连接sql查询来分析值

发布于 2024-12-05 14:17:24 字数 348 浏览 3 评论 0原文

我有一个包含三列的表 entriesid、name、display_name
我需要使用 join 编写一个查询,它将检查每个名称有多少个 display_names 并仅显示那些具有多个 display_names 的名称。
使用连接编写起来应该相当简单。在我使用下一个查询之前:

SELECT e1.name 
FROM entries e1 
WHERE (
  SELECT COUNT(DISTINCT e2.display_name) 
  FROM entries e2 
  WHERE e2.name = r1.name
) > 1;

I have a table entries with three columns: id, name, display_name.
I need to write a query with join, which will check how many display_names there is for each name and display only those names, which have multiple display_names.
This should be fairly simple to write with joins. Before I used next query:

SELECT e1.name 
FROM entries e1 
WHERE (
  SELECT COUNT(DISTINCT e2.display_name) 
  FROM entries e2 
  WHERE e2.name = r1.name
) > 1;

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评论(3

睫毛上残留的泪 2024-12-12 14:17:24

你不需要加入。

SELECT e1.name 
FROM entries e1 
GROUP BY e1.name
HAVING COUNT(*) > 1

请注意,您可以在having子句中使用别名,因此您也可以借用@Scorpi0和@Widor:

SELECT e1.name, COUNT(DISTINCT e1.display_name) as occurrences
FROM entries e1 
GROUP BY e1.name
HAVING occurrences > 1

允许这样做的原因是having在选择中的所有其他内容之后进行评估完成了。到那时别名的内容就会被知道。
您不能在 WHERE 子句中使用此别名,因为它 MySQL 知道别名中的内容之前运行(或可能运行)。

You don't need a join.

SELECT e1.name 
FROM entries e1 
GROUP BY e1.name
HAVING COUNT(*) > 1

Note that you can use an alias in a having clause, so borrowing from @Scorpi0 and @Widor you can also do:

SELECT e1.name, COUNT(DISTINCT e1.display_name) as occurrences
FROM entries e1 
GROUP BY e1.name
HAVING occurrences > 1

The reason this is allowed is that having is evaluated after all the other stuff in the select is done. By that time the contents of the alias will be known.
You cannot use this alias in the WHERE clause because that runs (or may run) before MySQL knows what in the alias.

弥枳 2024-12-12 14:17:24

您不需要联接,HAVING 子句可以为您进行过滤:

SELECT e1.name 
FROM entries e1 
GROUP BY e1.name
HAVING COUNT(DISTINCT e1.display_name) > 1

You don't need a join, the HAVING clause could do the filter for you:

SELECT e1.name 
FROM entries e1 
GROUP BY e1.name
HAVING COUNT(DISTINCT e1.display_name) > 1
冷情妓 2024-12-12 14:17:24
SELECT name, COUNT(display_name)
FROM entries
GROUP BY name
HAVING COUNT(display_name) > 1
SELECT name, COUNT(display_name)
FROM entries
GROUP BY name
HAVING COUNT(display_name) > 1
~没有更多了~
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