如果您有一个日历表，那么您的问题就解决了：

SELECT DATE(`date`) AS RegistrationDate, COUNT(u.id) AS NumberOfRegistrations
FROM Calendar c LEFT JOIN
     Users u
     ON DATE(u.date) = c.date
WHERE c.date >= '2018-05-01' AND c.date < '2018-05-14'
GROUP BY RegistrationDate;

一 Calendar 表可以是非常有用的，你可以找到一个谷歌搜索。
你可以用一个 numbers table，如果你手边有一张的话。
如果没有这些，您可以生成所需的日期：

SELECT DATE(`date`) AS RegistrationDate, COUNT(u.id) AS NumberOfRegistrations
FROM (SELECT DATE('2018-05-01') as dte UNION ALL
             DATE('2018-05-02') as dte UNION ALL
             DATE('2018-05-03') as dte UNION ALL
             DATE('2018-05-04') as dte UNION ALL
             DATE('2018-05-05') as dte UNION ALL
             DATE('2018-05-06') as dte UNION ALL
             DATE('2018-05-07') as dte UNION ALL
             DATE('2018-05-08') as dte UNION ALL
             DATE('2018-05-09') as dte UNION ALL
             DATE('2018-05-10') as dte UNION ALL
             DATE('2018-05-11') as dte UNION ALL
             DATE('2018-05-12') as dte UNION ALL
             DATE('2018-05-13') as dte UNION ALL
             DATE('2018-05-14') as dte
     ) c LEFT JOIN
     Users u
     ON DATE(u.date) = c.dte
WHERE c.dte >= '2018-05-01' AND c.dte < '2018-05-14'
GROUP BY RegistrationDate;

笔记：
对日期yyyy-mm-dd使用标准日期格式。
sql中的标准字符串分隔符是单引号。使用它，除非你有充分的理由使用双引号。
不要使用 between 有日期。目前还不清楚您是否打算包括2018-05-13，但您的查询可能会排除该日期的任何数据（由于时间成分）。

在我的评论中，这是我处理你问题的方法。
（可能还有其他方法）
联接中的第一个表创建日期范围。

SELECT
  DATE(`gen_date`) AS RegistrationDate, 
  COUNT(`id`) AS NumberOfRegistrations 
FROM 
  (
    SELECT * 
    FROM
      (SELECT adddate('1970-01-01', t4 * 10000 + t3 * 1000 + t2 * 100 + t1 * 10 + t0) gen_date 
        FROM 
          (select 0 t0 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t0, 
          (select 0 t1 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9)  t1, 
          (select 0 t2 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9)  t2,  
          (select 0 t3 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9)  t3, 
          (select 0 t4 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9)  t4
      ) v 
    WHERE 
      gen_date BETWEEN "2018-05-01" AND "2018-05-13"
  ) d 
LEFT OUTER JOIN users u ON Date(`gen_date`) = Date(`date`)
GROUP BY
  gen_date

看到它在这里工作了吗