我有两个表tbl_user1和tbl_user2这两个字段的字段名相同，但现在这些表之间没有任何关系我希望从两个表中找到总引用计数。 ..
tbl_user1
-----------------------
UID | referenceBy | firstName | lastName | emailAddress
----------------------------------------------------------------------------
1 | NULL | aa1 | ab1 | aa1@email.com
2 | aa1@email.com | aa2 | ab2 | aa2@email.com
3 | NULL | aa3 | ab3 | aa3@email.com
4 | aa2@email.com | aa4 | ab4 | aa4@email.com
5 | aa2@email.com | aa5 | ab5 | aa5@email.com
6 | bb1@email.com | aa6 | ab6 | aa6@email.com
7 | bb2@email.com | aa7 | ab7 | aa7@email.com
8 | bb3@email.com | aa8 | ab8 | aa8@email.com
9 | bb3@email.com | aa9 | ab9 | aa9@email.com
和第二张桌子就像......左右。
tbl_user2
-----------------------
UID | referenceBy | firstName | lastName | emailAddress
----------------------------------------------------------------------------
1 | NULL | bb1 | bc1 | bb1@email.com
2 | bb1@email.com | bb2 | bc2 | bb2@email.com
3 | NULL | bb3 | bc3 | bb3@email.com
4 | bb3@email.com | bb4 | bc4 | bb4@email.com
5 | bb2@email.com | bb5 | bc5 | bb5@email.com
6 | bb1@email.com | bb6 | bc6 | bb6@email.com
7 | aa2@email.com | bb7 | bc7 | bb7@email.com
8 | aa3@email.com | bb8 | bc8 | bb8@email.com
9 | bb5@email.com | bb9 | bc9 | bb9@email.com
现在，你可以看到这两个表之间没有任何关系，我希望结果如下......
MAIN_RESULT_THAT_I_WANT
-----------------------
referenceEmail | referenceEmailCount
----------------------------------------------------------------------------
aa1@email.com | 1
aa2@email.com | 3
aa3@email.com | 1
aa4@email.com | 0
aa5@email.com | 0
aa6@email.com | 0
aa7@email.com | 0
aa8@email.com | 0
aa9@email.com | 0
bb1@email.com | 3
bb2@email.com | 2
bb3@email.com | 3
bb4@email.com | 0
bb5@email.com | 1
bb6@email.com | 0
bb7@email.com | 0
bb8@email.com | 0
bb9@email.com | 0
这里结果是所有用户的所有emailAddress以及该特定emailAddress注册的用户总数。
答案 0 :(得分：2)
我猜你想要的结果只是复制和粘贴，因为它似乎不准确。就像HoneyBadger说的那样，aa6缺失并且结果仍然很奇怪，这表明你有另一个你没有告诉我们的清单？或者你只是把结果写在记事本......
如果您只想要一个电子邮件列表并计算，这将有效：
select referenceBy, count(1) as referenceEmailCount from (
select referenceBy from tbl_user1
union all
select referenceBy from tbl_user2
) as t
group by referenceBy
如果不是您需要的话，请提供更多信息。
答案 1 :(得分：0)
由于2个表的模式相同，因此您可以执行<div id="myJershy">
<img src="image/Pngs/collar/White.png" id="grey" style="position: fixed; top: 8px; left: 12px; width: 935px;">
<img class="orange" id="orange" src="image/Pngs/body/White.png" style="position: fixed; left: 10px; top: 8px;">
<img src="image/Pngs/sleeves/White.png" id="gold" style="position: fixed; top: 12px; left: 22px;">
<img class="back" id="black" src="image/Pngs/cuffs/White.png" style="position: fixed; left: 11px; top: 9px;">
</div>以获得组合结果，并可以执行外部查询以获得总计数。
union