如何在 mysql 查询中获取最新的商品价格？答案

【问题标题】：How can I get the latest item price in mysql query?如何在 mysql 查询中获取最新的商品价格？
【发布时间】：2018-03-12 03:01:25
【问题描述】：

我设法获得了该商品的最低、最高和平均价格，但无法获得最新价格。下面是我通过加入 item 和 item_price 表使用的选择查询。我该如何解决这个问题？

$sql = 'SELECT *, MIN(ip_price) AS lowest_price, MAX(ip_price) AS highest_price, 
        AVG(ip_price) AS average_price, MAX(ip_price_date) AS latest_date,
        (SELECT ip_price FROM cnf_item_price WHERE ip_price_date = "latest_date") AS latest_price
        FROM cnf_item
        INNER JOIN cnf_item_price ON cnf_item_price.ip_item_id = cnf_item.it_id
        WHERE 1 AND cnf_item_price.ip_supp_id=?
        GROUP BY cnf_item.it_id
        ORDER BY cnf_item.it_name ASC';
$stmt = $DB->prepare($sql);
$stmt->bindValue(1,$supplier_id);
$stmt->execute();

【问题讨论】：

1.不要使用邪恶的SELECT *。修复它。然后回复我们。
要获得最后一次更新返回，您可以使用受 PHP 函数影响的行查看stackoverflow.com/questions/7368225/…
哦，看看Why should I provide an MCVE for what seems to me to be a very simple SQL query

标签： php mysql

【解决方案1】：

SELECT it_id, it_code, it_name, it_desc, 
            ip_id, ip_item_id, ip_supp_id, ip_price, ip_price_date, ip_ref_no, ip_remarks, 
            MIN(ip_price) AS lowest_price, MAX(ip_price) AS highest_price,
            AVG(ip_price) AS average_price, MAX(ip_price_date) AS latest_date,
            (SELECT ip_price FROM cnf_item_price WHERE cnf_item_price.ip_item_id = cnf_item.it_id
            ORDER BY ip_price_date DESC LIMIT 1) AS latest_price
            FROM cnf_item
            INNER JOIN cnf_item_price ON cnf_item_price.ip_item_id = cnf_item.it_id
            WHERE 1 
            GROUP BY cnf_item.it_id
            ORDER BY cnf_item.it_name ASC;

【讨论】：

【解决方案2】：

谢谢#ahmet kamaran。它在这里适用于我的情况。我还将 * 替换为其他人建议所需的那些列。因此，在对上述建议进行了一些测试之后，这是我的代码，用于检索我需要的那些数据。

$sql = 'SELECT it_id, it_code, it_name, it_desc, 
        ip_id, ip_item_id, ip_supp_id, ip_price, ip_price_date, ip_ref_no, ip_remarks, 
        MIN(ip_price) AS lowest_price, MAX(ip_price) AS highest_price,
        AVG(ip_price) AS average_price, MAX(ip_price_date) AS latest_date,
        (SELECT ip_price FROM cnf_item_price ORDER BY ip_price_date DESC LIMIT 1) AS latest_price
        FROM cnf_item
        INNER JOIN cnf_item_price ON cnf_item_price.ip_item_id = cnf_item.it_id
        WHERE 1 AND cnf_item_price.ip_supp_id=?
        GROUP BY cnf_item.it_id
        ORDER BY cnf_item.it_name ASC';

$stmt = $DB->prepare($sql);
$stmt->bindValue(1,$supplier_id);
$stmt->execute();

http://sqlfiddle.com/#!9/ca6234/2

谢谢。

【讨论】：

【解决方案3】：

您可以尝试以下查询：

$sql = 'SELECT *, MIN(ip_price) AS lowest_price, MAX(ip_price) AS highest_price, 
        AVG(ip_price) AS average_price, MAX(ip_price_date) AS latest_date,
        (SELECT ip_price FROM cnf_item_price order by ip_price_date desc limit 1) AS latest_price
        FROM cnf_item
        INNER JOIN cnf_item_price ON cnf_item_price.ip_item_id = cnf_item.it_id
        WHERE 1 AND cnf_item_price.ip_supp_id=?
        GROUP BY cnf_item.it_id
        ORDER BY cnf_item.it_name ASC';
$stmt = $DB->prepare($sql);
$stmt->bindValue(1,$supplier_id);
$stmt->execute();

【讨论】：

子选择应该是相关的。并且您不应该在带有 GROUP BY 的语句中使用 SELECT *。