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直接从SQL语句问题贴子数据建表并生成建_sqlserver

sqlserver 搞代码 3年前 (2018-06-16) 155次浏览 已收录 0个评论

下面的存储过程,可帮你在回答SQL语句问题时,直接从贴子的样本数据建表并生成建表语句,省去大量的手工输入数据的工作。

/*Create Table from your web page data
* 2004-JAN-1, OpenVMS,V0.1
* 2004-JAN-2, V0.5, add tab & blank values logical

http://www.gaodaima.com/35492.html直接从SQL语句问题贴子数据建表并生成建_sqlserver

* 2004-JAN-3, V1.0, add SQL Statement generation
* 2004-JAN-4, V1.1, fix datatype like decimal(4,2) bug
* 2004-JAN-4, V1.2, fix field name bug
*
* Sample Call: in SQL Query Analyzer
exec dbo.create_table ‘##t2′,’varchar(20),datetime k’,’
ID                   AnDate            
99101                2002-11-24 00:00:00.000
99101                2003-11-15 00:00:00.000
99101                2003-11-29 00:00:00.000
99101                2003-12-20 00:00:00.000′

注意:
1 如用临时表名,只能用全局临时表 ##,否则不可访问
2 如果没有列名,则需要在第一行数据手动加上列名
3 字段名称不允许含空格
4 至少一行数据,否则没有意义
5 字段值为空需要写上NULL,字段值中的任何符号作为值的一部分
6 没有对定义类型和值的类型匹配检查
7 可指定值中含有空格,方法为在该类型定义中的尾部加字母 k, 如 datatime k,
8 如过值中含有单引号,需要复写 ‘ -》”
*/

IF EXISTS (SELECT name
    FROM   sysobjects
    WHERE  name = N’create_table’
    AND    type = ‘P’)
    DROP PROCEDURE create_table
go

create proc dbo.create_table
@table_name varchar(60),— Table name
@datatype varchar(1000),— separated by comma ‘,’
@str nvarchar(3000)     — input string pasted from web page
AS
BEGIN
declare @dt table(id int identity(1,1),fld_name varchar(30),fld_type varchar(20),blank int)
declare @sqlt table(sql_statement varchar(8000))
declare @tmp varchar(1000),@num1 int,@num2 int,@sql nvarchar(4000)
declare @a nvarchar(3000),@i int,@j int,@k int,@m int,@x nvarchar(1000)

SET NOCOUNT ON
if object_id(@table_name) is not null
   begin
    set @a=’TABLE ‘[email protected]_name+’ exists,choose a new one!’
    RAISERROR (@a,16,1)
    return
   end

–提取类型名
set @datatype=lower(replace(@datatype,’ ‘,”))
set @[email protected]
set @i=1
set @num1=0

while @i>0
begin
 select @i=charindex(‘,’,@datatype)
        –check datatype like decimal(10,4)
 if @i>charindex(‘(‘,@datatype) and @i<charindex(‘)’,@datatype)
           set @i=charindex(‘)’,@datatype)+1
 select @j=charindex(‘k’,@datatype)
 set @m=0
 if (@j>1 and @j<@i) or (@i=0 and @j=len(@datatype)) set @m=-1
 if @i>1
 begin
    insert into @dt(fld_type,blank)
  values(left(@datatype,@[email protected]),case when @m=-1 then 1 else 0 end)
    select @datatype=right(@datatype,len(@datatype)[email protected])
 end
 if @i=0 and len(@datatype)>0
    insert into @dt(fld_type,blank) values(left(@datatype,len(@datatype)[email protected]),
   case when @m=-1 then 1 else 0 end)
 if @i=1 or len(@datatype)=0 
 begin
 RAISERROR (‘error data type,comma sign can not be a prefix or surfix’,16,1)
 return
 end
 
 set @[email protected]+1
end

–检查类型
if exists (select fld_type from @dt
   where (case when charindex(‘(‘,fld_type)>0 then
               left(fld_type,charindex(‘(‘,fld_type)-1)
               else fld_type end) not in (select name from systypes) or
          charindex(‘(‘,fld_type)*charindex(‘)’,fld_type)=0 and
          charindex(‘(‘,fld_type)+charindex(‘)’,fld_type)>0)
   begin
    RAISERROR (‘error data type.’, 16, 1)
    return
   end

–提取字段和数据
set @a=replace(@str,char(9),’ ‘) — TAB char
set @a=rtrim(ltrim(@a))
if charindex(char(13)+char(10),right(@a,len(@a)-1))=0 or len(@a)=0
   begin
    RAISERROR (‘input data error,check your data.’, 16, 1)
    return
   end

if object_id(‘tempdb.dbo.#xx’) is not null drop table #xx
select identity(int,1,1) ID,space(50) val into #xx where 1=2
set @k=0
set @num2=0
set @m=0
while len(@a)>0
begin
 set @i=1
 set @x=left(@a,1)

 if @x=char(10) begin
    if @m>@num2 and @num2>0 and charindex(‘k’,@datatype)=0 begin
              RAISERROR (‘number of data is greater than the columns,you should add k in data type difinition.’, 16, 1)
              return
    end 
    set @m=0
 end

 if @x not in (‘ ‘,char(13),char(10))
 begin
          set @i=charindex(‘ ‘,@a)
          set @j=charindex(char(13)+char(10),@a)
   set @[email protected]+1
   if @k<>-1 set @[email protected]+1
   if @j>0 and (@j<@i or @j>@i and substring(@a,@i,@[email protected])=space(@[email protected])) begin
     set @[email protected]
     if @k>@num2 and @k<>-1 set @[email protected]
     set @k=-1
   end
          if @i=0 set @i=(case when @j>0 then @j else len(@a)+1 end)

   select @j=max(ID) from #xx
   if @m=1 or @j<[email protected] or (select blank from @dt where [email protected]) <> 1
      begin
        if @j<@num1 set @x='[‘+replace(rtrim(left(@a,@i-1)),’]’,’]]’)+’]’
        else set @x=rtrim(left(replace(@a,””,”””),@i-1)) 
               insert into #xx(val) values(@x)
      end
   else
     begin
       update #xx set val=val+’ ‘+rtrim(left(@a,@i-1)) where [email protected]
       set @[email protected]
     end
 end
 if @i<len(@a) set @a=ltrim(right(@a,len(@a)[email protected]))
 else set @a=”
end

update #xx set val=” where val=’NULL’
update #xx set val=””+val+”” where ID>@num2

if @num1<>@num2
begin
RAISERROR (‘datatype dismatch the columns’,16,1)
return
end

— if use the exists template table,drop it
if object_id(‘tempdb.dbo.’[email protected]_name) is not null
   exec(‘drop table ‘[email protected]_name)

— 建表
update a
set a.fld_name=b.val
from @dt a,#xx b
where a.ID=b.ID and a.ID<[email protected]

set @a=”
select @[email protected]+fld_name+’ ‘+fld_type+’,’ from @dt where ID<[email protected]
set @a=left(@a,len(@a)-1)
set @sql=’create table ‘[email protected]_name+'(‘[email protected]+’)’
exec(@sql)
insert into @sqlt select @sql

–插入数据
set @[email protected]+1
while @i<=(select max(ID) from #xx)
begin
set @a=”
set @sql=’select @[email protected]+val+”,”’+’ from (select top ‘+convert(varchar(10),@num1)
         +’ val from #xx where ID>=’+convert(varchar(10),(@i))+’) a’
exec sp_executesql @sql,N’@s nvarchar(3000) output’,@a output

set @a=left(@a,len(@a)-1)

set @sql=’insert into ‘[email protected]_name+’ select ‘[email protected]
if len(@a)>0 exec(@sql)
insert into @sqlt select @sql
 
set @[email protected][email protected]
end

select * from @sqlt
–select * from @dt
exec(‘select * from ‘[email protected]_name)
SET NOCOUNT OFF
END

测试
exec dbo.create_table ‘##t2′,’varchar(20),datetime k’,’
ID                   AnDate            
99101                2002-11-24 00:00:00.000
99101                2003-11-15 00:00:00.000
99101                2003-11-29 00:00:00.000
99101                2003-12-20 00:00:00.000′

结果
sql_statement 
——————————————————–
create table ##t2(ID varchar(20),AnDate datetime)
insert into ##t2 select ‘99101’,’2002-11-24 00:00:00.000′
insert into ##t2 select ‘99101’,’2003-11-15 00:00:00.000′
insert into ##t2 select ‘99101’,’2003-11-29 00:00:00.000′
insert into ##t2 select ‘99101’,’2003-12-20 00:00:00.000′

ID                   AnDate                                                
——————– —————————
99101                2002-11-24 00:00:00.000
99101                2003-11-15 00:00:00.000
99101                2003-11-29 00:00:00.000
99101                2003-12-20 00:00:00.000

 

oracle的写法在测试中。

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