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问这个问题的目的只是为了学习如何在 MYSQL 中进行嵌套查询。

1）以下查询有什么问题？

"SELECT tblwriter.writer_alias, tblwriter.writer_first_name, COUNT(tblordersub.suborder_alias) AS totalOrders FROM tblwriter, tblordersub WHERE tblwriter.writer_isactive = 1 AND tblordersub.writer_alias = tblwriter.writer_alias"

2）对于下面的MYSQL查询嵌套是否有更好的解决方案？

$currentorders = $db->rawQuery("SELECT order_title,order_alias FROM tblorder WHERE company_id=? AND YEAR(order_date) = YEAR(CURDATE()) AND MONTH(order_date) = MONTH(CURDATE())",$params);
$orderssummary = array();
if(!empty($currentorders)){
    foreach($currentorders as $corder){
        $param = array($corder["order_alias"]);
        $oprice = $db->rawQuery("SELECT payment_amount FROM tblpayment WHERE payment_status = 1 AND writer_alias IS NULL AND order_alias=?",$param);
        $itssuborders = $db->rawQuery("SELECT suborder_alias FROM tblordersub WHERE order_alias=?",$param);
        $thesuborders = array_implode("",",",$itssuborders);
        $cost = $db->rawQuery("SELECT SUM(payment_amount) AS total_subtotal FROM tblpayment WHERE writer_alias IS NOT NULL AND suborder_alias IN (".$thesuborders.")");
        $orderssummary[] = array("title"=>$corder["order_title"],"price"=>$oprice[0]["payment_amount"],"cost"=>$cost[0]["total_subtotal"]);
    }
}

3）有没有办法将以下三个查询合并为1？

"SELECT SUM(payment_amount) AS totalAmount FROM tblpayment WHERE company_id=? AND payment_status = 1 AND order_alias IS NOT NULL AND YEAR(payment_add_datetime) = YEAR(CURDATE()) AND MONTH(payment_add_datetime) = MONTH(CURDATE() - INTERVAL 2 MONTH)"
"SELECT SUM(payment_amount) AS totalAmount FROM tblpayment WHERE company_id=? AND payment_status = 0 AND writer_alias IS NOT NULL AND YEAR(payment_add_datetime) = YEAR(CURDATE()) AND MONTH(payment_add_datetime) = MONTH(CURDATE() - INTERVAL 2 MONTH)"
"SELECT SUM(payment_amount) AS totalAmount FROM tblpayment WHERE company_id=? AND payment_status = 1 AND writer_alias IS NOT NULL AND YEAR(payment_add_datetime) = YEAR(CURDATE()) AND MONTH(payment_add_datetime) = MONTH(CURDATE() - INTERVAL 2 MONTH)"

4）与第二个问题相同，有没有办法将以下查询组合成1个查询？

$biggest_customers = $db->rawQuery("SELECT payment_user_id, SUM(payment_amount) AS totalEARNED FROM tblpayment WHERE writer_alias IS NULL AND order_alias IS NOT NULL GROUP BY payment_user_id ORDER BY totalEARNED DESC LIMIT 10");
for($i=0;$i<count($biggest_customers);$i++){
    $params = array($biggest_customers[$i]["payment_user_id"]);
    $customerinformation = $db->rawQuery('SELECT customer_alias FROM tblcustomer WHERE user_id=?',$params);
    $biggest_customers[$i]["customer_alias"] = $customerinformation[0]["customer_alias"];
    unset($biggest_customers[$i]["payment_user_id"]);
}

希望能学到东西！！

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2 回答 2

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  1. 查看聚合查询
  2. 查看 JOIN
  3. 请参阅 CASE 语句/数据透视表查询
  4. 再次查看加入
于 2013-10-11T16:36:43.560 回答
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you can not use the COUNT() statement (aggregation query) without a GROUP BY when selecting multiple columns. You also should rather use a join statement to select values from different tables. Something like this maybe:

SELECT tblwriter.writer_alias, tblwriter.writer_first_name, COUNT(tblordersub.suborder_alias) AS totalOrders 
FROM tblwriter 
INNER JOIN tblordersub ON (tblordersub.writer_alias = tblwriter.writer_alias) 
WHERE tblwriter.writer_isactive = 1
GROUP BY tblordersub.suborder_alias

By just using FROM tblwriter, tblordersub in your query you are creating a cross join, which is the cartesian product of the data in your tables. You probably don't want that.

于 2013-10-11T16:55:03.627 回答