Question

这是我的疑问：

select * from (
    select u.id, 'u' as [type], u.firstName, u.lastName, c.name as companyName,
        u.lastName + u.firstName + isNull(c.name, '') as sortName
    from users as u
        left outer join companies as c on c.id = u.company_id

    union all

    select id, 'c' as [type], null as firstName, null as lastName, name as 
        companyName, name as sortName
    from companies
) as result
where sortName like '%a%'
order by sortName
OFFSET 0 ROWS FETCH NEXT 10 ROWS ONLY

它从表users和companies中返回10行，这些行匹配搜索字符串＆＃34; a＆＃34;。使用offset / fetch，我可以轻松地在我的应用程序中进行分页。但我还需要知道的是，数据库中有多少行匹配搜索字符串＆＃34; a＆＃34;没有获取限制。

我的第一个方法是结果查询count(*)，但这不起作用（因为我猜的是联盟）。

Answer 1

最好在提供VIEW查询的情况下创建UNION。然后，您可以COUNT(*)记录并进一步显示前面的

Answer 2

你可以尝试在你选择*语句中添加COUNT（*）OVER（），如下所示。总行数将位于以下示例中名为“total_rows”的新列中。

select *, total_rows=COUNT(*) OVER() from (
    select u.id, 'u' as [type], u.firstName, u.lastName, c.name as companyName,
        u.lastName + u.firstName + isNull(c.name, '') as sortName
    from users as u
        left outer join companies as c on c.id = u.company_id

    union all

    select id, 'c' as [type], null as firstName, null as lastName, name as 
        companyName, name as sortName
    from companies
) as result
where sortName like '%a%'
order by sortName
OFFSET 0 ROWS FETCH NEXT 10 ROWS ONLY

与SQL Server联合的分页

2 个答案: