Question

我有一个数据库可以有类似的行，但有不同的键和不同的布尔列。这是数据库的样子：

列：_id，name，img，address，available

两个条目可能如下所示：

_id | name | img                | address   | available
-------------------------------------------------------
1   | John | http://img.com/222 | 1 Main St | 1
2   | John | http://img.com/222 | 1 Main St | 0

我想要一个查询，它会给我所有具有不同键的结果，如果有重复的条目（忽略_id会有所不同的事实），它将只返回第一个。这是我的查询：

SELECT p1.* 
FROM (SELECT DISTINCT _id, available FROM people) p
    INNER JOIN people p1
    ON p1._id=p._id
ORDER BY p1.available DESC;

我知道这不对，但也许它解释了我正在寻找的东西。我想在这里使用GROUP BY吗？

Answer 1

I want a query that will give me all results that have a distinct key, and if there are duplicate entries(ignoring the fact that _id would be different), it will give back only the first one.....the _id isn't what I want to be distinct, as they [the ids] are already unique. ... . Ideally it will order by 'available' in descending order so that if there are two columns with the same data(aside from _id and available), it will return the row with '1' for the available column

             select name, image, address, max(availability) as avail
              from T
             group by name, image, address

然后，您可以将上面查询返回的集合作为内联视图加入到您的表中：

            select * from T
            inner join
            (

             select name, image, address, max(availability) avail
              from T
             group by name, image, address
            ) as foo

            on T.name = foo.name and T.image = foo.image and T.address = foo.address and T.availability = foo.avail

如果有一个复合索引会有所帮助:(名称，图像，地址）。

警告：如果特定{name，image，address}三元组的可用性为1的行有多行，则查询将为三元组返回多行：

             2   | John | http://img.com/222 | 1 Main St | 1
             6   | John | http://img.com/222 | 1 Main St | 1

P.S。听起来好像你希望在你的表中创建三元组（名称，图像，地址）一个备用的UNIQUE键。

Answer 2

这个sql可以解决你的问题：

选择b。* from（从人们选择不同的_id）a，人们b其中a._id = b._id按b.available排序

Answer 3

我实际上刚刚问了一个类似的问题，并从一位经验丰富的用户那里得到了很好的答案：

SQL Populating with distinct data and a sequence

根据他告诉我的内容，这个查询可能会为您提供所需内容：

SELECT p1.* 
FROM (SELECT DISTINCT _id, name from people) p
   INNER JOIN people p1
   ON p1._id=p._id
ORDER BY p1.available desc

如果失败并且不起作用，请道歉！

编辑：我刚想到我不知道哪个名称+ _id组合这将提取..可用= 1或者可用= 0或随机选择..！让我知道无论如何会发生什么......

Answer 4

如果你想要_id中available值最低的_id的第一行，那么你可以“记录”available里面的_id分组生成的汇总值。

要比较的值的构建方式是按照_id字段的降序排列记录，然后按降序排列select people.* from people inner join ( select name, img, address, min((1-available)*100000000 + _id) avail_id from people group by name, img, address ) as foo on people._id = foo.avail_id % 100000000;字段，并允许轻松检索select people.* from people left outer join people as other on other.name = people.name and other.img = people.img and people.address=other.address and (1 - people.available) * 100000000 + people._id > (1 - other.available) * 100000000 + other._id where other.available is null;使用模运算符（假设可用的最大值为1，且id永远不会超过100000000）。

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我改编了Tim's query。

您也可以执行without subquery：

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如何在SQL中的非键上运行DISTINCT

4 个答案: