我有2个表'table1'和'table2'。
table1
有10,000条记录,table2
有5,000条记录。
两个表都有“RECORD_ID”列。 table2
中的所有RECORD_ID都可以在table1
中找到。
我想根据table1
的“RECORD_ID”列更新table2
的“PRICE”列。
update table1 set PRICE = table2.PRICE where RECORD_ID = table2.RECORD_ID
我收到此错误消息:
SQL0206N“table2.PRICE”在使用它的上下文中无效 SQLSTATE = 42703
我正在使用DB2。
答案 0 :(得分:2)
UPDATE table1 SET table1.price = (SELECT table2.price FROM table2 WHERE table2.record_id = table1.record_id)
答案 1 :(得分:1)
你必须使用这样的连接:
UPDATE
Table1
SET
Table1.Column = T2.Column
FROM
Table1 T1
INNER JOIN
Table2 T2
ON
T1.PK = T2.FK;
答案 2 :(得分:1)
试试这个:
<div class="set-size charts-container">
<div class="pie-wrapper progress-30">
<span class="label">30<span class="smaller">%</span></span>
<div class="pie">
<div class="left-side half-circle"></div>
<div class="right-side half-circle"></div>
</div>
</div>
<div class="pie-wrapper progress-60">
<span class="label">60<span class="smaller">%</span></span>
<div class="pie">
<div class="left-side half-circle"></div>
<div class="right-side half-circle"></div>
</div>
</div>
<div class="pie-wrapper progress-90">
<span class="label">90<span class="smaller">%</span></span>
<div class="pie">
<div class="left-side half-circle"></div>
<div class="right-side half-circle"></div>
</div>
</div>
<div class="pie-wrapper progress-45 style-2">
<span class="label">45<span class="smaller">%</span></span>
<div class="pie">
<div class="left-side half-circle"></div>
<div class="right-side half-circle"></div>
</div>
<div class="shadow"></div>
</div>
<div class="pie-wrapper progress-75 style-2">
<span class="label">75<span class="smaller">%</span></span>
<div class="pie">
<div class="left-side half-circle"></div>
<div class="right-side half-circle"></div>
</div>
<div class="shadow"></div>
</div>
</div>