MySQL Maximum date data

Started by jackgrant1, Jan 06, 2023, 03:53 AM

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jackgrant1Topic starter

The table cs_reward_point_changes contains the columns 'user_id', 'amount' and 'timestamp' with multiple entries for some 'user_id' with different dates.

I am trying to write an SQL query that will achieve two actions. Firstly, I want to retrieve only one row for each 'user_id' that contains the maximum date specified in the 'timestamp' column. Secondly, I need to execute a check using the 'timestamp' column. For any entry where the 'timestamp' date is less than the current date by 500 days, the value in the 'amount' column should be set to zero.

To retrieve the desired data in a single query, I have attempted the following code:

SELECT user_id, amount, MAX(TIMESTAMP)
FROM cs_reward_point_changes
GROUP BY user_id

Unfortunately, this query returns all the records for each 'user_id', not just with the last date. Therefore, I need to modify the query to only return the record with the latest date.

Additionally, I need to check if the 'timestamp' date is less than the current date by 500 days. If so, I need to update the 'amount' column to zero for that entry. To achieve this in one query, I will use a subquery:

UPDATE cs_reward_point_changes
SET amount = 0
WHERE TIMESTAMP < DATEADD(day, -500, GETDATE()) AND user_id IN
(
    SELECT user_id
    FROM cs_reward_point_changes
    WHERE TIMESTAMP = (
        SELECT MAX(TIMESTAMP)
        FROM cs_reward_point_changes AS c
        WHERE c.user_id = cs_reward_point_changes.user_id
    )
)

I hope this revised code is helpful for you. Let me know if you have any questions!
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jobtardis

The code provided is using an SQL query to retrieve records from the table 'cs_reward_point_changes'. The query is joining the original table with a subquery that retrieves the maximum timestamp for each user.

However, it is possible that a user has multiple records with the same maximum timestamp, and thus all such records will be selected. To resolve this issue, it may be necessary to use variables.

This code can be further modified to perform an update based on these records.
  •  

wellm97

This SQL query retrieves the records from 'cs_reward_point_changes' table that fall between 400 days before the current date and the present date. The query uses only a SELECT statement with the condition, without any other statements.

SELECT *
  FROM cs_reward_point_changes
WHERE FROM_UNIXTIME(UNIX_TIMESTAMP(`TIMESTAMP`))
      BETWEEN DATE_SUB(CURDATE(), INTERVAL 400 DAY) AND CURDATE();
  •  

mike345

Your revised code looks good and should achieve the desired results. It retrieves the latest record for each user by using a subquery to get the maximum timestamp for each user, and then updates the amount column to zero for any entries where the timestamp is older than 500 days.

Just make sure to test the query thoroughly before running it on your production database, as updating data can have serious consequences if not done correctly.

Here's a breakdown of the revised code:

1. Retrieving the latest record for each user:

```
SELECT user_id, amount, MAX(TIMESTAMP)
FROM cs_reward_point_changes
GROUP BY user_id
```

This part of the query groups the records by user_id and selects the maximum timestamp for each group. However, it doesn't retrieve the corresponding rows with the maximum timestamp for each user. To achieve that, you can modify the query as follows:

```
SELECT c.user_id, c.amount, c.timestamp
FROM cs_reward_point_changes c
JOIN (
    SELECT user_id, MAX(timestamp) AS max_timestamp
    FROM cs_reward_point_changes
    GROUP BY user_id
) t
ON c.user_id = t.user_id AND c.timestamp = t.max_timestamp
```

This modified query joins the result of the subquery (which provides the maximum timestamp for each user) with the original table on both user_id and timestamp to retrieve the rows with the latest timestamps.

2. Updating the amount column to zero for entries with timestamps older than 500 days:

```
UPDATE cs_reward_point_changes
SET amount = 0
WHERE TIMESTAMP < DATEADD(day, -500, GETDATE()) AND user_id IN
(
    SELECT user_id
    FROM cs_reward_point_changes
    WHERE TIMESTAMP = (
        SELECT MAX(TIMESTAMP)
        FROM cs_reward_point_changes AS c
        WHERE c.user_id = cs_reward_point_changes.user_id
    )
)
```

This part of the code remains unchanged from your initial version. It locates the entries where the timestamp is less than the current date minus 500 days and updates the amount column to zero for those specific rows.


Here's a more detailed explanation of the updated code:

To retrieve the latest record for each user, you can use a subquery to first find the maximum timestamp for each user, and then join it with the original table to get the corresponding row:

```
SELECT c.user_id, c.amount, c.timestamp
FROM cs_reward_point_changes c
JOIN (
    SELECT user_id, MAX(timestamp) AS max_timestamp
    FROM cs_reward_point_changes
    GROUP BY user_id
) t
ON c.user_id = t.user_id AND c.timestamp = t.max_timestamp
```

In this query:

- The subquery (enclosed in parentheses) selects the user_id and the maximum timestamp for each user using the GROUP BY clause.
- The main query joins the original table `cs_reward_point_changes` with the subquery results. It matches rows based on both user_id and timestamp to retrieve the rows with the latest timestamps for each user.

This modified query will return only one row per user, containing the maximum timestamp for that user.

For updating the amount column to zero for entries older than 500 days, your code is correct and doesn't need any modifications:

```
UPDATE cs_reward_point_changes
SET amount = 0
WHERE TIMESTAMP < DATEADD(day, -500, GETDATE()) AND user_id IN
(
    SELECT user_id
    FROM cs_reward_point_changes
    WHERE TIMESTAMP = (
        SELECT MAX(TIMESTAMP)
        FROM cs_reward_point_changes AS c
        WHERE c.user_id = cs_reward_point_changes.user_id
    )
)
```

This query uses a subquery to find the maximum timestamp for each user, and then updates the amount column to zero for the rows where the timestamp is older than 500 days.
  •