4

我有下表用于存储用户数据：

例如

TABLE: users
COLUMNS:
...
maritalStatus (INT)   - FK
gender        (CHAR)
occupation    (INT)   - FK
...

现在我想比较这个表中的两个用户，看看有多少列匹配任何两个给定用户（比如用户 X 和用户 Y）

我通过 mySQL 存储过程分别获取每个值然后比较它们

例如

    SELECT maritalStatus from users where userID = X INTO myVar1;
    SELECT maritalStatus from users where userID = Y INTO myVar2;

    IF myVar1 = myVar2 THEN

    ...

    END IF;

有没有更短的方法使用 SQL 查询，我可以比较表中的两行并查看哪些列不同？我不需要知道它们实际上有多大不同，只需要知道它们是否包含相同的值。此外，我只会比较选定的列，而不是用户表中的每一列。

4

6 回答 6

12

这将为user和 user选择不同的列数：xy

SELECT ( u1.martialStatus <> u2.martialStatus )
     + ( u1.gender        <> u2.gender        )
     + ( u1.occupation    <> u2.occupation    )
FROM
  users u1,
  users u2
WHERE u1.id = x
  AND u2.id = y
于 2012-05-08T07:00:48.030 回答
3

你也可以使用这个：

select 

   -- add other columns as needed
   (a.lastname,a.gender) 
=  (b.lastname,a.gender) as similar,


  a.lastname as a_lastname,
  a.firstname as a_firstname,
  a.age as a_age,

  'x' as x,

  b.lastname as b_lastname,
  b.firstname as b_firstname,
  b.age as b_age



from person a
cross join person b
where a.id = 1 and b.id = 2

输出：

SIMILAR A_LASTNAME A_FIRSTNAME A_AGE X B_LASTNAME B_FIRSTNAME B_AGE
1       Lennon     John        40    x Lennon     Julian      15

现场测试：http ://www.sqlfiddle.com/#!2/840a1/2

于 2012-05-08T07:36:14.113 回答
0

您可以使用以下方法计算具有相同列的用户数group by：

select  u1.maritalStatus
,       u1.gender
,       u1.occupation
,       count(*)
from    users u1
group by
        u1.maritalStatus
,       u1.gender
,       u1.occupation
于 2012-05-08T06:52:05.227 回答
0

只是 PHP 中 Peter Langs 建议的一个继续示例：

$arr_cols   = array('martialStatus', 'gender', 'occupation');
$arr_where = array();
$arr_select = array();
foreach($arr_cols as $h) {

    $arr_having[] = "compare_{$h}";
    $arr_select[] = "(u1.{$h} != u2.{$h}) AS compare_{$h}";

}

$str_having  = implode(' + ', $arr_where);
$str_select = implode(', ', $arr_where);

$query = mysql_query("
SELECT {$str_select}
FROM users AS u1, users AS u2
WHERE u1.userid = {$int_userid_1} AND u2.userid = {$int_userid_2}
HAVING {$str_having} > 0
");

/* Having case can be removed if you need the row regardless. */

/* Afterwards you check these values: */

$row = mysql_fetch_assoc($query);
foreach($arr_cols as $h)
    if ($row["compare_{$h}"])
         echo "Found difference in column {$h}!";
于 2012-05-08T07:15:58.333 回答
0

我认为，这可能对某人有所帮助。目标：查找具有相同名称的行并用旧记录更新新记录日期。这可能是您必须为不同国家/地区复制新闻项目并保持与原始日期相同的情况。

CREATE TABLE `t` (
  `id` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `locale` varchar(10) DEFAULT 'en',
  `title` varchar(255) DEFAULT NULL,
  `slug` varchar(255) DEFAULT NULL,
  `body` text,
  `image` varchar(255) DEFAULT NULL,
  `thumb` varchar(255) DEFAULT NULL,
  `slug_title` varchar(255) DEFAULT NULL,
  `excerpt` text,
  `meta_title` varchar(200) DEFAULT NULL,
  `meta_description` varchar(160) DEFAULT NULL,
  `other_meta_tags` text,
  `read_count` int(10) DEFAULT '0',
  `status` varchar(20) DEFAULT NULL,
  `revised` text,
  `created` datetime DEFAULT NULL,
  `modified` datetime DEFAULT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=20 DEFAULT CHARSET=utf8;


INSERT INTO `t` (`id`, `locale`, `title`, `slug`, `body`, `image`, `thumb`, `slug_title`, `excerpt`, `meta_title`, `meta_description`, `other_meta_tags`, `read_count`, `status`, `revised`, `created`, `modified`)
VALUES
	(2, 'en', 'A title once again', '/news/title-one-again', 'And the article body follows.', '/uploads/2014/11/telecommunications100x100.jpg', NULL, NULL, NULL, '', '', NULL, 0, 'Draft', NULL, '2014-09-22 12:26:17', '2014-10-23 10:13:21'),
	(3, 'en', 'A title once again', '/news/title-strikes-back', 'This is really exciting! Not.', '/uploads/2014/11/telecommunications100x100.jpg', NULL, NULL, NULL, '', '', NULL, 0, 'Unpublished', NULL, '2014-09-23 12:26:17', '2014-10-31 11:12:55'),
	(4, 'en_GB', 'test', '/news/test', 'test', '/uploads/2014/11/telecommunications100x100.jpg', NULL, NULL, NULL, '', '', NULL, 0, 'Published', NULL, '2014-10-23 10:14:30', '2014-10-23 10:14:30');


update t join
       t t2
       on t.title = t2.title 
    set t2.created = t.created
    where t.title = t2.title ;

更新 t join t t2 on t.title = t2.title set t2.created = t.created where t.title = t2.title ;

于 2015-01-30T16:18:00.853 回答
0

如果另一个 Magento 开发人员在这里找到了自己的方式，则此 Q/A 的一个特定用途是比较表中的两个地址条目。“Magento 1”会将相同的地址放入两次，唯一的区别是键entity_id列和address_type列（帐单或运输）。

已经知道订单的entity_id，使用它来获取与订单关联的帐单和送货地址 ID：
SELECT entity_id FROM sales_flat_order_address WHERE parent_id = 3137;

然后查看它们是否对该订单有所不同：

SELECT a1.parent_id AS 'order_id'
,  ( a1.street    <> a2.street    )
 + ( a1.city      <> a2.city      )
 + ( a1.postcode  <> a2.postcode  )
 + ( a1.region_id <> a2.region_id )
 AS 'diffs'
FROM
  sales_flat_order_address a1,
  sales_flat_order_address a2
WHERE a1.entity_id = 6273
  AND a2.entity_id = 6274
;

给出输出：

+----------+-------+
| order_id | diffs |
+----------+-------+
|     3137 |     0 |
+----------+-------+

如果有办法大规模地做到这一点，那就太棒了。

于 2018-05-24T16:12:26.617 回答