Question

我有来自不同请求的结果作为例子。

第一次要求：

SELECT SUM(PRODUCT_QTY) FROM TABLEBLA
WHERE STATE='DONE' and  location_dest_id in (12,18) and location_id=5

给出：

sum(product_qty)
----------------
1200

第二次要求：

SELECT SUM(PRODUCT_QTY) FROM TABLEBLA
WHERE STATE='CANCEL' and location_id (12,18) and location_dest_id=5

给出：

sum(product_qty)
----------------
900

我希望将结果显示为1200-900 = 300

Answer 1

一种简单的方法是条件聚合：

WHERE

或者，如果这是较大查询的一部分，并且您不希望在SELECT (SUM(CASE WHEN STATE = 'DONE' THEN PRODUCT_QTY ELSE 0 END) - SUM(CASE WHEN STATE = 'CANCEL' THEN PRODUCT_QTY END) ) as diff FROM TABLEBLA WHERE location_dest_id in (12, 18) and location_id = 5;子句中进行过滤：

$print_projects .= <<<HTML
    <div id="$id" class="project">
        $projectname  has the number $projectnumber
        <div class="specific-informations">
            Projectname: $projectname<br>
            Projectnumber: $projectnumber
        </div>
    </div>
HTML;

Answer 2

SELECT SUM(CASE WHEN STATE = 'DONE'  THEN PRODUCT_QTY ELSE 0 END) - 
SUM(CASE WHEN STATE = 'CANCEL' THEN PRODUCT_QTY ELSE 0 END) 
 FROM TABLEBLA WHERE location_id IN (12,18) AND location_dest_id=5

在单个查询中获得总和的差异

2 个答案: