Question

我尝试使用相同的“pocde”列减去2个不同的表。我已经说明了下面的代码，但它对我不起作用。任何的想法？

$result = mysql_query("
    SELECT 
        productlist.*, 
        SUM(rreturn.total)-SUM(rreturn.total) as totals, 
        SUM(rsales.tax)-SUM(rreturn.tax) as tax
    FROM productlist
    LEFT JOIN rsales ON rsales.pcode = productlist.pcode
    LEFT JOIN rreturn ON rreturn.pcode = productlist.pcode
    GROUP BY pcode
    ORDER BY total ASC");

    while($row = mysql_fetch_array($result)) {
            echo '<tr>';
            echo '<td style = "background-image:url(images/buts1.png)">'.($row['totals'].'</td>';   
            echo '<td style = "background-image:url(images/buts1.png)">'.($row['tax'].'</td>';              
            echo '</tr>';
    }

Answer 1

由于分组依据，您无法执行此查询。您可以像这样更改您的查询：

SELECT pl.*, plagg.totals, plagg.tax
from productlist pl
inner join (
    SELECT 
        productlist.pcode, 
        SUM(rreturn.total)-SUM(rreturn.total) as totals, 
        SUM(rsales.tax)-SUM(rreturn.tax) as tax 
    FROM productlist
    LEFT JOIN rsales ON rsales.pcode = productlist.pcode
    LEFT JOIN rreturn ON rreturn.pcode = productlist.pcode
    GROUP BY productlist.pcode
) plagg on (plagg.pcode = pl.pcode)

Answer 2

PHP代码中存在错误，因为$row = mysql_fetch_array($result)会创建数字数组。您必须使用$row[0]，$row[1]等。如果您想尝试在代码中使用关联的数组（$row['totals']等），则必须使用

mysql_fetch_array($result, MYSQL_ASSOC)

或您使用

mysql_fetch_assoc($result)

。请注意，两者都已弃用，您应该使用PDO。

加入两个不同的表

2 个答案: