果你担心闰年，你可以改变 DAYOFYEAR('2018-12-31') 至 DAYOFYEAR(CONCAT(YEAR(NOW), '-12-31'))

我不知道为什么你最初的查询 Date_Format() 在柱子上( service_date )，已存储在mysql日期类型中，用于范围比较。它也不能使用任何索引（如果已定义）。
你只需要加/减 Interval 截止日期：

SELECT
   p.*,
   c.company 
FROM
   products p 
   LEFT JOIN
      customers c 
      ON c.id = p.id 
WHERE
   p.service_date >= CURDATE() 
   AND p.service_date <= (CURDATE() + INTERVAL 45 DAY) 
   AND c.email_service = 1 
ORDER BY
   p.service_date ASC