我的sql代码执行但我不知道如何获得purchase_request total_qty和purchase_order数量的区别。
表Purchase_Order
counter | qty |
---------------
100001 | 10 |
100001 | 10 |
100001 | 10 |
100004 | 30 |
表Purchase_Request
counter | total_qty |
---------------------
100001 | 50 |
100002 | 100 |
100003 | 50 |
100004 | 70 |
我想像这样编码,但我不知道如何在我的代码中混合它。
a.total_qty-b.qty as balance
这是我的代码
<?php
$mysqli = new mysqli("localhost", "root", "", "test");
$result = $mysqli->query("
select a.counter,a.total_qty from purchase_request a inner join purchase_order b on a.counter= b.counter group by a.counter
");
echo'<table id="tfhover" cellspacing="0" class="tablesorter" style="text-transform:uppercase;" border="1px">
<thead>
<tr>
<th></th>
<th>counter</th>
<th>QTY</th>
<th>balance</th>
</tr>
</thead>';
echo'<tbody>';
$i=1;
while($row = $result->fetch_assoc()){
echo'<tr>
<td>'.$i++.'</td>
<td>'.$row['counter'].'</td>
<td>'.$row['total_qty'].'</td>
<td>'.$row['balance'].'</td>
</tr>';
}
echo "</tbody></table>";
?>
答案 0 :(得分:0)
你试过这个吗?
select a.counter,
a.total_qty,
a.total_qty - b.qty balance
from (select counter,
sum(total_qty) total_qty
form purchase_request
group by counter) a
inner join (select counter,
sum(qty) qty
from purchase_order
group by counter) b
on a.counter= b.counter
group by a.counter
编辑:我知道了,你需要的是汇总你的数量,然后做数学
答案 1 :(得分:-1)
select a.counter,
a.total_qty,
sum(a.total_qty) - sum(b.qty) as balance
from purchase_request a
left inner join purchase_order b
on a.counter= b.counter
group by a.counter