我有2张桌子:
表1. options_ethnicity具有以下条目:
options_ethnicity
ethnicity_id ethnicity_name 1 White 2 Hispanic 3 African/American
表2. inquiries带有以下条目:
inquiries
inquiry_id ethnicity_id 1 1 2 1 3 1 4 2 5 2
我想生成一个表,按种族显示查询数量。到目前为止,我的查询看起来像这样:
SELECT options_ethnicity.ethnicity_name, COUNT('inquiries.ethnicity_id') AS count FROM (inquiries LEFT JOIN options_ethnicity ON options_ethnicity.ethnicity_id = inquiries.ethnicity_id) GROUP BY options_ethnicity.ethnicity_id
该查询给出了正确的答案,但没有针对非洲/美洲的列,其结果为0。
White 3 Hispanic 2
如果我用RIGHT JOIN代替LEFT JOIN,我会得到所有3个种族的名字,但是非洲裔/美洲裔的计数是错误的。
White 3 Hispanic 2 African/American 1
任何帮助,将不胜感激。
这是此帖子的更新,其中似乎有一个有效的查询:
SELECT options_ethnicity.ethnicity_name, COALESCE(COUNT(inquiries.ethnicity_id), 0) AS count FROM options_ethnicity LEFT JOIN inquiries ON inquiries.ethnicity_id = options_ethnicity.ethnicity_id GROUP BY options_ethnicity.ethnicity_id UNION ALL SELECT 'NULL Placeholder' AS ethnicity_name, COUNT(inquiries.inquiry_id) AS count FROM inquiries WHERE inquiries.ethnicity_id IS NULL
由于您使用的是LEFT JOIN,因此对LEFT JOIN中定义的表的引用可以为空。这意味着您需要将此NULL值转换为零(在这种情况下):
SELECT oe.ethnicity_name, COALESCE(COUNT(i.ethnicity_id), 0) AS count FROM OPTIONS_ETHNICITY oe LEFT JOIN INQUIRIES i ON i.ethnicity_id = oe.ethnicity_id GROUP BY oe.ethnicity_id
本示例使用COALESCE(一种处理NULL值的ANSI标准方法)。它将返回第一个非null值,但是如果找不到,则将返回null。 IFNULL在MySQL上是有效的替代方法,但是当COALESCE可以使用时,它不能移植到其他数据库。
在实际数据库表中,查询表中有一些条目,其中ethnicity_id为NULL,即未记录种族。关于如何获取这些空值以进行显示的任何想法吗?
我想我了解您面临的问题:
SELECT oe.ethnicity_name, COALESCE(COUNT(i.ethnicity_id), 0) AS count FROM (SELECT t.ethnicity_name, t.ethnicity_id FROM OPTIONS_ETHNICITY t UNION ALL SELECT 'NULL placeholder' AS ethnicity_name, NULL AS ethnicity_id) oe LEFT JOIN INQUIRIES i ON i.ethnicity_id = oe.ethnicity_id GROUP BY oe.ethnicity_id
这将拾取所有NULL ethncity_id实例,但会将计数归因于“ NULL占位符”组。IE:
ethnicity_name | COUNT ------------------------ White | 3 Hispanic | 2 NULL placeholder | ?