将查询从 Oracle 转换为 SQL Server

问题描述

我正在尝试将以下查询用于 HPALM。

但是这是 Oracle 语法，我需要将其转换为 sql Server。

确实，const sidebar = document.querySelector('.sidebar'); const closebtn = document.querySelector('.closebtn'); closebtn.addEventListener('click',() => { sidebar.classList.toggle('show'); }); 是 Oracle 的一个特性。

SYS_CONNECT_BY_PATH

可以翻译吗？如果是，怎么办？

谢谢

解决方法

您可以为此使用递归 CTE（也可以在 Oracle 中），因此您可以先在当前的 Oracle 数据库中检查它。

@(Html.Kendo().Grid(Model)
        .Name("marketWatchGrid")
        .Columns(columns =>
        {
            columns.Bound(p => p.Clients).ClientTemplate("# if(Clients.indexOf(''')>-1) { # #= Clients # #} else {# #: Clients # #}#")
        })
    );

下面的例子

with a (ls_father_id,ls_id,path_) as (
  select
    ls_father_id,ls_name as path_
  from lists
  where ls_name in (select f.ls_name from lists f where f.ls_father_id = 0)

  union all

  select 
    b.ls_father_id,b.ls_id,a.path_ || '\' || b.ls_name
  from b
    join a
      on b.ls_father_id = a.ls_id
)
select *
from a

LS_ID | LS_FATHER_ID | LS_NAME | PATH_         
----: | -----------: | :------ | :-------------
    1 |            0 | aaa     | aaa           
    2 |            1 | bb      | aaa\bb        
    3 |            0 | c       | c             
    4 |            2 | cca     | aaa\bb\cca    
    5 |            4 | vvv     | aaa\bb\cca\vvv
    6 |            3 | f       | c\f           
    7 |            1 | ee      | aaa\ee        

create table t
as
with a (ls_id,ls_father_id,ls_name) as (
  select 1,'aaa' from dual union all
  select 2,1,'bb'  from dual union all
  select 3,'c'   from dual union all
  select 4,2,'cca' from dual union all
  select 5,4,'vvv' from dual union all
  select 6,3,'f'   from dual union all
  select 7,'ee'  from dual
)
select *
from a

select
  t.*,substr(sys_connect_by_path(ls_name,'\'),2) as path_
from t
start with ls_name in (select f.ls_name from t f where f.ls_father_id = 0)
connect by prior ls_id = ls_father_id
order by ls_id

LS_ID | LS_FATHER_ID | PATH_         
----: | -----------: | :-------------
    1 |            0 | aaa           
    2 |            1 | aaa\bb        
    3 |            0 | c             
    4 |            2 | aaa\bb\cca    
    5 |            4 | aaa\bb\cca\vvv
    6 |            3 | c\f           
    7 |            1 | aaa\ee        

dbfiddle here