Question

我目前正在研究更长的MySQL查询，我会逐步检查它，因为它的长度可以找到问题。现在我达到了一个我无法向自己解释的地步。

在我的查询中，我调用一个子查询，其中Select与Select（子查询）相似。我曾经＆＃34;建立＆＃34;子查询之前和它的工作。但是在调用它时 - 为了进一步计算 - 出现了一个错误＃1064 - 据我所知 - 与子查询无关，因为打开的括号全部都是关闭的，而我常常以我的方式处理子查询之前做过。但问题必须与名为avat的查询部分和以下Select(avat)有关。错误是＃1064，解析错误。对我来说，还是其他什么新事物？

SELECT sr.usr_id_give, 
       sr.usr_id_get, 
       IFNULL((sr.value / srgive.value),0) as asv,
       IF((SELECT asv) IS NULL,0.1,1)  AS sv,

(SELECT IF(EXISTS(SELECT y.usr_id2, x.usr_id2 FROM usr_connx x 
LEFT JOIN usr_connx y ON y.usr_id1 = x.usr_id2 AND y.usr_id2 <> x.usr_id1 
LEFT JOIN usr_connx z ON z.usr_id2 = y.usr_id2 AND z.usr_id1 = x.usr_id1 
WHERE x.usr_id1 = 3 AND z.usr_id1 IS NULL GROUP BY y.usr_id2 HAVING COUNT(*) >= 2),1.5,
IF(EXISTS(SELECT y.usr_id2, x.usr_id2 FROM usr_connx x 
LEFT JOIN usr_connx y ON y.usr_id1 = x.usr_id2 AND y.usr_id2 <> x.usr_id1 
LEFT JOIN usr_connx z ON z.usr_id2 = y.usr_id2 AND z.usr_id1 = x.usr_id1 
WHERE x.usr_id1 = 3 AND z.usr_id1 IS NULL GROUP BY y.usr_id2 HAVING COUNT(*) >= 2),0.5,0))) AS grade,
       ((SELECT sv) * (SELECT grade)) AS vat,

(SELECT AVG(value)
FROM (SELECT srr.value
FROM skill_ratings AS srr
WHERE srr.usr_id_get = 221 
AND srr.skill_id = 1
LIMIT 5) skill_ratings) as avat,

SELECT(avat)

FROM skill_ratings AS sr
LEFT JOIN skill_ratings srgive ON srgive.usr_id_get = sr.usr_id_give
WHERE sr.usr_id_get = 1

我想要完成的是它可以正常工作，我可以调用子查询，因为它在我尝试调用之前就已经工作了。

提前感谢您的帮助。也许，如果这是一个困难的问题，欢迎提供解释，以便我可以学习并避免将来出现类似的错误。

这是错误：

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'SELECT(avat)

FROM skill_ratings AS sr
LEFT JOIN skill_ratings srgive ON srgi' at line 23

MySQL查询在调用子查询时显示错误

0 个答案: