blocks|key|1550802|text|select+coalesce(t1.name,t2.name,t3.name,t4.name)+name,++t1.closeid,+t2.openid,+t3.assign,+t4.unassign
from+table-1+t1
full+outer+join+table-2+t2+on+t1.name+=+t2.name
full+outer+join+table-3+t3+on+t2.name+=+t3.name
full+outer+join+table-4+t4+on+t3.name+=+t4.name|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1550803|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>select coalesce(t1.name,t2.name,t3.name,t4.name) name, t1.closeid, t2.openid, t3.assign, t4.unassign
from table-1 t1
full outer join table-2 t2 on t1.name = t2.name
full outer join table-3 t3 on t2.name = t3.name
full outer join table-4 t4 on t3.name = t4.name
</code></pre>

blocks|key|5682884|text|我会抓取所有不同的名称，然后与该列表进行左连接，这将使理解发生了什么变得更容易。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5682885|;WITH+cte(name)
AS+(
++++SELECT+name+FROM+table1
++++UNION
++++SELECT+name+FROM+table2
++++UNION
++++SELECT+name+FROM+table3
++++UNION
++++SELECT+name+FROM+table4
)
SELECT
++cte.name,
++table1.closedid,
++table2.openid,
++table3.assign,
++table4.unassign
FROM
++cte
++LEFT+JOIN+table1+ON+cte.name+=+table1.name
++LEFT+JOIN+table2+on+cte.name+=+table2.name
++LEFT+JOIN+table3+on+cte.name+=+table3.name
++LEFT+JOIN+table4+on+cte.name+=+table4.name|code-block|syntax|javascript|5682886|SQLFiddle|offset|length|5682887|entityMap|0|LINK|mutability|MUTABLE|url|http://sqlfiddle.com/#!6/97f05/12^0|0|0|0|9|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

I would grab all the distinct names and then do left joins with that list, would make it easier to understand what's going on.

<pre><code>;WITH cte(name)
AS (
 SELECT name FROM table1
 UNION
 SELECT name FROM table2
 UNION
 SELECT name FROM table3
 UNION
 SELECT name FROM table4
)
SELECT
 cte.name,
 table1.closedid,
 table2.openid,
 table3.assign,
 table4.unassign
FROM
 cte
 LEFT JOIN table1 ON cte.name = table1.name
 LEFT JOIN table2 on cte.name = table2.name
 LEFT JOIN table3 on cte.name = table3.name
 LEFT JOIN table4 on cte.name = table4.name
</code></pre>

<a href="http://sqlfiddle.com/#!6/97f05/12" rel="nofollow">SQLFiddle</a>

blocks|key|1898121|text|我想，这就是你要找的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1898122|++SELECT+coalesce(t1.NAME,+t2.NAME,+t3.NAME,+t4.NAME)+NAME
++++,t1.closedid
++++,t2.openedid
++++,t3.assign
++++,t4.unassign
FROM+[table-1]+t1
FULL+OUTER+JOIN+[table-2]+t2+ON+t1.NAME+=+t2.NAME
FULL+OUTER+JOIN+[table-3]+t3+ON+isnull(t1.NAME,+t2.NAME)+=+t3.NAME
FULL+OUTER+JOIN+[table-4]+t4+ON+coalesce(t1.NAME,+t2.NAME,+t3.NAME)+=+t4.NAME|code-block|syntax|javascript|1898123|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I think, this is what you are looking for:

<pre><code> SELECT coalesce(t1.NAME, t2.NAME, t3.NAME, t4.NAME) NAME
 ,t1.closedid
 ,t2.openedid
 ,t3.assign
 ,t4.unassign
FROM [table-1] t1
FULL OUTER JOIN [table-2] t2 ON t1.NAME = t2.NAME
FULL OUTER JOIN [table-3] t3 ON isnull(t1.NAME, t2.NAME) = t3.NAME
FULL OUTER JOIN [table-4] t4 ON coalesce(t1.NAME, t2.NAME, t3.NAME) = t4.NAME
</code></pre>

blocks|key|5682921|text|select
n.name,+t1.closedid,+t2.openedid,+t3.assign,+t4.unassign
from+
(select+name+from+[table-1]
union
select+name+from+[table-2]
union
select+name+from+[table-3]
union
select+name+from+[table-4]
)+as+n+
left+join+[table-1]+t1+on+n.name++=+t1.name
left+join+[table-2]+t2+on+n.name+=+t2.name
left+join+[table-3]+t3+on+n.name+=+t3.name
left+join+[table-4]+t4+on+n.name+=+t4.name|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|5682922|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>select
n.name, t1.closedid, t2.openedid, t3.assign, t4.unassign
from 
(select name from [table-1]
union
select name from [table-2]
union
select name from [table-3]
union
select name from [table-4]
) as n 
left join [table-1] t1 on n.name = t1.name
left join [table-2] t2 on n.name = t2.name
left join [table-3] t3 on n.name = t3.name
left join [table-4] t4 on n.name = t4.name
</code></pre>

blocks|key|4214561|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4214562|使用UNION获取您独特的名称列表|ordered-list-item|offset|length|style|CODE|4214563|使用LEFT+JOIN检索您的data|4214564|4214565|示例：|4214566|SELECT+list.name,+t1.assigned,+t2.unsassigned,+t3.closed,+t4.open
FROM+
+++(select+name+from+table1
++++union+select+name+from+table2
++++union+select+name+from+table3
++++union+select+name+from+table4
+++)+list
LEFT+JOIN+table1+t1+ON+(list.name+=+t1.name)
LEFT+JOIN+table2+t2+ON+(list.name+=+t2.name)
LEFT+JOIN+table3+t3+ON+(list.name+=+t3.name)
LEFT+JOIN+table4+t4+ON+(list.name+=+t4.name)|code-block|syntax|javascript|4214567|entityMap^0|0|2|5|0|2|9|0|0|0|0^^$0|@$1|2|3|-4|4|5|6|T|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|U|7|@$D|V|E|W|F|G]]|8|@]|9|$]]|$1|H|3|I|4|C|6|X|7|@$D|Y|E|Z|F|G]]|8|@]|9|$]]|$1|J|3|-4|4|5|6|10|7|@]|8|@]|9|$]]|$1|K|3|L|4|5|6|11|7|@]|8|@]|9|$]]|$1|M|3|N|4|O|6|12|7|@]|8|@]|9|$P|Q]]|$1|R|3|-4|4|5|6|13|7|@]|8|@]|9|$]]]|S|$]]

<ol>
<li>Use <code>UNION</code> to get your distinct list of names</li>
<li>Use <code>LEFT JOIN</code> to retrieve your data</li>
</ol>

Example:

<pre><code>SELECT list.name, t1.assigned, t2.unsassigned, t3.closed, t4.open
FROM 
 (select name from table1
 union select name from table2
 union select name from table3
 union select name from table4
 ) list
LEFT JOIN table1 t1 ON (list.name = t1.name)
LEFT JOIN table2 t2 ON (list.name = t2.name)
LEFT JOIN table3 t3 ON (list.name = t3.name)
LEFT JOIN table4 t4 ON (list.name = t4.name)
</code></pre>

table-1

<pre><code>name closedid
rere 4
trtr 5
ewew 6
</code></pre>

And
table-2

<pre><code>name openedid 
rere 6
trtr 7
ytyt 8 
uyuy 5
</code></pre>

And

<pre><code>table-3
name assign 
rere 6
ytyt 8 
uyuy 5
rtyy 9
</code></pre>

And
table-4

<pre><code>name unassign 
rere 6
trtr 7
errt 5
hdtg 9
</code></pre>

I want final output like this:

<pre><code>name closedid opened assign unassign
rere 4 6 6 6
trtr 5 7 null 7
ytyt null 8 8 null
uyuy null 5 5 null
ewew 6 null null null
rtyy null null 9 null
errt null null null 5
hdtg null null null 9
</code></pre>

how to join 4 table into single table?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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表-1name  closedidrere    4trtr    5ewew    6和表2name  openedid rere    6trtr    7ytyt    8 uyuy    5和table-3name  assign rere    6ytyt    8 uyuy    5rtyy    9和表4...

问如何将4个表连接成一个表？
EN

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问如何将4个表连接成一个表？EN