标签: mysql-error-1241

在子查询中选择两列

我有以下SQL语句:

SELECT  Office, 
COUNT(Office) As Tot_Part, 
(SELECT Office, COUNT(Office) 
      FROM trespondent 
      WHERE completion_status= 'Started' 
      OR completion_status = 'Complete' 
      GROUP BY Office ORDER BY Office
) As Total_Resp 
FROM trespondent 
WHERE completion_status <> 'New' 
GROUP BY Office 
ORDER BY Office
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我所遇到的问题是SELECT括号中的陈述 - 我要做的是将total整体人员和total已经开始或已完成的人员带回来.

1241 - 操作数应包含1列

如果我包含 COUNT(Office),则两列不匹配.

如何改进查询以实现我正在寻找的结果集?

Office    Tot   Resp
London    20    2
Leeds     30    17
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mysql mysql-error-1241

2
推荐指数
1
解决办法
88
查看次数

MySQL INSERT 错误操作数应包含 1 列

我有这个 MySQL 插入查询,它给了我一个Error Code: 1241. Operand should contain 1 column(s).

INSERT INTO esjp_content
    ( esjp_content.primary_key, esjp_content.template_id, esjp_content.section_ref,
    esjp_content.position, esjp_content.indent, esjp_content.content,
    esjp_content.summary_id, esjp_content.role_ref, esjp_content.author_id,
    esjp_content.event_id, esjp_content.sys_time )
VALUES
    ( ( 3, 1, 1, 0, 0, 'Some test content.', 0, 1, 1, 69, UNIX_TIMESTAMP(NOW()) ),
    ( 4, 1, 1, 1, 1, 'Some test content2.', 0, 1, 1, 69, UNIX_TIMESTAMP(NOW()) ) )
ON DUPLICATE KEY UPDATE
    esjp_content.primary_key=VALUES(esjp_content.primary_key),
    esjp_content.template_id=VALUES(esjp_content.template_id),
    esjp_content.section_ref=VALUES(esjp_content.section_ref),
    esjp_content.position=VALUES(esjp_content.position),
    esjp_content.indent=VALUES(esjp_content.indent),
    esjp_content.content=VALUES(esjp_content.content),
    esjp_content.summary_id=VALUES(esjp_content.summary_id),
    esjp_content.role_ref=VALUES(esjp_content.role_ref),
    esjp_content.author_id=VALUES(esjp_content.author_id),
    esjp_content.event_id=VALUES(esjp_content.event_id),
    esjp_content.sys_time=VALUES(esjp_content.sys_time);
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它工作正常,如果我一次只尝试插入 1 条记录,但我认为这INSERT …

mysql insert mysql-error-1241

2
推荐指数
1
解决办法
8753
查看次数

#1241 - 操作数应包含1列

在以下查询中,我尝试为每个类别获取一个图像

SELECT  l.*
FROM    (
   SELECT  id AS category_id,
           COALESCE(
           (
           SELECT  *
           FROM    images li
           WHERE li.category_id = dlo.id

           ORDER BY
                       li.save_status DESC, li.category_id DESC, li.id DESC
           LIMIT 1
           ), 0) AS mid
   FROM    cats_imgs dlo where id='1'
   ) lo
JOIN    images l
ON      l.save_status = '3'
   AND l.category_id >= lo.category_id
   AND l.category_id <= lo.category_id
   AND l.id >= lo.mid
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但不断收到以下错误:

#1241 - Operand should contain 1 column(s)
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有任何想法吗 ?

php mysql mysql-error-1241

1
推荐指数
1
解决办法
8635
查看次数

在MySQL中不起作用

我有一个SQL查询的问题

这是我的查询:

SELECT `rooms`.*  
FROM `rooms`  
WHERE (`rooms`.`room_id` NOT IN (
    SELECT * 
    FROM `reservations` `res` 
    WHERE ('2012-05-03' NOT BETWEEN res.expected_checkin_date 
    AND res.expected_checkout_date) 
    AND ('2012-05-08' NOT BETWEEN res.expected_checkin_date 
    AND res.expected_checkout_date)))
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问题是当第二个select返回null时,它显示以下错误:

#1241 - Operand should contain 1 column(s)

有谁能够帮我?

EDIT

当我运行这个查询时,它显示所有行甚至包含第二个内部的那个select 我不知道为什么?

SELECT `rooms`.*, `type`.*, `status`.* FROM `rooms` INNER JOIN `room_types` AS `type` ON `type`.`room_type_id` = `rooms`.`room_type_id` AND `type`.`num_beds` >= 1 AND type.room_type_id=2 INNER JOIN `room_statuses` AS `status` ON `status`.`room_status_id` = `rooms`.`room_status_id` WHERE (`rooms`.`room_id` NOT IN (SELECT …
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mysql sql mysql-error-1241

1
推荐指数
1
解决办法
1244
查看次数

MySQL查询SELECT a.foo AS吧 - 多个?

我有一个查询,我正在尝试将多个表列分组为"文本",看看我的查询:

SELECT a.id as id, 
       a.name AS title, 
       (a.address1, a.address2, a.suburb, a.state, a.pcode) AS text, 
       a.suburb AS suburb
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尝试以这种方式使用时出现此错误:

1241 - 操作数应包含1列

我想做什么不可能?有没有解决这个问题?

php mysql mysql-error-1241

0
推荐指数
1
解决办法
275
查看次数

#1241 - 操作数应包含 1 列 - 为什么我会得到这个?

我正在尝试运行查询,但收到上述错误。谁能告诉我为什么我会得到它。

UPDATE `catalog_eav_attribute`
SET `used_for_sort_by` = 1
WHERE attribute_id = (
SELECT * FROM `eav_attribute` WHERE `entity_type_id` = (SELECT `entity_type_id` FROM `eav_entity_type` WHERE `entity_model` = "catalog/product") AND `attribute_code` = "created_at");
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mysql sql mysql-error-1241

0
推荐指数
1
解决办法
2713
查看次数

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