将来自多个表的类似数据分组并添加结果

时间:2015-02-28 02:27:02

标签: php mysqli sum average

我希望能够显示客户拥有的商品数量和总数,并为每个商品和总价格添加价格。我无法解决如何做到这一点的逻辑。到目前为止,我可以显示客户的所有项目,但有些项目会重复。

$clientid = 24;

$query = "SELECT items.itemid, prices.priceid FROM items 
JOIN prices ON items.itemid = prices.itemid 
WHERE items.clientid = '$clientid'";


$result =  $db_connect -> query($query);
$row_cnt = $result -> num_rows; 

echo $joinrow_cnt."<br/><br/><br/>";

while($assocresult = mysqli_fetch_assoc($result)){
    $price = $assocresult['priceid'];
    $item = $assocresult['itemid'];
    $realprice = ($price/100)*80;        //after 80% discount
    echo "ItemId: ".$item." Price : " .$price." real price: " .$realprice."<br/>";  
}

到目前为止,我得到了:

ItemId: 1 Price : 100 real price: 80
ItemId: 3 Price : 25 real price: 20
ItemId: 1 Price : 100 real price: 80
ItemId: 5 Price : 200 real price: 160
ItemId: 3 Price : 25 real price: 20
ItemId: 1 Price : 100 real price: 80

了解ItemId 1和3的重复方式。有很多这样的行,其中重复了项ID。

我希望输出如下:

ItemId: 1x3 Price: 100 real price = 80x3=240
ItemId: 3x2 Price: 25 real price = 20x2=50
ItemId: 5   Price: 200 real price: 160 //since this item is listed only once.

请帮忙。非常感谢!!

2 个答案:

答案 0 :(得分:2)

检查解决方案如下:

$clientid = 24;

$query = "SELECT items.itemid, prices.priceid, count(items.itemid) as itemcnt FROM items
JOIN prices ON items.itemid = prices.itemid 
WHERE items.clientid = '$clientid' group by items.itemid";


$result =  $db_connect -> query($query);
$row_cnt = $result -> num_rows; 

echo $row_cnt."<br/><br/><br/>";

while($assocresult = mysqli_fetch_assoc($result)){
$price = $assocresult['priceid'];
$item = $assocresult['itemid'];
$itemcnt = $assocresult['itemcnt'];
$realprice = ($price/100)*80;        //after 80% discount
echo "ItemId: ".$item."x".$itemcnt." Price: " .$price." real price = " .$realprice."x".$itemcnt."=".($realprice*$itemcnt)."<br/>";
}

我在第10行上没有收到任何错误。您的代码中$db_connect的价值是多少?

让我知道。谢谢!

答案 1 :(得分:1)

您可以获取所有itemsprices数据并将其与WHERE匹配,然后使用GROUP BY将重复数据收集为一行。使用COUNT(..),您可以获得每个项目的行数。

这应该可以解决问题:

SELECT
    COUNT(items.itemid) as cnt,
    items.itemid,
    prices.priceid
FROM items, prices
WHERE 
    items.itemid = prices.itemid
AND
    items.clientid = '$clientid'
GROUP BY items.itemid