SELECT COUNT(*) FROM help_withdrawal WHERE (user_id = ? AND payer_type = ? AND create_time BETWEEN ? AND ?)

I'm sorry, as an AI language model, I do not have access to the database or the context of the query. Please provide more information or context for me to better understand the query.

相关问题

SELECT t1.supplier_id, t1.ky_count, t1.ky_amount, IFNULL(t2.ky_refund_count,0) as ky_refund_count, IFNULL(t2.ky_refund_amount,0) as ky_refund_amount FROM ( SELECT a.supplier_id, count( DISTINCT c.order_no ) AS ky_count, SUM( IFNULL( c.record_amount, 0 )) / 100 AS ky_amount FROM settle_order a INNER JOIN settle_order_receipt b ON a.order_id = b.order_id INNER JOIN cash_withdrawal_record c ON b.third_trade_no = c.order_no WHERE a.is_del = 0 AND a.order_time >= '2023-05-28 00:00:00' AND a.order_time < '2023-05-29 00:00:00' AND a.order_type in (70,75) AND a.supplier_id IN (78,63,58,57,64,72,71,74,83,77,70,69,67,82,65,87,73,59,66,60,86,85,79,80,84,90) AND b.channel_code = 61 AND c.con_bank_account_no IN ( 247, 325 ) AND c.record_status = 1 AND c.record_time > '2023-05-01 00:00:00' GROUP BY a.supplier_id ) t1 LEFT JOIN ( SELECT a.supplier_id, count( DISTINCT b.order_no ) AS ky_refund_count, SUM( IFNULL( b.record_amount, 0 )) / 100 AS ky_refund_amount FROM settle_order_refund a INNER JOIN cash_withdrawal_record b ON a.third_refund_id = b.order_no WHERE a.is_del = 0 AND a.order_type in (70,75) AND a.apply_time >= '2023-05-28 00:00:00' AND a.apply_time < '2023-05-29 00:00:00' AND a.supplier_id IN (78,63,58,57,64,72,71,74,83,77,70,69,67,82,65,87,73,59,66,60,86,85,79,80,84,90) AND a.channel_code = 61 AND b.con_bank_account_no IN ( 247, 325 ) AND b.record_status = 5 AND b.record_time > '2023-05-01 00:00:00' GROUP BY a.supplier_id ) t2 ON t1.supplier_id = t2.supplier_id order by t1.supplier_id asc

以下是对该 SQL 语句的一些优化建议：

1. 使用 EXISTS 替代 DISTINCT + JOIN

在子查询中，使用了 DISTINCT 关键词去重，然后使用 JOIN 连接表。这样的方式效率较低，可以考虑使用 EXISTS 替代。具体来说，应该将连接条件改为 EXISTS 子查询的 WHERE 子句，如下所示：

SELECT
    a.supplier_id,
    COUNT(DISTINCT c.order_no) AS ky_count,
    SUM(IFNULL(c.record_amount, 0)) / 100 AS ky_amount
FROM
    settle_order AS a
    INNER JOIN settle_order_receipt AS b ON a.order_id = b.order_id
    INNER JOIN cash_withdrawal_record AS c ON EXISTS (
        SELECT 1
        FROM cash_withdrawal_record AS d
        WHERE b.third_trade_no = d.order_no
            AND d.con_bank_account_no IN (247, 325)
            AND d.record_status = 1
            AND d.record_time > '2023-05-01 00:00:00'
    )
WHERE
    a.is_del = 0
    AND a.order_time >= '2023-05-28 00:00:00'
    AND a.order_time < '2023-05-29 00:00:00'
    AND a.order_type IN (70, 75)
    AND a.supplier_id IN (78, 63, 58, 57, 64, 72, 71, 74, 83, 77, 70, 69, 67, 82, 65, 87, 73, 59, 66, 60, 86, 85, 79, 80, 84, 90)
    AND b.channel_code = 61
GROUP BY
    a.supplier_id

同样的，对于另一个子查询也可以使用 EXISTS 进行优化。

2. 使用 UNION ALL 代替 LEFT JOIN

在原 SQL 中，使用了 LEFT JOIN 来将两个子查询合并，可以考虑使用 UNION ALL 代替。使用 UNION ALL 的好处是它不会去重，而我们已经在子查询中使用了 GROUP BY 进行了去重，因此可以放心使用 UNION ALL。

SELECT
    t.supplier_id,
    t.ky_count,
    t.ky_amount,
    t.ky_refund_count,
    t.ky_refund_amount
FROM (
    SELECT
        a.supplier_id,
        COUNT(DISTINCT c.order_no) AS ky_count,
        SUM(IFNULL(c.record_amount, 0)) / 100 AS ky_amount,
        0 AS ky_refund_count,
        0 AS ky_refund_amount
    FROM
        settle_order AS a
        INNER JOIN settle_order_receipt AS b ON a.order_id = b.order_id
        INNER JOIN cash_withdrawal_record AS c ON EXISTS (
            SELECT 1
            FROM cash_withdrawal_record AS d
            WHERE b.third_trade_no = d.order_no
                AND d.con_bank_account_no IN (247, 325)
                AND d.record_status = 1
                AND d.record_time > '2023-05-01 00:00:00'
        )
    WHERE
        a.is_del = 0
        AND a.order_time >= '2023-05-28 00:00:00'
        AND a.order_time < '2023-05-29 00:00:00'
        AND a.order_type IN (70, 75)
        AND a.supplier_id IN (78, 63, 58, 57, 64, 72, 71, 74, 83, 77, 70, 69, 67, 82, 65, 87, 73, 59, 66, 60, 86, 85, 79, 80, 84, 90)
        AND b.channel_code = 61
    GROUP BY
        a.supplier_id
    UNION ALL
    SELECT
        a.supplier_id,
        0 AS ky_count,
        0 AS ky_amount,
        COUNT(DISTINCT b.order_no) AS ky_refund_count,
        SUM(IFNULL(b.record_amount, 0)) / 100 AS ky_refund_amount
    FROM
        settle_order_refund AS a
        INNER JOIN cash_withdrawal_record AS b ON EXISTS (
            SELECT 1
            FROM settle_order_refund AS c
            WHERE c.third_refund_id = b.order_no
                AND c.is_del = 0
                AND c.order_type IN (70, 75)
                AND c.apply_time >= '2023-05-28 00:00:00'
                AND c.apply_time < '2023-05-29 00:00:00'
                AND c.supplier_id IN (78, 63, 58, 57, 64, 72, 71, 74, 83, 77, 70, 69, 67, 82, 65, 87, 73, 59, 66, 60, 86, 85, 79, 80, 84, 90)
                AND c.channel_code = 61
        )
    WHERE
        b.con_bank_account_no IN (247, 325)
        AND b.record_status = 5
        AND b.record_time > '2023-05-01 00:00:00'
    GROUP BY
        a.supplier_id
) AS t
ORDER BY
    t.supplier_id ASC

3. 使用索引优化查询

在子查询中，涉及到多个表的连表查询，需要使用到合适的索引才能提高查询效率。具体来说，可以考虑在以下字段上建立索引：

- settle_order.is_del
- settle_order.order_time
- settle_order.order_type
- settle_order.supplier_id
- settle_order_receipt.order_id
- settle_order_receipt.channel_code
- settle_order_receipt.third_trade_no
- cash_withdrawal_record.order_no
- cash_withdrawal_record.con_bank_account_no
- cash_withdrawal_record.record_status
- cash_withdrawal_record.record_time
- settle_order_refund.is_del
- settle_order_refund.order_type
- settle_order_refund.apply_time
- settle_order_refund.supplier_id
- settle_order_refund.channel_code
- cash_withdrawal_record.order_no
- cash_withdrawal_record.con_bank_account_no
- cash_withdrawal_record.record_status
- cash_withdrawal_record.record_time

以上是一些可能的优化建议，具体的优化方案需要根据实际情况进行调整。

def withdrawal(): while True: balance = input("请输入金额:") if balance == "q": break elif balance.isdigit(): a = int(user_list[account]["余额"]) user_list[account]["余额"] = a - int(balance) print("取款成功") print("余额:" + str(user_list[account]["余额"])) break else: print("输入有误,请重新输入")

这是一个 Python 函数，用于实现取款的功能。

函数的流程如下：

1. 使用 `while` 循环，不断询问用户输入金额。

2. 如果用户输入 `q`，则退出循环。

3. 如果用户输入的是数字，则将账户余额减去输入的金额，并输出 "取款成功" 和当前的余额。然后退出循环。

4. 否则，输出 "输入有误，请重新输入"，继续循环。