提出这个问题的目的只是学习如何在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)可以为以下MYSQL查询嵌套提供更好的解决方案吗?
$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合并到一个查询中?
$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"]);
}
希望学习一些事情!!
答案 0 :(得分:2)
答案 1 :(得分:0)
在选择多个列时,如果没有GROUP BY,则无法使用COUNT()语句(聚合查询)。您还应该使用join语句从不同的表中选择值。这样的事情可能是:
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
只需在查询中使用FROM tblwriter, tblordersub
即可创建交叉联接,即表中数据的cartesian product。你可能不希望这样。