MYSQL 复杂查询 500 万行，错误 1317 [已关闭]

Question

我有一个包含 500 万行的表。 示例

+----+-----------+---------+-------
|  product    | date       |  count    | 
+-------------+------------+-----------+
|  платье     | 04-01-2024 | 186574    | 
|  штаны      | 04-01-2024 | 20564     |
|  кепка      | 07-01-2024 | 104443    | 
|  штаны      | 06-01-2024 | 10574     | 
|  платье     | 06-01-2024 | 223001    |
+-------------+------------+-----------+

我需要选择具有 2 个自定义日期的行，然后如果产品 name1 = name2 获得计数之间的差异，然后获得此差异的前 100 名。

现在我创建 2 个选择，然后循环遍历数组来获取它。但我有关于记忆的错误。我如何在 mysql 中创建 1 个查询？

我的询问

    $get_ex = $connection->prepare("SELECT * FROM product WHERE date = :date");
    $get_ex->execute(array(':date' => $first_date));
    $ex_list_2 = $get_ex->fetchAll(PDO::FETCH_ASSOC);

    
    $get_ex_3 = $connection->prepare("SELECT * FROM product WHERE date = :date");
    $get_ex_3->execute(array(':date' => $second_date));
    $ex_list_3 = $get_ex_3->fetchAll(PDO::FETCH_ASSOC);

    
    $result = Array();
    $i=0;
    foreach($ex_list_2 as $key => $value)
    {
        foreach($ex_list_3 as $key2 => $value2)
        {
            if($value['product_name'] == $value2['product_name'])
            {
                $result[$i]['product_name'] = $value['product_name'];
                $result[$i]['count_1'] = $value['count'];
                $result[$i]['count_2'] = $value2['count'];
                $result[$i]['res'] = $value['count']- $value2['count'];
                $i++;
            }
        }
        
    }

和结果-

致命错误：允许的内存大小 1073741824 字节已耗尽

内存限制= 2048mb

对于结果，我需要带有下一个信息的数组

+-------------+-------------+-------------+-----------+
|  product    | 04-01-2024  |  06-01-2024 |  diff     | 
+-------------+-------------+-------------+-----------+
|  платье     | 186574      | 223001      | 36427     |
|  штаны      | 20564       | 10574       | 99000     |
+-------------+-------------+-------------+-----------+

无法使用 order by 并且在 mysql 中出现错误 1317

Answer 1

不要执行两个查询，使用单个

JOIN

查询。

SELECT t1.product_name, t1.count AS count1, t2.count AS count2, t1.count - t2.count AS res
FROM product AS t1
JOIN product AS t2 ON t1.product_name = t2.product_name
WHERE t1.date = :first_date AND t2.date = :second_date
ORDER BY res DESC
LIMIT 100

Answer 2

为了优化查询并避免内存问题，您可以使用单个 SQL 查询来执行所有必要的操作。目的是将表与产品名称本身连接起来，按两个特定日期过滤行，计算计数差异，然后对结果进行排序以获得前 100 个差异。这种方法效率更高，因为它利用数据库引擎进行计算，减少应用程序的内存负载。

这是一个示例 SQL 查询，应该可以实现您正在寻找的内容：

SELECT 
    t1.product,
    t1.count AS date1,
    t2.count AS date2,
    t2.count - t1.count AS diff
FROM 
    product AS t1
JOIN 
    product AS t2 ON t1.product = t2.product
WHERE 
    t1.date = '2024-01-04' AND 
    t2.date = '2024-01-06'
ORDER BY 
    ABS(diff) DESC
LIMIT 100;

以下是如何调整 PHP 代码以使用新的 SQL 查询：

try {
    // Assuming $connection is your PDO connection
    $first_date = '2024-01-04';  // Set your first date
    $second_date = '2024-01-06'; // Set your second date

    // Prepare the optimized SQL query
    $sql = "
        SELECT 
            t1.product,
            t1.count AS date1,
            t2.count AS date2,
            t2.count - t1.count AS diff
        FROM 
            product AS t1
        JOIN 
            product AS t2 ON t1.product = t2.product
        WHERE 
            t1.date = :first_date AND 
            t2.date = :second_date
        ORDER BY 
            ABS(diff) DESC
        LIMIT 100";

    $stmt = $connection->prepare($sql);

    // Bind the parameters
    $stmt->bindParam(':first_date', $first_date);
    $stmt->bindParam(':second_date', $second_date);

    // Execute the query
    $stmt->execute();

    // Fetch the results
    $result = $stmt->fetchAll(PDO::FETCH_ASSOC);

    // Output or process the results as needed
    foreach ($result as $row) {
        echo "Product: " . $row['product'] . ", Count on " . $first_date . ": " . $row['date1'] . ", Count on " . $second_date . ": " . $row['date2'] . ", Difference: " . $row['diff'] . "<br/>";
    }
} catch (Exception $e) {
    echo "Error: " . $e->getMessage();
}