Mysql 不带WHERE语句的整数比较

我正在尝试编写一个MySQL语句，它将返回以下结果：

## Name | Day 0 | Day 1 | Day 2 | Day 3 | Day 4 | Day 5 |
##Jeff  |   0   |    3  |     1 |     2 |    1  |   1   |
##Larry |   1   |    1  |     4 |     4 |    1  |   0   |

基于每个员工每天执行的任务数量

我的数据库表如下所示：

员工

id

（INT）、

number

（VARCHAR）、

name

（VARCHAR）、

datestart

（VARCHAR）

项目

id

（INT）、

number

（VARCHAR）、

dateend

（DATETIME）

现在我用的是这样一句话：

SELECT 
a.name AS "Name",
count(abs(datediff(STR_TO_DATE(a.dateStarted, '%Y-%m-%d %H:%i:%s'), b.dateEnded))) AS "Day 0",
count(abs(datediff(STR_TO_DATE(a.dateStarted, '%Y-%m-%d %H:%i:%s'), b.dateEnded))) AS "Day 1",
count(abs(datediff(STR_TO_DATE(a.dateStarted, '%Y-%m-%d %H:%i:%s'), b.dateEnded))) AS "Day 2",
count(abs(datediff(STR_TO_DATE(a.dateStarted, '%Y-%m-%d %H:%i:%s'), b.dateEnded))) AS "Day 3",
count(abs(datediff(STR_TO_DATE(a.dateStarted, '%Y-%m-%d %H:%i:%s'), b.dateEnded))) AS "Day 4",
count(abs(datediff(STR_TO_DATE(a.dateStarted, '%Y-%m-%d %H:%i:%s'), b.dateEnded))) AS "Day 5"
FROM employee a, project b
WHERE b.number=a.number 
AND "Day 0" = 0
AND "Day 1" = 1
AND "Day 2" = 2
AND "Day 3" = 3
AND "Day 4" = 4
AND "Day 5" >= 5

电流输出

上述声明有效，但由于某些原因，它不能提供上述要求中提到的预期结果。关于如何修复/更改它，有什么想法吗

编辑

如果我取出：

AND "Day 0" = 0
AND "Day 1" = 1
AND "Day 2" = 2
AND "Day 3" = 3
AND "Day 4" = 4
AND "Day 5" >= 5

然后打印出：

## Name | Day 0 | Day 1 | Day 2 | Day 3 | Day 4 | Day 5 |
##Jeff  |   9   |    9  |     9 |     9 |    9  |   9   |

试试这个：

SELECT
  emp.name AS '## Name',
  (SELECT COUNT(*)     
   FROM project p JOIN employee e ON p.number = e.number
   WHERE e.name = emp.name 
     AND datediff(str_to_date(p.dateEnded, '%Y-%m-%d'), e.dateStarted) = 0
   ) AS 'Day 0'
  ,(SELECT COUNT(*)
    FROM project p JOIN employee e ON p.number = e.number
    WHERE e.name = emp.name 
      AND datediff(str_to_date(p.dateEnded, '%Y-%m-%d'), e.dateStarted) = 1
   ) AS 'Day 1'
  ,(SELECT COUNT(*)
    FROM project p JOIN employee e ON p.number = e.number
    WHERE e.name = emp.name 
      AND datediff(str_to_date(p.dateEnded, '%Y-%m-%d'), e.dateStarted) = 2
   ) AS 'Day 2'
  ,(SELECT COUNT(*)
    FROM project p JOIN employee e ON p.number = e.number
    WHERE e.name = emp.name 
      AND datediff(str_to_date(p.dateEnded, '%Y-%m-%d'), e.dateStarted) = 3
   ) AS 'Day 3'
  ,(SELECT COUNT(*)
    FROM project p JOIN employee e ON p.number = e.number
    WHERE e.name = emp.name 
      AND datediff(str_to_date(p.dateEnded, '%Y-%m-%d'), e.dateStarted) = 4
   ) AS 'Day 4'
  ,(SELECT COUNT(*)
    FROM project p JOIN employee e ON p.number = e.number
    WHERE e.name = emp.name 
      AND datediff(str_to_date(p.dateEnded, '%Y-%m-%d'), e.dateStarted) >= 5
   ) AS 'Day 5'
FROM employee emp
GROUP BY emp.name

请参阅（根据提供的信息对您的数据进行了一些假设）。

尝试以下方法：

SELECT
  emp.name AS '## Name',
  (SELECT COUNT(*)     
   FROM project p JOIN employee e ON p.number = e.number
   WHERE e.name = emp.name 
     AND datediff(str_to_date(p.dateEnded, '%Y-%m-%d'), e.dateStarted) = 0
   ) AS 'Day 0'
  ,(SELECT COUNT(*)
    FROM project p JOIN employee e ON p.number = e.number
    WHERE e.name = emp.name 
      AND datediff(str_to_date(p.dateEnded, '%Y-%m-%d'), e.dateStarted) = 1
   ) AS 'Day 1'
  ,(SELECT COUNT(*)
    FROM project p JOIN employee e ON p.number = e.number
    WHERE e.name = emp.name 
      AND datediff(str_to_date(p.dateEnded, '%Y-%m-%d'), e.dateStarted) = 2
   ) AS 'Day 2'
  ,(SELECT COUNT(*)
    FROM project p JOIN employee e ON p.number = e.number
    WHERE e.name = emp.name 
      AND datediff(str_to_date(p.dateEnded, '%Y-%m-%d'), e.dateStarted) = 3
   ) AS 'Day 3'
  ,(SELECT COUNT(*)
    FROM project p JOIN employee e ON p.number = e.number
    WHERE e.name = emp.name 
      AND datediff(str_to_date(p.dateEnded, '%Y-%m-%d'), e.dateStarted) = 4
   ) AS 'Day 4'
  ,(SELECT COUNT(*)
    FROM project p JOIN employee e ON p.number = e.number
    WHERE e.name = emp.name 
      AND datediff(str_to_date(p.dateEnded, '%Y-%m-%d'), e.dateStarted) >= 5
   ) AS 'Day 5'
FROM employee emp
GROUP BY emp.name

请参阅（根据提供的信息对您的数据进行了一些假设）。

尝试以下方法：

SELECT
  emp.name AS '## Name',
  (SELECT COUNT(*)     
   FROM project p JOIN employee e ON p.number = e.number
   WHERE e.name = emp.name 
     AND datediff(str_to_date(p.dateEnded, '%Y-%m-%d'), e.dateStarted) = 0
   ) AS 'Day 0'
  ,(SELECT COUNT(*)
    FROM project p JOIN employee e ON p.number = e.number
    WHERE e.name = emp.name 
      AND datediff(str_to_date(p.dateEnded, '%Y-%m-%d'), e.dateStarted) = 1
   ) AS 'Day 1'
  ,(SELECT COUNT(*)
    FROM project p JOIN employee e ON p.number = e.number
    WHERE e.name = emp.name 
      AND datediff(str_to_date(p.dateEnded, '%Y-%m-%d'), e.dateStarted) = 2
   ) AS 'Day 2'
  ,(SELECT COUNT(*)
    FROM project p JOIN employee e ON p.number = e.number
    WHERE e.name = emp.name 
      AND datediff(str_to_date(p.dateEnded, '%Y-%m-%d'), e.dateStarted) = 3
   ) AS 'Day 3'
  ,(SELECT COUNT(*)
    FROM project p JOIN employee e ON p.number = e.number
    WHERE e.name = emp.name 
      AND datediff(str_to_date(p.dateEnded, '%Y-%m-%d'), e.dateStarted) = 4
   ) AS 'Day 4'
  ,(SELECT COUNT(*)
    FROM project p JOIN employee e ON p.number = e.number
    WHERE e.name = emp.name 
      AND datediff(str_to_date(p.dateEnded, '%Y-%m-%d'), e.dateStarted) >= 5
   ) AS 'Day 5'
FROM employee emp
GROUP BY emp.name

请参阅（根据提供的信息对您的数据进行了一些假设）。

尝试以下方法：

SELECT
  emp.name AS '## Name',
  (SELECT COUNT(*)     
   FROM project p JOIN employee e ON p.number = e.number
   WHERE e.name = emp.name 
     AND datediff(str_to_date(p.dateEnded, '%Y-%m-%d'), e.dateStarted) = 0
   ) AS 'Day 0'
  ,(SELECT COUNT(*)
    FROM project p JOIN employee e ON p.number = e.number
    WHERE e.name = emp.name 
      AND datediff(str_to_date(p.dateEnded, '%Y-%m-%d'), e.dateStarted) = 1
   ) AS 'Day 1'
  ,(SELECT COUNT(*)
    FROM project p JOIN employee e ON p.number = e.number
    WHERE e.name = emp.name 
      AND datediff(str_to_date(p.dateEnded, '%Y-%m-%d'), e.dateStarted) = 2
   ) AS 'Day 2'
  ,(SELECT COUNT(*)
    FROM project p JOIN employee e ON p.number = e.number
    WHERE e.name = emp.name 
      AND datediff(str_to_date(p.dateEnded, '%Y-%m-%d'), e.dateStarted) = 3
   ) AS 'Day 3'
  ,(SELECT COUNT(*)
    FROM project p JOIN employee e ON p.number = e.number
    WHERE e.name = emp.name 
      AND datediff(str_to_date(p.dateEnded, '%Y-%m-%d'), e.dateStarted) = 4
   ) AS 'Day 4'
  ,(SELECT COUNT(*)
    FROM project p JOIN employee e ON p.number = e.number
    WHERE e.name = emp.name 
      AND datediff(str_to_date(p.dateEnded, '%Y-%m-%d'), e.dateStarted) >= 5
   ) AS 'Day 5'
FROM employee emp
GROUP BY emp.name

请参阅（根据提供的信息对您的数据进行了一些假设）。

尝试以下方法：

SELECT a.name AS "Name", 
       SUM(noOfDays = 0) AS "Day 0", SUM(noOfDays = 1) AS "Day 1", 
       SUM(noOfDays = 2) AS "Day 2", SUM(noOfDays = 3) AS "Day 3", 
       SUM(noOfDays = 4) AS "Day 4", SUM(noOfDays >= 5) AS "Day 5", 
       COUNT(1) AS "Total Days"
FROM (SELECT a.name, DATEDIFF(DATE(b.dateEnded), DATE(a.dateStarted)) noOfDays
      FROM employee a INNER JOIN project b ON b.number = a.number 
      WHERE b.dateEnded IS NOT NULL
    ) AS A
GROUP BY a.name;

SELECT a.name AS "Name", 
       SUM(noOfDays = 0) AS "Day 0", SUM(noOfDays = 1) AS "Day 1", 
       SUM(noOfDays = 2) AS "Day 2", SUM(noOfDays = 3) AS "Day 3", 
       SUM(noOfDays = 4) AS "Day 4", SUM(noOfDays >= 5) AS "Day 5" 
FROM (SELECT a.number, a.name, DATEDIFF(DATE(b.dateEnded), DATE(a.dateStarted)) noOfDays
      FROM employee a INNER JOIN project b ON (b.number = a.number and b.dateEnded>a.dateStarted)
    ) AS a
GROUP BY a.name

检查

试试这个：

SELECT a.name AS "Name", 
       SUM(noOfDays = 0) AS "Day 0", SUM(noOfDays = 1) AS "Day 1", 
       SUM(noOfDays = 2) AS "Day 2", SUM(noOfDays = 3) AS "Day 3", 
       SUM(noOfDays = 4) AS "Day 4", SUM(noOfDays >= 5) AS "Day 5", 
       COUNT(1) AS "Total Days"
FROM (SELECT a.name, DATEDIFF(DATE(b.dateEnded), DATE(a.dateStarted)) noOfDays
      FROM employee a INNER JOIN project b ON b.number = a.number 
      WHERE b.dateEnded IS NOT NULL
    ) AS A
GROUP BY a.name;

SELECT a.name AS "Name", 
       SUM(noOfDays = 0) AS "Day 0", SUM(noOfDays = 1) AS "Day 1", 
       SUM(noOfDays = 2) AS "Day 2", SUM(noOfDays = 3) AS "Day 3", 
       SUM(noOfDays = 4) AS "Day 4", SUM(noOfDays >= 5) AS "Day 5" 
FROM (SELECT a.number, a.name, DATEDIFF(DATE(b.dateEnded), DATE(a.dateStarted)) noOfDays
      FROM employee a INNER JOIN project b ON (b.number = a.number and b.dateEnded>a.dateStarted)
    ) AS a
GROUP BY a.name

检查

试试这个：

SELECT a.name AS "Name", 
       SUM(noOfDays = 0) AS "Day 0", SUM(noOfDays = 1) AS "Day 1", 
       SUM(noOfDays = 2) AS "Day 2", SUM(noOfDays = 3) AS "Day 3", 
       SUM(noOfDays = 4) AS "Day 4", SUM(noOfDays >= 5) AS "Day 5", 
       COUNT(1) AS "Total Days"
FROM (SELECT a.name, DATEDIFF(DATE(b.dateEnded), DATE(a.dateStarted)) noOfDays
      FROM employee a INNER JOIN project b ON b.number = a.number 
      WHERE b.dateEnded IS NOT NULL
    ) AS A
GROUP BY a.name;

SELECT a.name AS "Name", 
       SUM(noOfDays = 0) AS "Day 0", SUM(noOfDays = 1) AS "Day 1", 
       SUM(noOfDays = 2) AS "Day 2", SUM(noOfDays = 3) AS "Day 3", 
       SUM(noOfDays = 4) AS "Day 4", SUM(noOfDays >= 5) AS "Day 5" 
FROM (SELECT a.number, a.name, DATEDIFF(DATE(b.dateEnded), DATE(a.dateStarted)) noOfDays
      FROM employee a INNER JOIN project b ON (b.number = a.number and b.dateEnded>a.dateStarted)
    ) AS a
GROUP BY a.name

检查

试试这个：

SELECT a.name AS "Name", 
       SUM(noOfDays = 0) AS "Day 0", SUM(noOfDays = 1) AS "Day 1", 
       SUM(noOfDays = 2) AS "Day 2", SUM(noOfDays = 3) AS "Day 3", 
       SUM(noOfDays = 4) AS "Day 4", SUM(noOfDays >= 5) AS "Day 5", 
       COUNT(1) AS "Total Days"
FROM (SELECT a.name, DATEDIFF(DATE(b.dateEnded), DATE(a.dateStarted)) noOfDays
      FROM employee a INNER JOIN project b ON b.number = a.number 
      WHERE b.dateEnded IS NOT NULL
    ) AS A
GROUP BY a.name;

SELECT a.name AS "Name", 
       SUM(noOfDays = 0) AS "Day 0", SUM(noOfDays = 1) AS "Day 1", 
       SUM(noOfDays = 2) AS "Day 2", SUM(noOfDays = 3) AS "Day 3", 
       SUM(noOfDays = 4) AS "Day 4", SUM(noOfDays >= 5) AS "Day 5" 
FROM (SELECT a.number, a.name, DATEDIFF(DATE(b.dateEnded), DATE(a.dateStarted)) noOfDays
      FROM employee a INNER JOIN project b ON (b.number = a.number and b.dateEnded>a.dateStarted)
    ) AS a
GROUP BY a.name

检查

---

可以通过执行以下join语句或where来改进以前的解决方案

确保项目上有索引（唯一？）（编号、日期结束） 您还可以尝试添加员工索引（编号、日期开始）

空值通常不是任何索引的一部分（因为它们是空的），您可以尝试将该字段的默认值更新为“0000-00-00”，这样它将被索引，并可能增加速度

试试这个：

SELECT a.name AS "Name", 
       SUM(noOfDays = 0) AS "Day 0", SUM(noOfDays = 1) AS "Day 1", 
       SUM(noOfDays = 2) AS "Day 2", SUM(noOfDays = 3) AS "Day 3", 
       SUM(noOfDays = 4) AS "Day 4", SUM(noOfDays >= 5) AS "Day 5", 
       COUNT(1) AS "Total Days"
FROM (SELECT a.name, DATEDIFF(DATE(b.dateEnded), DATE(a.dateStarted)) noOfDays
      FROM employee a INNER JOIN project b ON b.number = a.number 
      WHERE b.dateEnded IS NOT NULL
    ) AS A
GROUP BY a.name;

SELECT a.name AS "Name", 
       SUM(noOfDays = 0) AS "Day 0", SUM(noOfDays = 1) AS "Day 1", 
       SUM(noOfDays = 2) AS "Day 2", SUM(noOfDays = 3) AS "Day 3", 
       SUM(noOfDays = 4) AS "Day 4", SUM(noOfDays >= 5) AS "Day 5" 
FROM (SELECT a.number, a.name, DATEDIFF(DATE(b.dateEnded), DATE(a.dateStarted)) noOfDays
      FROM employee a INNER JOIN project b ON (b.number = a.number and b.dateEnded>a.dateStarted)
    ) AS a
GROUP BY a.name

检查

---

可以通过执行以下join语句或where来改进以前的解决方案

确保项目上有索引（唯一？）（编号、日期结束） 您还可以尝试添加员工索引（编号、日期开始）

空值通常不是任何索引的一部分（因为它们是空的），您可以尝试将该字段的默认值更新为“0000-00-00”，这样它将被索引，并可能增加速度

试试这个：

SELECT a.name AS "Name", 
       SUM(noOfDays = 0) AS "Day 0", SUM(noOfDays = 1) AS "Day 1", 
       SUM(noOfDays = 2) AS "Day 2", SUM(noOfDays = 3) AS "Day 3", 
       SUM(noOfDays = 4) AS "Day 4", SUM(noOfDays >= 5) AS "Day 5", 
       COUNT(1) AS "Total Days"
FROM (SELECT a.name, DATEDIFF(DATE(b.dateEnded), DATE(a.dateStarted)) noOfDays
      FROM employee a INNER JOIN project b ON b.number = a.number 
      WHERE b.dateEnded IS NOT NULL
    ) AS A
GROUP BY a.name;

SELECT a.name AS "Name", 
       SUM(noOfDays = 0) AS "Day 0", SUM(noOfDays = 1) AS "Day 1", 
       SUM(noOfDays = 2) AS "Day 2", SUM(noOfDays = 3) AS "Day 3", 
       SUM(noOfDays = 4) AS "Day 4", SUM(noOfDays >= 5) AS "Day 5" 
FROM (SELECT a.number, a.name, DATEDIFF(DATE(b.dateEnded), DATE(a.dateStarted)) noOfDays
      FROM employee a INNER JOIN project b ON (b.number = a.number and b.dateEnded>a.dateStarted)
    ) AS a
GROUP BY a.name

检查

---

可以通过执行以下join语句或where来改进以前的解决方案

确保项目上有索引（唯一？）（编号、日期结束） 您还可以尝试添加员工索引（编号、日期开始）

空值通常不是任何索引的一部分（因为它们是空的），您可以尝试将该字段的默认值更新为“0000-00-00”，这样它将被索引，并可能增加速度

试试这个：

SELECT a.name AS "Name", 
       SUM(noOfDays = 0) AS "Day 0", SUM(noOfDays = 1) AS "Day 1", 
       SUM(noOfDays = 2) AS "Day 2", SUM(noOfDays = 3) AS "Day 3", 
       SUM(noOfDays = 4) AS "Day 4", SUM(noOfDays >= 5) AS "Day 5", 
       COUNT(1) AS "Total Days"
FROM (SELECT a.name, DATEDIFF(DATE(b.dateEnded), DATE(a.dateStarted)) noOfDays
      FROM employee a INNER JOIN project b ON b.number = a.number 
      WHERE b.dateEnded IS NOT NULL
    ) AS A
GROUP BY a.name;

SELECT a.name AS "Name", 
       SUM(noOfDays = 0) AS "Day 0", SUM(noOfDays = 1) AS "Day 1", 
       SUM(noOfDays = 2) AS "Day 2", SUM(noOfDays = 3) AS "Day 3", 
       SUM(noOfDays = 4) AS "Day 4", SUM(noOfDays >= 5) AS "Day 5" 
FROM (SELECT a.number, a.name, DATEDIFF(DATE(b.dateEnded), DATE(a.dateStarted)) noOfDays
      FROM employee a INNER JOIN project b ON (b.number = a.number and b.dateEnded>a.dateStarted)
    ) AS a
GROUP BY a.name

检查

---

可以通过执行以下join语句或where来改进以前的解决方案

确保项目上有索引（唯一？）（编号、日期结束） 您还可以尝试添加员工索引（编号、日期开始）

空值通常不是任何索引的一部分（因为它们是空的），您可以尝试将该字段的默认值更新为“0000-00-00”，这样它将被索引，并可能增加速度

试试这个：

SELECT a.name AS "Name", 
       SUM(noOfDays = 0) AS "Day 0", SUM(noOfDays = 1) AS "Day 1", 
       SUM(noOfDays = 2) AS "Day 2", SUM(noOfDays = 3) AS "Day 3", 
       SUM(noOfDays = 4) AS "Day 4", SUM(noOfDays >= 5) AS "Day 5", 
       COUNT(1) AS "Total Days"
FROM (SELECT a.name, DATEDIFF(DATE(b.dateEnded), DATE(a.dateStarted)) noOfDays
      FROM employee a INNER JOIN project b ON b.number = a.number 
      WHERE b.dateEnded IS NOT NULL
    ) AS A
GROUP BY a.name;

SELECT a.name AS "Name", 
       SUM(noOfDays = 0) AS "Day 0", SUM(noOfDays = 1) AS "Day 1", 
       SUM(noOfDays = 2) AS "Day 2", SUM(noOfDays = 3) AS "Day 3", 
       SUM(noOfDays = 4) AS "Day 4", SUM(noOfDays >= 5) AS "Day 5" 
FROM (SELECT a.number, a.name, DATEDIFF(DATE(b.dateEnded), DATE(a.dateStarted)) noOfDays
      FROM employee a INNER JOIN project b ON (b.number = a.number and b.dateEnded>a.dateStarted)
    ) AS a
GROUP BY a.name

检查

---

使用前面一个答案中的SQL FIDLE数据，我尝试了以下方法：-

SELECT Sub0.name AS "Name", 
    SUM(IF(DayDesc = 0 AND DaysCount IS NOT NULL, 1, 0)) AS "Day 0",
    SUM(IF(DayDesc = 1 AND DaysCount IS NOT NULL, 1, 0)) AS "Day 1",
    SUM(IF(DayDesc = 2 AND DaysCount IS NOT NULL, 1, 0)) AS "Day 2",
    SUM(IF(DayDesc = 3 AND DaysCount IS NOT NULL, 1, 0)) AS "Day 3",
    SUM(IF(DayDesc = 4 AND DaysCount IS NOT NULL, 1, 0)) AS "Day 4",
    SUM(IF(DayDesc = 5 AND DaysCount IS NOT NULL, 1, 0)) AS "Day 5"
FROM
(
    SELECT DISTINCT name FROM employee
)Sub0
CROSS JOIN
(
    SELECT 0 AS DayMin, 0 AS DayMax, 0 AS DayDesc
    UNION
    SELECT 1 AS DayMin, 1 AS DayMax, 1 AS DayDesc
    UNION
    SELECT 2 AS DayMin, 2 AS DayMax, 2 AS DayDesc
    UNION
    SELECT 3 AS DayMin, 3 AS DayMax, 3 AS DayDesc
    UNION
    SELECT 4 AS DayMin, 4 AS DayMax, 4 AS DayDesc
    UNION
    SELECT 5 AS DayMin, 999999999 AS DayMax, 5 AS DayDesc
) Sub1
LEFT OUTER JOIN 
(
    SELECT a.name, ABS(DATEDIFF(STR_TO_DATE(a.dateStarted, '%Y-%m-%d %H:%i:%s'), b.dateEnded)) DaysCount
    FROM employee a
    INNER JOIN project b
    ON b.number = a.number 
) Sub2
ON Sub2.DaysCount BETWEEN Sub1.DayMin AND Sub1.DayMax
AND Sub2.name = Sub0.name
GROUP BY Sub0.name

似乎给出了正确的结果，而且速度相当快

SQL fiddle在此：-

效率更高一点（但不能处理没有项目的员工）：-

SQL fiddle在此：-

---

使用前面一个答案中的SQL FIDLE数据，我尝试了以下方法：-

SELECT Sub0.name AS "Name", 
    SUM(IF(DayDesc = 0 AND DaysCount IS NOT NULL, 1, 0)) AS "Day 0",
    SUM(IF(DayDesc = 1 AND DaysCount IS NOT NULL, 1, 0)) AS "Day 1",
    SUM(IF(DayDesc = 2 AND DaysCount IS NOT NULL, 1, 0)) AS "Day 2",
    SUM(IF(DayDesc = 3 AND DaysCount IS NOT NULL, 1, 0)) AS "Day 3",
    SUM(IF(DayDesc = 4 AND DaysCount IS NOT NULL, 1, 0)) AS "Day 4",
    SUM(IF(DayDesc = 5 AND DaysCount IS NOT NULL, 1, 0)) AS "Day 5"
FROM
(
    SELECT DISTINCT name FROM employee
)Sub0
CROSS JOIN
(
    SELECT 0 AS DayMin, 0 AS DayMax, 0 AS DayDesc
    UNION
    SELECT 1 AS DayMin, 1 AS DayMax, 1 AS DayDesc
    UNION
    SELECT 2 AS DayMin, 2 AS DayMax, 2 AS DayDesc
    UNION
    SELECT 3 AS DayMin, 3 AS DayMax, 3 AS DayDesc
    UNION
    SELECT 4 AS DayMin, 4 AS DayMax, 4 AS DayDesc
    UNION
    SELECT 5 AS DayMin, 999999999 AS DayMax, 5 AS DayDesc
) Sub1
LEFT OUTER JOIN 
(
    SELECT a.name, ABS(DATEDIFF(STR_TO_DATE(a.dateStarted, '%Y-%m-%d %H:%i:%s'), b.dateEnded)) DaysCount
    FROM employee a
    INNER JOIN project b
    ON b.number = a.number 
) Sub2
ON Sub2.DaysCount BETWEEN Sub1.DayMin AND Sub1.DayMax
AND Sub2.name = Sub0.name
GROUP BY Sub0.name

似乎给出了正确的结果，而且速度相当快

SQL fiddle在此：-

效率更高一点（但不能处理没有项目的员工）：-

SQL fiddle在此：-

---

使用前面一个答案中的SQL FIDLE数据，我尝试了以下方法：-

SELECT Sub0.name AS "Name", 
    SUM(IF(DayDesc = 0 AND DaysCount IS NOT NULL, 1, 0)) AS "Day 0",
    SUM(IF(DayDesc = 1 AND DaysCount IS NOT NULL, 1, 0)) AS "Day 1",
    SUM(IF(DayDesc = 2 AND DaysCount IS NOT NULL, 1, 0)) AS "Day 2",
    SUM(IF(DayDesc = 3 AND DaysCount IS NOT NULL, 1, 0)) AS "Day 3",
    SUM(IF(DayDesc = 4 AND DaysCount IS NOT NULL, 1, 0)) AS "Day 4",
    SUM(IF(DayDesc = 5 AND DaysCount IS NOT NULL, 1, 0)) AS "Day 5"
FROM
(
    SELECT DISTINCT name FROM employee
)Sub0
CROSS JOIN
(
    SELECT 0 AS DayMin, 0 AS DayMax, 0 AS DayDesc
    UNION
    SELECT 1 AS DayMin, 1 AS DayMax, 1 AS DayDesc
    UNION
    SELECT 2 AS DayMin, 2 AS DayMax, 2 AS DayDesc
    UNION
    SELECT 3 AS DayMin, 3 AS DayMax, 3 AS DayDesc
    UNION
    SELECT 4 AS DayMin, 4 AS DayMax, 4 AS DayDesc
    UNION
    SELECT 5 AS DayMin, 999999999 AS DayMax, 5 AS DayDesc
) Sub1
LEFT OUTER JOIN 
(
    SELECT a.name, ABS(DATEDIFF(STR_TO_DATE(a.dateStarted, '%Y-%m-%d %H:%i:%s'), b.dateEnded)) DaysCount
    FROM employee a
    INNER JOIN project b
    ON b.number = a.number 
) Sub2
ON Sub2.DaysCount BETWEEN Sub1.DayMin AND Sub1.DayMax
AND Sub2.name = Sub0.name
GROUP BY Sub0.name

似乎给出了正确的结果，而且速度相当快

SQL fiddle在此：-

效率更高一点（但不能处理没有项目的员工）：-

SQL fiddle在此：-

---

使用前面一个答案中的SQL FIDLE数据，我尝试了以下方法：-

SELECT Sub0.name AS "Name", 
    SUM(IF(DayDesc = 0 AND DaysCount IS NOT NULL, 1, 0)) AS "Day 0",
    SUM(IF(DayDesc = 1 AND DaysCount IS NOT NULL, 1, 0)) AS "Day 1",
    SUM(IF(DayDesc = 2 AND DaysCount IS NOT NULL, 1, 0)) AS "Day 2",
    SUM(IF(DayDesc = 3 AND DaysCount IS NOT NULL, 1, 0)) AS "Day 3",
    SUM(IF(DayDesc = 4 AND DaysCount IS NOT NULL, 1, 0)) AS "Day 4",
    SUM(IF(DayDesc = 5 AND DaysCount IS NOT NULL, 1, 0)) AS "Day 5"
FROM
(
    SELECT DISTINCT name FROM employee
)Sub0
CROSS JOIN
(
    SELECT 0 AS DayMin, 0 AS DayMax, 0 AS DayDesc
    UNION
    SELECT 1 AS DayMin, 1 AS DayMax, 1 AS DayDesc
    UNION
    SELECT 2 AS DayMin, 2 AS DayMax, 2 AS DayDesc
    UNION
    SELECT 3 AS DayMin, 3 AS DayMax, 3 AS DayDesc
    UNION
    SELECT 4 AS DayMin, 4 AS DayMax, 4 AS DayDesc
    UNION
    SELECT 5 AS DayMin, 999999999 AS DayMax, 5 AS DayDesc
) Sub1
LEFT OUTER JOIN 
(
    SELECT a.name, ABS(DATEDIFF(STR_TO_DATE(a.dateStarted, '%Y-%m-%d %H:%i:%s'), b.dateEnded)) DaysCount
    FROM employee a
    INNER JOIN project b
    ON b.number = a.number 
) Sub2
ON Sub2.DaysCount BETWEEN Sub1.DayMin AND Sub1.DayMax
AND Sub2.name = Sub0.name
GROUP BY Sub0.name

似乎给出了正确的结果，而且速度相当快

SQL fiddle在此：-

效率更高一点（但不能处理没有项目的员工）：-

SQL fiddle在此：-

---

---

员工和项目之间没有连接条件？

STR\u TO\u日期（a.datestart，%Y-%m-%d%H:%i:%s）的意义是什么？你的日期是以字符串形式存储的吗？@hashbrown，如果这两个id是两个独立的东西（个人id，项目id），那么加入它们就没有意义了。如果它们确实引用了相同的东西（名称非常糟糕），那么您可以将它们合并到一个表中。@PhilPerry。这就是为什么我要问这两个表之间的关系/连接条件是什么？您如何知道哪个员工与哪个项目相关？仍在挣扎？考虑提供适当的DDL（和/或SqLoFIDLE）以及在<代码>雇员>代码>和<代码>项目< /代码>之间的连接结果条件吗？<代码> StruthtoSyd（A.DestestARTE，'%Y-%M%%D%H:%I:%S'）< /C> >的意义是什么？你的日期是以字符串形式存储的吗？@hashbrown，如果这两个id是两个独立的东西（个人id，项目id），那么加入它们就没有意义了。如果它们确实引用了相同的东西（名称非常糟糕），那么您可以将它们合并到一个表中。@PhilPerry。这就是为什么我要问这两个表之间的关系/连接条件是什么？您如何知道哪个员工与哪个项目相关？仍在挣扎？考虑提供适当的DDL（和/或SqLFIDLE），连同期望的结果，在<代码>雇员< /代码>和<代码>项目< /代码>之间的连接条件？“代码> Struthtox日期的意义是什么？（A.DAT）