Question

如何编写MYSQL函数以基于列的连续值（int，日期或时间戳记）返回记录。

例如：

// id连续记录4次，id：5,6,7,8

InvoiceRecyclerViewAdapter adapter = new InvoiceRecyclerViewAdapter(
        this ,cart_productslist, new InvoiceRecyclerViewAdapter.OnQuantityChangeListener(){

    @Override
    void onQuantityChange( float difference ){

        total += difference;
        total_textview.setText("Rs "+ total);

    }

} );

//日期连续4个日期记录

select * from employee where consecutive(id,4);

///只要id是连续的，就全部返回。

select * from employee where consecutive(date,4);

Answer 1

在ISO标准的SQL 2003或更高版本中，我将结合使用ROW_NUMBER() OVER ( ORDER BY id )和OFFSET。该查询可在SQL Server 2012或更高版本，Oracle 10g或更高版本，PostgreSQL或MySQL 8 or later中运行-但几乎没有人运行MySQL 8 ...

第一个示例：

SELECT
    employee.*
FROM
    (
        SELECT
            id,
            ROW_NUMBER() OVER ( ORDER BY id ) AS rn
        FROM
            employee
    ) AS sq

    INNER JOIN employee ON employee.id = sq.id
ORDER BY
    id
OFFSET
    0 ROWS FETCH 4

幸运的是，MySQL使您可以递增变量并在SELECT中对其他不纯表达式进行评估，就好像它们是迭代求值的（这是在“ SQL思维方式”中做出的错误假设。 “）。

So here's a trick to get a row-rank in MySQL, using an inline initialization：

SELECT
    *,
    @rank := @rank + 1 AS rn
FROM
    employee,
    ( SELECT @rank := 0 )
ORDER BY
    employee.id

因此我们可以将其用作子查询：

SELECT
    employee.*
FROM
    (
        SELECT
            employee.id,
            @rank := @rank + 1 AS rn
        FROM
            employee,
            ( SELECT @rank := 0 )
        ORDER BY
            employee.id
    ) AS sq

    INNER JOIN employee ON employee.id = sq.id

ORDER BY
    id
LIMIT
    4 OFFSET 0

MYSQL函数根据列的连续值返回记录

1 个答案: