Question

不知道为什么我只得错误统计一行！

以下是工作代码：

SELECT project_name, sub_project_name
FROM projects, sub_projects
WHERE projects.project_id = sub_projects.projects_project_id

结果：

project_name         sub_project_name
Bakken               Ghost fracture Study
Bakken               Bakken Mylo QAQC
Bossier              Doyle Boles K No.1
Eagleford            Kennedy Unit#1H
Eagleford            Wehmeyer Unit #1
Niobrara             Crow Valley 7-62-32-1M
Poland               Poland
Woodford             Ridenour Phase 2
Woodford             Teague 1-14H

每个sub_project都有多个表，我基本上试图查看其中一个表中的列是否包含任何包含非Null值的行，然后在每个sub_project的另一列中的sub_project旁边显示该计数。

以下是我查询以下语句时得到的内容：

SELECT projects.project_name, sub_projects.sub_project_name, COUNT( bl.bl_por ) AS porosity
FROM projects, sub_projects
LEFT JOIN bl ON sub_projects.sub_project_id = bl.sub_project_id
WHERE projects.project_id = sub_projects.projects_project_id

结果：

project_name         sub_project_name          porosity
Bakken               Ghost fracture Study      99

第一行中每个sub_project的所有行，没有别的。

这里有什么问题？

编辑：

Erwin的解决方案钉了它。但是我的表中没有sub_project_id。我只是将它添加到一个表中，以便更容易进行测试。

所以我使用Erwin的GROUP BY建议来编辑我的陈述，我得到了正确的矩阵形状但不是正确的数字。计算结束了。

SELECT p.project_name, sp.sub_project_name, COUNT( bl.bl_por ) AS porosity
FROM projects p, sub_projects sp, wells w, cores c, samples s, inputs i
LEFT JOIN bl ON i.inputs_id = bl.inputs_inputs_id
WHERE p.project_id = sp.projects_project_id
AND s.sample_id = i.samples_sample_id
AND c.core_id = s.cores_core_id
AND sp.sub_project_id = c.sub_projects_has_wells_sub_projects_sub_project_id
GROUP BY p.project_name, sp.sub_project_name

结果：

project_name    sub_project_name        porosity
Bakken          Bakken Mylo QAQC        147
Bakken          Ghost fracture Study    252
Bossier         Doyle Boles K No.1      189
Eagleford       Kennedy Unit#1H         294
Eagleford       Wehmeyer Unit #1        0
Niobrara        Crow Valley 7-62-32-1M  0
Poland          Poland                  714
Woodford        Ridenour Phase 2        483
Woodford        Teague 1-14H            0

正确的结果应该是：

Bakken          Bakken Mylo QAQC        7
Bakken          Ghost fracture Study    12
Bossier         Doyle Boles K No.1      9
Eagleford       Kennedy Unit#1H         14
Eagleford       Wehmeyer Unit #1        0
Niobrara        Crow Valley 7-62-32-1M  0
Poland          Poland                  34
Woodford        Ridenour Phase 2        23
Woodford        Teague 1-14H            0

Answer 1

尝试显式JOIN条件和显式GROUP BY（尽管mysql允许稍后跳过）。

SELECT p.project_name, sp.sub_project_name, COUNT(bl.bl_por) AS porosity
FROM   projects p
JOIN   sub_projects sp ON p.project_id = sp.projects_project_id
LEFT   JOIN bl ON sp.sub_project_id = bl.sub_project_id
GROUP  BY p.project_name, sp.sub_project_name;

回答其他问题：

再次尝试使用正确的SQL语法whit explicit JOIN和join-conditions：

SELECT p.project_name, sp.sub_project_name, COUNT(bl.bl_por) AS porosity
FROM   projects p
JOIN   sub_projects sp ON p.project_id = sp.projects_project_id
JOIN   wells w -- no JOIN condition? Results in cross join.
JOIN   cores c ON sp.sub_project_id = c.sub_projects_has_wells_sub_projects_sub_project_id
JOIN   samples s ON c.core_id = s.cores_core_id
JOIN   inputs i ON s.sample_id = i.samples_sample_id
LEFT   JOIN bl ON i.inputs_id = bl.inputs_inputs_id
GROUP  BY p.project_name, sp.sub_project_name

通过这种方式，您可以立即注意到表wells无条件地加入。这会导致交叉连接：左侧的每一行都会延伸到右侧的每一行，从而产生很多行。可能是过度计数的来源。

另请注意，计数仅计算非空值。 bl.bl_por IS NULL的任何行都不计算在内。如果您确实要计算行数，可以使用count(bl.*)。

在MYSQL中使用COUNT后只有一行

编辑：

1 个答案:

回答其他问题：