Question

我有两张桌子：表1具有5列（col1，col2，col3，col4，val1）和表2，表1具有5列（col1，col2，col3，col4，val2）。 1. Table1.val1不包含任何值。 2.对于Table1和Table2，col1，col2，col3和col4是相同的，它们是主键。

我想做的是，当Table1.col1 = Table2.col1和Table1.col2 = Table2.col2和Table1.col3 = Table2.col3时，用sum（Table2.val2）更新Table1.val1 Table1.col4 = Table2.col4。

我做了类似的事情：

UPDATE Table1
SET val1 = (

select t_sommevbrute.sumvalbrute from (
Select  col1,col2,col3,col4,SUM(val2) sumvalbrute
From Table2
Where col3 = 2014 And col2=51 
GROUP BY col1, col2, col3, col4) t_sommevbrute

WHERE Table1.col1 = t_sommevbrute.col1
and Table1.col2 = t_sommevbrute.col2
and Table1.col3 = t_sommevbrute.col3
and Table1.col4 = t_sommevbrute.col4)

但与此问题相关：Oracle SQL: Update a table with data from another table，我应该有WHERE EXISTS条款。

任何帮助请!! 感谢。

Answer 1

您可以通过以下方式执行此操作： -

使用临时表： - 首先创建临时表： -

Create table temp1 as
Select  col1,col2,col3,col4,SUM(val2) as sumvalbrute
     From table2
     Where col3 = 3 And col2=2 
     GROUP BY col1, col2, col3, col4;

然后使用Temp表更新Main Table1： -

UPDATE table1 SET table1.val1 = (SELECT temp1.sumvalbrute
                                  FROM temp1 
                                  WHERE Table1.col1 = temp1.Col1
                                  AND Table1.col2 = temp1.Col2
                                  AND Table1.col3 = temp1.Col3
                                  AND Table1.col4 = temp1.Col4);

SQL小提琴： - http://sqlfiddle.com/#!4/4864d/5

使用临时表： -

UPDATE table1 SET table1.val1 = (SELECT temp1.sumvalbrute
                                 FROM 
                                 (Select  col1,col2,col3,col4,SUM(val2) as sumvalbrute
                                  From table2
                                  Where col3 = 3 And col2=2 
                                  GROUP BY col1, col2, col3, col4) temp1 
                                  WHERE Table1.col1 = temp1.Col1
                                  AND Table1.col2 = temp1.Col2
                                  AND Table1.col3 = temp1.Col3
                                  AND Table1.col4 = temp1.Col4);

SQL小提琴： - http://sqlfiddle.com/#!4/c9286/2

Answer 2

您可以大大简化查询：

UPDATE Table1
    SET val1 = (select SUM(val2)
                from Table2 t
                where Table1.col1 = t.col1 and
                      Table1.col2 = t.col2 and
                      Table1.col3 = t.col3 and
                      Table1.col4 = t.col4
               )
    Where col3 = 2014 And col2 = 51;

你可以使用where exists。目的是在没有匹配时阻止NULL值。但是，我认为将where子句拉到外面可能解决了这个问题。我注意到where子句使用的值与相关条件相同。

Answer 3

你可以试试这个：

<?php
$url = "https://www.indiegogo.com/projects/over-and-above-africa-the-solution-to-end-poaching-drones";
$timeout = 30;
$http = curl_init($url);
curl_setopt($http, CURLOPT_RETURNTRANSFER, true);
curl_setopt($http, CURLOPT_COOKIEJAR, 'cookie.txt');
curl_setopt($http, CURLOPT_USERAGENT, 'Mozilla/5.0 (Windows; U; Windows NT 6.1; fr; rv:1.9.2.13) Gecko/20101203 Firefox/3.6.13');
curl_setopt($http, CURLOPT_FRESH_CONNECT, true);
curl_setopt($http, CURLOPT_FOLLOWLOCATION, true);
curl_setopt($http, CURLOPT_TIMEOUT, $timeout);
curl_setopt($http, CURLOPT_CONNECTTIMEOUT, $timeout);

if (preg_match('`^https://`i', $url))
{
    curl_setopt($http, CURLOPT_SSL_VERIFYPEER, false);
    curl_setopt($http, CURLOPT_SSL_VERIFYHOST, 0);
}

        curl_setopt($http, CURLOPT_HEADER, true);
        curl_setopt($http,CURLOPT_HTTPHEADER,array('Accept-Language', 'nl,en;q=0.7;en-us;q=0.3'));
$sContenu = curl_exec($http);

echo curl_getinfo($http, CURLINFO_HTTP_CODE);

curl_close($http);
echo $sContenu;
?>

Answer 4

只能使用合并语句来实现这种更新：

Merge into Table1 using ( select SUM(val2) as val2, t.col1, t.col2, t.col3, t.col4 from Table2 t group by t.col1, t.col2, t.col3, t.col4 ) t -- This is the alias for table 2 on ( Table1.col1 = t.col1 and Table1.col2 = t.col2 and Table1.col3 = t.col3 and Table1.col4 = t.col4 ) when matched then UPDATE Table1.val1 = t.val2 --HERE IS YOUR **UPDATE** where Table1.col3 = 2004 and Table1.col2 = 51 ; commit;

请不要忘记将别名添加到表中，并在此代码后添加一个良好的COMMIT 此解决方案避免您创建临时表，并且效果很好

最诚挚的问候。

Oracle SQL：使用另一个表的SUM查询更新列

4 个答案: