Question

我仍然是SQL的新手，我还没有完全理解我的代码中的问题来自哪里。下面的代码主要来自我的工作，所以我没有从头开始编写代码。该代码基于它收集了大量不同的信息和过滤器。如果您查看代码，您将看到学生有许多与其相关的观察学生。代码的第一个版本返回所有拥有observation_student且observe_id = 2567的学生的信息。这似乎与以下代码一起正常工作：

SELECT DISTINCT
SUBSTRING(s.osis_id,INSTR(s.osis_id,'-')+1) AS osid,
s.id AS student_id,
CONCAT(s.last_name, ' ',s.first_name) AS sname
FROM students s

# course info
INNER JOIN 
(
    SELECT c.id AS cid,
    c.description AS cname,
    cs.date_end,
    cs.student_id,
    gl.description AS grade,
    c.gradelevel_id
    FROM courses_students cs
    INNER JOIN courses c ON c.id = cs.course_id
    INNER JOIN gradelevels gl ON gl.id = c.gradelevel_id
    WHERE
    IFNULL(cs.date_end, NOW()) >= NOW()
    AND IFNULL(c.date_end, NOW()) >= NOW()
    AND c.school_id = 1509
    AND c.subject_id = 24
) AS cs ON cs.student_id = s.id

# RTI flag info
INNER JOIN 
(
    SELECT os.id,
    os.student_id
    FROM observations o
    INNER JOIN observations_students os ON os.observation_id = 2567
    WHERE
    o.school_id = 1509
) AS os ON os.student_id = s.id

LEFT JOIN schools_students ss ON ss.student_id = s.id
WHERE s.active = 1
AND ss.school_id = 1509
AND IFNULL(ss.date_end,NOW()) >= NOW()
AND cs.gradelevel_id BETWEEN 10 AND 16

在此之后我想做的是对于每个有2567观察的学生，我想找到学生所拥有的2009年观察数。为此，我将添加另一个LEFT JOIN，完成的代码如下所示：

SELECT DISTINCT
SUBSTRING(s.osis_id,INSTR(s.osis_id,'-')+1) AS osid,
s.id AS student_id,
CONCAT(s.last_name, ' ',s.first_name) AS sname,
COUNT(fdos.id) AS fd_count
FROM students s

# course info
INNER JOIN 
(
    SELECT c.id AS cid,
    c.description AS cname,
    cs.date_end,
    cs.student_id,
    gl.description AS grade,
    c.gradelevel_id
    FROM courses_students cs
    INNER JOIN courses c ON c.id = cs.course_id
    INNER JOIN gradelevels gl ON gl.id = c.gradelevel_id
    WHERE
    IFNULL(cs.date_end, NOW()) >= NOW()
    AND IFNULL(c.date_end, NOW()) >= NOW()
    AND c.school_id = 1509
    AND c.subject_id = 24
) AS cs ON cs.student_id = s.id

# RTI flag info
INNER JOIN 
(
    SELECT os.id,
    os.student_id
    FROM observations o
    INNER JOIN observations_students os ON os.observation_id = 2567
    WHERE
    o.school_id = 1509
) AS os ON os.student_id = s.id

LEFT JOIN
(
    SELECT fdos.id,
    fdos.student_id
    FROM observations o
    INNER JOIN observations_students fdos ON fdos.observation_id = 2009
    WHERE
    o.school_id = 1509
) AS fdos ON fdos.student_id = s.id

LEFT JOIN schools_students ss ON ss.student_id = s.id
WHERE s.active = 1
AND ss.school_id = 1509
AND IFNULL(ss.date_end,NOW()) >= NOW()
AND cs.gradelevel_id BETWEEN 10 AND 16

如果我将“COUNT（fdos.id）AS fd_count”更改为“fdos.id AS fdosid”，则返回正确的条目数。但是，从COUNT返回的数字不是相同的数字，并且不正确。谁能理解这里发生了什么，足以解释我做错了什么？

感谢您的时间。

Answer 1

似乎os的INNER JOIN已经过滤了你的结果，只显示observe_id = 2567.因此你不能获得不同的observation_id的任何其他记录。您可以将INNER JOIN更改为LEFT JOIN，看看它是如何进行的。

SELECT DISTINCT
SUBSTRING(s.osis_id,INSTR(s.osis_id,'-')+1) AS osid,
s.id AS student_id,
CONCAT(s.last_name, ' ',s.first_name) AS sname,
COUNT(fdos.id) AS fd_count
FROM students s

# course info
INNER JOIN 
(
    SELECT c.id AS cid,
    c.description AS cname,
    cs.date_end,
    cs.student_id,
    gl.description AS grade,
    c.gradelevel_id
    FROM courses_students cs
    INNER JOIN courses c ON c.id = cs.course_id
    INNER JOIN gradelevels gl ON gl.id = c.gradelevel_id
    WHERE
    IFNULL(cs.date_end, NOW()) >= NOW()
    AND IFNULL(c.date_end, NOW()) >= NOW()
    AND c.school_id = 1509
    AND c.subject_id = 24
) AS cs ON cs.student_id = s.id

# RTI flag info
LEFT JOIN  #change this to LEFT JOIN
(
    SELECT os.id,
    os.student_id
    FROM observations o
    INNER JOIN observations_students os ON os.observation_id = 2567
    WHERE
    o.school_id = 1509
) AS os ON os.student_id = s.id

LEFT JOIN
(
    SELECT fdos.id,
    fdos.student_id
    FROM observations o
    INNER JOIN observations_students fdos ON fdos.observation_id = 2009
    WHERE
    o.school_id = 1509
) AS fdos ON fdos.student_id = s.id

LEFT JOIN schools_students ss ON ss.student_id = s.id
WHERE s.active = 1
AND ss.school_id = 1509
AND IFNULL(ss.date_end,NOW()) >= NOW()
AND cs.gradelevel_id BETWEEN 10 AND 16

Answer 2

我敢打赌你在使用MySQL。

如果您使用以下任何内容：

GROUP BY条款;
HAVING条款;
聚合函数，count()是

然后您的查询将被聚合一个。

这意味着，数据将按照GROUP BY子句中指定的字段进行分组，这些字段应按原样保留在选择列表和查询中的其他位置。所有其他字段应该是聚合函数的参数，否则数据库不知道该集合中应该返回的哪个值匹配。

所有主要数据库都会为您的查询构成错误，因为对于一堆字段没有GROUP BY子句：s.osis_id，s.id，{{1 }和s.last_name。 MySQL不会。相反，它会隐式地对数据进行分组。我不知道什么是分组标准，我不想，因为这种行为容易出错且不可靠。

相反，您的查询应该被重写。最简单的方法是：

使用不具有s.first_name功能的现有查询，即获取count()的列表;
将整个查询用作另一个子查询，省略fdos.id子句;
数学生。

这样的事情：

DISTINCT

Answer 3

快速解决方法似乎是将COUNT(fdos.id)更改为COUNT(*)。

这是一个解释。 fdos结果是外部 -joined，因此对于连接左侧的某些行，可能不会返回fdos行。如果未返回它们，则相应的列（包括fdos.id）将返回NULL。但是COUNT()省略了NULL，这意味着COUNT(fdos.id)将省略连接结果集的某些行。无论匹配，NULL等计算所有行的标准方法都是COUNT(*)。

MySQL - Count没有返回我正在寻找的值

3 个答案: