Question

在使用group by，rank和where时遇到sql问题我试图从我的数据库中的表进行分组和排名，并且已经能够解决这个问题，所以现在我想使用列等于值的地方。这是我的代码

SELECT g1.adnumber , g1.subject_id , g1.score ,
COUNT(*) AS rank FROM tbl_result AS g1 
JOIN tbl_result AS g2 
ON (g2.score, g2.adnumber) >= (g1.score, g1.adnumber) 
AND g1.subject_id = g2.subject_id 
GROUP BY g1.adnumber , g1.subject_id , g1.score 
ORDER BY g1.subject_id , rank 
where g1.adnumber = 'demonumber'

这是我得到的错误

＃1064 - 您的SQL语法出错;检查与MySQL服务器版本对应的手册，以获得正确的语法靠近'哪里g1.adnumber ='ctydemo'LIMIT 0,25'在第1行

Answer 1

你不能只在最后放一个where条款。您需要拥有正确的syntax

粗略地解释说，你必须有这个选择的顺序：

SELECT
    *
[FROM]
[WHERE where_condition]
[GROUP BY {col_name | expr | position}
  [ASC | DESC], ... [WITH ROLLUP]]
[HAVING where_condition]
[ORDER BY {col_name | expr | position}
  [ASC | DESC], ...]
[LIMIT {[offset,] row_count | row_count OFFSET offset}]

在您的情况下，查询可能会这样工作。查看更改JOIN, ON, WHERE, AND

SELECT
    g1.adnumber,
    g1.subject_id,
    g1.score,
    COUNT(*) AS rank
FROM tbl_result AS g1 
    JOIN tbl_result AS g2 
    ON g1.subject_id = g2.subject_id 
WHERE g1.adnumber = 'demonumber'
   # not sure what you are trying with the next line
   # AND (g2.score, g2.adnumber) >= (g1.score, g1.adnumber) 
   # maybe something like this:
   AND g2.score >= g1.score
   AND g2.adnumber >= g1.adnumber
GROUP BY 
    g1.adnumber,
    g1.subject_id,
    g1.score 
ORDER BY 
    g1.subject_id,
    g1.rank

Answer 2

哪些陈述应该在分组之前？＆amp;按陈述排序。

这应该有效（未经测试）

SELECT 
    g1.adnumber , g1.subject_id , g1.score , COUNT(*) AS rank 
FROM 
    tbl_result AS g1 JOIN tbl_result AS g2 ON (g2.score, g2.adnumber) >= (g1.score, g1.adnumber) AND g1.subject_id = g2.subject_id 
WHERE 
    g1.adnumber = 'demonumber'
GROUP BY 
    g1.adnumber , g1.subject_id , g1.score 
ORDER BY 
    g1.subject_id , rank

Answer 3

您使用了错误的执行语法。在您的SQL where子句应放在group by 之前。请尝试以下SQL

SELECT g1.adnumber , g1.subject_id , g1.score ,
COUNT(*) AS rank FROM tbl_result AS g1 
JOIN tbl_result AS g2 
ON (g2.score, g2.adnumber) >= (g1.score, g1.adnumber) 
AND g1.subject_id = g2.subject_id
where g1.adnumber = 'demonumber'
GROUP BY g1.adnumber , g1.subject_id , g1.score 
ORDER BY g1.subject_id , rank;

在使用group by，rank和

3 个答案: