我已经编写了查询来获取数据。
select Terr, Product
from sdata, md
where sdata.PSetID = md.PsetID;
我想将此获取的数据复制到新表中。
我尝试了以下查询,但它没有工作:
select *
into ttl
from
(select Terr, Product
from sdata, md
where sdata.PSetID = md.PsetID);
我做错了什么?
答案 0 :(得分:3)
您需要将sub-select
名称添加到SELECT *
INTO ttl
FROM (SELECT Terr,
Product
FROM sdata
INNER JOIN md
ON sdata.PSetID = md.PsetID) A; --Here
SELECT Terr,
Product
INTO ttl
FROM sdata
INNER JOIN md
ON sdata.PSetID = md.PsetID
或者只是
SELECT A.* FROM (
Select * FROM TABLE1
) A
WHERE (A.ColumnOne >= 18 OR ColumnOne IS NULL)
UNION
SELECT B.* FROM (
Select * FROM TABLE2
) B
WHERE (B.ColumnOne >= 18 OR ColumnOne IS NULL)
UNION
SELECT C.* FROM (
Select * FROM TABLE2
) C
WHERE (C.ColumnOne >= 18 OR ColumnOne IS NULL)
答案 1 :(得分:0)
另一个选项是使用以下语法:
SELECT column_name(s) INTO新表格 FROM table1;
在你的情况下:
<div class="form-group col-md-12 input_fields_wrap">
<label for="merk">Merk</label>
<div class="input-group">
<input type="text" class="form-control" name="merk[]">
<span class="input-group-btn">
<button class="btn btn-secondary add_field_button" type="button">add row</button>
</span>
</div>
<div id="1"></div>
</div>