Question

说，我有一张给定here的表格。该表具有以下结构（测试日，学生姓名和获得的标记）

D       NAME    MARKS
2001-01-01  a   1
2001-01-04  a   4
2001-01-06  a   3
2001-01-08  a   3
2001-01-01  b   1
2001-01-10  b   15
2001-01-01  c   1
2001-01-06  c   2
2001-01-08  c   5
2001-01-10  c   7

我想通过给那些没有参加每个考试的学生的0分来更新表格。更新表应该看起来像this

D   NAME    MARKS
2001-01-01  a   1
2001-01-02  a   0
2001-01-04  a   4
2001-01-06  a   3
2001-01-08  a   3
2001-01-02  a   0
2001-01-01  b   1
2001-01-02  b   0
2001-01-04  b   0
2001-01-06  b   0
2001-01-08  b   0
2001-01-10  b   15
2001-01-01  c   1
2001-01-02  c   2
2001-01-04  c   0
2001-01-06  c   0
2001-01-08  c   5
2001-01-10  c   7

到目前为止，我能想出的唯一解决方案（非常慢的查询）是：

SELECT DISTINCT(D) FROM tableA;
SELECT DISTINCT(NAME) FROM tableA;

使用PHP，在嵌套循环中进行SQL查询

INSERT IGNORE (D,NAME,MARKS)($D,$NAME,0);

然而，整个代码花费了太多时间（以分钟为单位），因为行数超过50k。

有什么更好的建议吗？

Answer 1

可能会对可能的日期和可能的名称进行交叉连接，并且与当前结果保持联接： -

INSERT INTO A (D, name, marks)
SELECT Dates.D, Names.name, 0
FROM (SELECT DISTINCT D FROM A) Dates
CROSS JOIN  (SELECT DISTINCT name FROM A) Names
LEFT OUTER JOIN A
ON Dates.D = A.D AND Names.name = A.name
WHERE A.name IS NULL

请注意，这确实假设每天至少有一个人有标记。

如果你有一个名字表和一个日期表，那就更好了，只是在表A中使用了这些行的ID

如果您想在一系列日期中执行此操作，即使当天没有人标记： -

INSERT INTO A (D, name, marks)
SELECT Dates.aDate, Names.name, 0
FROM 
(
    SELECT DATE_ADD('2001-01-01', INTERVAL units.i + tens.i * 10 + hundreds.i * 100 DAY) AS aDate
    FROM (SELECT 0 AS i 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) units
    CROSS JOIN (SELECT 0 AS i 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) tens
    CROSS JOIN (SELECT 0 AS i 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) hundreds
    HAVING aDate BETWEEN '2001-01-01' AND '2001-12-30'
) Dates
CROSS JOIN  
(
    SELECT DISTINCT name 
    FROM A
) Names
LEFT OUTER JOIN A
ON Dates.aDate = A.D AND Names.name = A.name
WHERE A.name IS NULL

加快PHP中的MySQL查询速度

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