Question

这是我的代码，我无法弄清楚为什么以下代码返回错误操作数应该包含1列。我知道这两个查询都与我下面的查询无关。任何帮助将不胜感激

SELECT * 
FROM 2014_summary, 2014_profiles 
WHERE 2014_summary.player_id = 2014_profiles.player_id AND tournament_id = 'bec904c4-b2a3-4eba-a2bc-335aca680a40' AND 2014_summary.player_id 
NOT IN
(SELECT 
    T1.*, 
    T2.player_id as primary_player_id 
FROM (SELECT id, lead_id, form_id, MAX(case when field_number = 1 then value end) display_name, 
    MAX(case when field_number = 7 then value end) email, 
    MAX(case when field_number = 6 then value end) tournament_name, 
    MAX(case when field_number = 3 then value end) primary_golfer, 
    MAX(case when field_number = 4 then value end) backup_golfer, 
    MAX(case when field_number = 5 then value end) date, 
    MAX(case when field_number = 8 then value end) tournament_id 
FROM `wp_rg_lead_detail` 
GROUP BY lead_id)T1 
INNER JOIN (SELECT CONCAT(first_name, ' ', last_name) as player_name, player_id FROM 2014_profiles)T2 ON T1.primary_golfer = T2.player_name
WHERE display_name = "Hosker" AND tournament_id != 'bec904c4-b2a3-4eba-a2bc-335aca680a40')
ORDER BY (last_name) ASC

Answer 1

您的not in是：

2014_summary.player_id NOT IN
    (SELECT  T1.*, T2.player_id as primary_player_id

MySQL不知道在select中使用哪个列与player_id进行比较。所以它返回一个错误。实际上，这是ANSI标准SQL功能，所有数据库都以这种方式工作。 in中的子查询只需要返回一列。

我想你想要：

2014_summary.player_id NOT IN
    (SELECT TT2.player_id
     . . .

我收到错误操作数应该包含1列

1 个答案: