Query: Fetch 3 Records Which Has Higher Value

query: fetch 3 records which has higher value

If you want the three distinct symbol_id having the highest buy rate then you may try the following:

public function HigherValue() {
    $higher_value = DB::table('finaltrade')
        ->select('symbol_id')
        ->groupBy('symbol_id')
        ->orderByRaw('MAX(buy_rate) DESC')
        ->limit(3)
        ->get();

    return response()->json($higher_value);
}

This would correpsond to the following raw MySQL query:

SELECT symbol_id
FROM finaltrade
GROUP BY symbol_id
ORDER BY MAX(buy_rate) DESC
LIMIT 3;

write a sql Query to fetch records with multiple values

In MySQL you can do next query:

select roll_no
from Student
group by roll_no
having count(distinct name) > 1;

MySQL group by

Result:

+=========+
| roll_no |
+=========+
| 2       |
+---------+
| 6       |
+---------+

Fetch the rows which have the Max value for a column for each distinct value of another column

This will retrieve all rows for which the my_date column value is equal to the maximum value of my_date for that userid. This may retrieve multiple rows for the userid where the maximum date is on multiple rows.

select userid,
       my_date,
       ...
from
(
select userid,
       my_date,
       ...
       max(my_date) over (partition by userid) max_my_date
from   users
)
where my_date = max_my_date

"Analytic functions rock"

Edit: With regard to the first comment ...

"using analytic queries and a self-join defeats the purpose of analytic queries"

There is no self-join in this code. There is instead a predicate placed on the result of the inline view that contains the analytic function -- a very different matter, and completely standard practice.

"The default window in Oracle is from the first row in the partition to the current one"

The windowing clause is only applicable in the presence of the order by clause. With no order by clause, no windowing clause is applied by default and none can be explicitly specified.

The code works.

SQL select only rows with max value on a column

At first glance...

All you need is a GROUP BY clause with the MAX aggregate function:

SELECT id, MAX(rev)
FROM YourTable
GROUP BY id

It's never that simple, is it?

I just noticed you need the content column as well.

This is a very common question in SQL: find the whole data for the row with some max value in a column per some group identifier. I heard that a lot during my career. Actually, it was one the questions I answered in my current job's technical interview.

It is, actually, so common that Stack Overflow community has created a single tag just to deal with questions like that: greatest-n-per-group.

Basically, you have two approaches to solve that problem:

Joining with simple group-identifier, max-value-in-group Sub-query

In this approach, you first find the group-identifier, max-value-in-group (already solved above) in a sub-query. Then you join your table to the sub-query with equality on both group-identifier and max-value-in-group:

SELECT a.id, a.rev, a.contents
FROM YourTable a
INNER JOIN (
    SELECT id, MAX(rev) rev
    FROM YourTable
    GROUP BY id
) b ON a.id = b.id AND a.rev = b.rev

Left Joining with self, tweaking join conditions and filters

In this approach, you left join the table with itself. Equality goes in the group-identifier. Then, 2 smart moves:

1. The second join condition is having left side value less than right value
2. When you do step 1, the row(s) that actually have the max value will have NULL in the right side (it's a LEFT JOIN, remember?). Then, we filter the joined result, showing only the rows where the right side is NULL.

So you end up with:

SELECT a.*
FROM YourTable a
LEFT OUTER JOIN YourTable b
    ON a.id = b.id AND a.rev < b.rev
WHERE b.id IS NULL;

Conclusion

Both approaches bring the exact same result.

If you have two rows with max-value-in-group for group-identifier, both rows will be in the result in both approaches.

Both approaches are SQL ANSI compatible, thus, will work with your favorite RDBMS, regardless of its "flavor".

Both approaches are also performance friendly, however your mileage may vary (RDBMS, DB Structure, Indexes, etc.). So when you pick one approach over the other, benchmark. And make sure you pick the one which make most of sense to you.

Get records with max value for each group of grouped SQL results

There's a super-simple way to do this in mysql:

select * 
from (select * from mytable order by `Group`, age desc, Person) x
group by `Group`

This works because in mysql you're allowed to not aggregate non-group-by columns, in which case mysql just returns the first row. The solution is to first order the data such that for each group the row you want is first, then group by the columns you want the value for.

You avoid complicated subqueries that try to find the max() etc, and also the problems of returning multiple rows when there are more than one with the same maximum value (as the other answers would do)

Note: This is a mysql-only solution. All other databases I know will throw an SQL syntax error with the message "non aggregated columns are not listed in the group by clause" or similar. Because this solution uses undocumented behavior, the more cautious may want to include a test to assert that it remains working should a future version of MySQL change this behavior.

Version 5.7 update:

Since version 5.7, the sql-mode setting includes ONLY_FULL_GROUP_BY by default, so to make this work you must not have this option (edit the option file for the server to remove this setting).

Select query to fetch same records more than once based on a column value

With a table of numbers (from 1 or 0 up to a high enough value) you can do a query like this.

select d.Content_ID, 0 as Download_Time
from downloads as d 
  inner join Numbers as N
    on N.number between 1 and d.Download_Count;

How can I SELECT rows with MAX(Column value), PARTITION by another column in MYSQL?

You are so close! All you need to do is select BOTH the home and its max date time, then join back to the topten table on BOTH fields:

SELECT tt.*
FROM topten tt
INNER JOIN
    (SELECT home, MAX(datetime) AS MaxDateTime
    FROM topten
    GROUP BY home) groupedtt 
ON tt.home = groupedtt.home 
AND tt.datetime = groupedtt.MaxDateTime

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