以下是我的表格示例：
项
+----------+---------------+
| txt_item | txt_unique_id |
+----------+---------------+
| Circle | 1 |
| Square | 2 |
| Triangle | 3 |
+----------+---------------+
tag_master
+---------+----------+
| txt_tag | opt_type |
+---------+----------+
| red | color |
| blue | color |
| yellow | color |
| large | size |
| medium | size |
| small | size |
+---------+----------+
item_tags
+---------+---------------+
| txt_tag | txt_unique_id |
+---------+---------------+
| red | 1 |
| blue | 1 |
| large | 1 |
| small | 1 |
| red | 2 |
| yellow | 2 |
| small | 2 |
| medium | 2 |
| red | 3 |
| yellow | 3 |
+---------+---------------+
我想要归还：
+----------+----------------------------+
| Circle | red, blue, large, small |
| Square | red, yellow, small, medium |
| Triangle | red, yellow |
+----------+----------------------------+
这就是我得到的：
+----------+---------------------------------------------+
| Circle | red, red, red, blue, large, small, small |
| Square | red, red, red, yellow, yellow, small, small |
| Triangle | red, red, red, yellow, yellow |
+----------+---------------------------------------------+
我在这里：
CREATE TABLE #screening_tags
(
txt_unique_id VARCHAR(36),
tags VARCHAR(1000)
)
INSERT INTO #screening_tags
(txt_unique_id,
tags)
(SELECT txt_unique_id,
( STUFF((SELECT ' , ' + t.txt_tag
FROM item_tags t
JOIN tag_master_ tm
ON t.txt_tag = tm.txt_tag
JOIN items i
ON t.txt_unique_id = i.txt_unique_id
ORDER BY opt_type,
txt_tage
FOR xml path('')), 1, 1, '') )
FROM item_tags t)
SELECT *
FROM #screening_tags
我也尝试过使用COALESCE，但我遗漏了一些东西。我需要一个DISTINCT或者其他东西，但我尝试过的所有东西都不起作用。如果我想通过opt_type进行ORDER BY，我就无法使用DISTINCT或TOP 1。感谢帮助。
答案 0 :(得分：0)
尝试将连接移到tag_master上 - 查看预期结果，您不需要该表中的任何内容。实际上，您似乎不需要任何连接。
CREATE TABLE #screening_tags
(
txt_unique_id VARCHAR(36),
tags VARCHAR(1000)
)
INSERT INTO #screening_tags
(txt_unique_id,
tags)
(SELECT txt_unique_id,
( STUFF((SELECT ' , ' + t.txt_tag
FROM items i
where t.txt_unique_id = i.txt_unique_id
ORDER BY opt_type,
txt_tag
FOR xml path('')), 1, 1, '') )
FROM item_tags t)
SELECT *
FROM #screening_tags
答案 1 :(得分：0)
不确定您使用的是什么数据库，但是如果它是Postgres，请尝试一下：
<nav class="navbar navbar-expand-lg navbar-dark bg-primary">
<a class="navbar-brand" href="#">Angular Events</a>
<button class="navbar-toggler" type="button" data-toggle="collapse" data-target="#navbarColor01"
aria-controls="navbarColor01" aria-expanded="false" aria-label="Toggle navigation">
<span class="navbar-toggler-icon"></span>
</button>
<div class="collapse navbar-collapse" id="navbarColor01">
<ul class="navbar-nav mr-auto">
<li class="nav-item active">
<a class="nav-link" [routerLink]="['/events']">All Events <span class="sr-only">(current)</span></a>
</li>
<li class="nav-item">
<a class="nav-link" [routerLink]="['/events/new']">Create Event</a>
</li>
<li class="nav-item dropdown">
<a class="nav-link dropdown-toggle" data-toggle="dropdown" href="#" routerLink="" role="button" aria-haspopup="true"
aria-expanded="false">Dropdown</a>
<div class="dropdown-menu" x-placement="bottom-start"
style="position: absolute; transform: translate3d(0px, 42px, 0px); top: 0px; left: 0px; will-change: transform;">
<a class="dropdown-item" href="#">Action</a>
<a class="dropdown-item" href="#">Another action</a>
<a class="dropdown-item" href="#">Something else here</a>
<div class="dropdown-divider"></div>
<a class="dropdown-item" href="#">Separated link</a>
</div>
</li>
</ul>
<form class="form-inline my-2 my-lg-0">
<input class="form-control mr-sm-2" placeholder="Search" type="text">
<button class="btn btn-secondary my-2 my-sm-0" type="submit">Search</button>
</form>
</div>
</nav>
SELECT i.item, string_agg(DISTINCT tm.text_tag,',') AS txt_tag_agg
FROM items i
INNER JOIN item_tags it ON i.txt_unique_id = it.txt_unique_id
INNER JOIN tag_master tm ON it.txt_tag = tm.text_tag
GROUP BY i.item
函数将构建由string_agg列分组的指定行值的聚合字符串。
答案 2 :(得分：0)
假设您使用的是SQL-server＆gt; = 2017，string_agg()现已可用。
MS SQL Server 2017架构设置：
查询1 ：
select i.txt_item , string_agg(it.txt_tag, ',')
from items i
inner join item_tags it
on i.txt_unique_id = it.txt_unique_id
inner join tag_master tm
on tm.txt_tag = it.txt_tag
where opt_type = 'color'
group by i.txt_item
<强> Results ：
| txt_item | |
|----------|------------|
| Circle | red,blue |
| Square | red,yellow |
| Triangle | red,yellow |
答案 3 :(得分：0)
使用子查询至少可以在sql server，mysql，sqlite和postgresql上工作：
select i.txt_item,
(select string_agg(txt_tag, ',')
from item_tags it where i.txt_unique_id=it.txt_unique_id) as txt_tag
from items i
返回：
txt_item txt_tag
Circle red,blue,large,small
Square red,yellow,small,medium
Triangle red,yellow
在mysql上使用group_concat（txt_tag）而不是string_agg。 在postgres上，使用array_agg（txt_tag）。
答案 4 :(得分：0)
我想通了 - 我必须将STUFF子查询链接到WHERE子句中的主查询
CREATE TABLE #screening_tags
(
txt_unique_id VARCHAR(36),
tags VARCHAR(1000)
)
INSERT INTO #screening_tags
(txt_unique_id,
tags)
(SELECT txt_unique_id,
( STUFF((SELECT ', ' + t.txt_tag
FROM item_tags t1
WHERE t.txt_unique_id = t2.txt_unique_id
FOR xml path('')), 1, 1, '') )
FROM item_tags t2)
SELECT *
FROM #screening_tags
+----------+----------------------------+
| Circle | red, blue, large, small |
| Square | red, yellow, small, medium |
| Triangle | red, yellow |
+----------+----------------------------+