Two databases compare and contrast view stored procedures and table structure differences
One, view and stored procedure comparison
The principle uses the system table "sysobjects" and the system table "syscomments" to compare views and stored procedures in the database. system table "sysobjects" has been described in detail before, it is interesting to see: SQL Server System table sysobjects introduction and use
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Code
/*--调用示例
exec p_compdb ‘DBNAME1‘,‘DBNAME2‘
exec p_compdb ‘DBNAME2‘,‘DBNAME3‘
--*/
CREATE
proc p_compdb
@db1 sysname,
--第一个库
@db2 sysname
--第二个库
as
exec
(
‘
select 类型=case isnull(a.xtype,b.xtype) when ‘
‘V‘
‘ then ‘
‘视图‘
‘ else ‘
‘存储过程‘
‘ end
,匹配情况=case
when a.name is null then ‘
‘库 [‘
[email protected]+
‘] 中无‘
‘
when b.name is null then ‘
‘库 [‘
[email protected]+
‘] 中无‘
‘
else ‘
‘结构不同‘
‘ end
,对象名称=isnull(a.name,b.name),a.text as atext, b.text as btext
from(
select a.name,a.xtype,b.colid,b.text
from [‘
[email protected]+
‘]..sysobjects a,[‘
[email protected]+
‘]..syscomments b
where a.id=b.id and a.xtype in(‘
‘V‘
‘,‘
‘P‘
‘) and a.status>=0
)a full join(
select a.name,a.xtype,b.colid,b.text
from [‘
[email protected]+
‘]..sysobjects a,[‘
[email protected]+
‘]..syscomments b
where a.id=b.id and a.xtype in(‘
‘V‘
‘,‘
‘P‘
‘) and a.status>=0
)b on a.name=b.name and a.xtype=b.xtype and a.colid=b.colid
where a.name is null
or b.name is null
or isnull(a.text,‘
‘‘
‘) <>isnull(b.text,‘
‘‘
‘)
--group by a.name,b.name,a.xtype,b.xtype
--order by 类型,匹配情况,对象名称‘
)
|
"Execution Results"
Comparison of data table structure
"Principle" uses the system table "sysobjects", "sysindexes", "Sysindexkeys", "syscomments", "Sysclumns", "systypes", "extended_properties", Compare the table structure in the database. (more system tables are involved.) Don't introduce them. directly on the code. )
Code
/*--比较两个数据库的表结构差异--*/
/*--调用示例
exec p_comparestructure ‘DBNAME1‘,‘DBNAME2‘
exec p_comparestructure ‘DBNAME2‘,‘DBNAME3‘
--*/
create
proc p_comparestructure
@dbname1
varchar
(250),
--要比较的数据库名1
@dbname2
varchar
(250)
--要比较的数据库名2
as
create
table
#tb1(表名1
varchar
(250),字段名
varchar
(250),序号
int
,标识
bit
,主键
bit
,类型
varchar
(250),
占用字节数
int
,长度
int
,小数位数
int
,允许空
bit
,默认值 sql_variant,字段说明 sql_variant)
create
table
#tb2(表名2
varchar
(250),字段名
varchar
(250),序号
int
,标识
bit
,主键
bit
,类型
varchar
(250),
占用字节数
int
,长度
int
,小数位数
int
,允许空
bit
,默认值 sql_variant,字段说明 sql_variant)
--得到数据库1的结构
exec
(
‘insert into #tb1 SELECT
表名=d.name,字段名=a.name,序号=a.colid,
标识=case when a.status=0x80 then 1 else 0 end,
主键=case when exists(SELECT 1 FROM ‘
[email protected]+
‘..sysobjects where xtype=‘
‘PK‘
‘ and parent_obj=a.id and name in (
SELECT name FROM ‘
[email protected]+
‘..sysindexes WHERE indid in(
SELECT indid FROM ‘
[email protected]+
‘..sysindexkeys WHERE id = a.id AND colid=a.colid
))) then 1 else 0 end,
类型=b.name,占用字节数=a.length,长度=a.prec,小数位数=a.scale,允许空=a.isnullable,
默认值=isnull(e.text,‘
‘‘
‘),字段说明=isnull(g.[value],‘
‘‘
‘)
FROM ‘
[email protected]+
‘..syscolumns a
left join ‘
[email protected]+
‘..systypes b on a.xtype=b.xusertype
inner join ‘
[email protected]+
‘..sysobjects d on a.id=d.id and d.xtype=‘
‘U‘
‘ and d.name <>‘
‘dtproperties‘
‘
left join ‘
[email protected]+
‘..syscomments e on a.cdefault=e.id
left join sys.extended_properties g
ON
a.ID=g.major_id AND a.COLID=g.minor_id
order by a.id,a.colorder‘
)
--得到数据库2的结构
exec
(
‘insert into #tb2 SELECT
表名=d.name,字段名=a.name,序号=a.colid,
标识=case when a.status=0x80 then 1 else 0 end,
主键=case when exists(SELECT 1 FROM ‘
[email protected]+
‘..sysobjects where xtype=‘
‘PK‘
‘ and parent_obj=a.id and name in (
SELECT name FROM ‘
[email protected]+
‘..sysindexes WHERE indid in(
SELECT indid FROM ‘
[email protected]+
‘..sysindexkeys WHERE id = a.id AND colid=a.colid
))) then 1 else 0 end,
类型=b.name,占用字节数=a.length,长度=a.prec,小数位数=a.scale,允许空=a.isnullable,
默认值=isnull(e.text,‘
‘‘
‘),字段说明=isnull(g.[value],‘
‘‘
‘)
FROM ‘
[email protected]+
‘..syscolumns a
left join ‘
[email protected]+
‘..systypes b on a.xtype=b.xusertype
inner join ‘
[email protected]+
‘..sysobjects d on a.id=d.id and d.xtype=‘
‘U‘
‘ and d.name <>‘
‘dtproperties‘
‘
left join ‘
[email protected]+
‘..syscomments e on a.cdefault=e.id
left join sys.extended_properties g
ON
a.ID=g.major_id AND a.COLID=g.minor_id
order by a.id,a.colorder‘
)
--and not exists(select 1 from #tb2 where 表名2=a.表名1)
select
比较结果=
case
when
a.表名1
is
null
and
b.序号=1
then
‘库1缺少表:‘
+b.表名2
when
b.表名2
is
null
and
a.序号=1
then
‘库2缺少表:‘
+a.表名1
when
a.字段名
is
null
and
exists(
select
1
from
#tb1
where
表名1=b.表名2)
then
‘库1 [‘
+b.表名2+
‘] 缺少字段:‘
+b.字段名
when
b.字段名
is
null
and
exists(
select
1
from
#tb2
where
表名2=a.表名1)
then
‘库2 [‘
+a.表名1+
‘] 缺少字段:‘
+a.字段名
when
a.标识 <>b.标识
then
‘标识不同‘
when
a.主键 <>b.主键
then
‘主键设置不同‘
when
a.类型 <>b.类型
then ‘字段类型不同‘
when
a.占用字节数 <>b.占用字节数
then
‘占用字节数‘
when
a.长度 <>b.长度
then
‘长度不同‘
when
a.小数位数 <>b.小数位数
then
‘小数位数不同‘
when
a.允许空 <>b.允许空
then
‘是否允许空不同‘
when
a.默认值 <>b.默认值
then
‘默认值不同‘
when
a.字段说明 <>b.字段说明
then
‘字段说明不同‘
else
‘‘
end
,
*
from
#tb1 a
full
join
#tb2 b
on
a.表名1=b.表名2
and
a.字段名=b.字段名
where
a.表名1
is
null
or a.字段名
is
null
or
b.表名2
is
null
or
b.字段名
is
null
or
a.标识 <>b.标识
or
a.主键 <>b.主键
or
a.类型 <>b.类型
or
a.占用字节数 <>b.占用字节数
or
a.长度 <>b.长度
or
a.小数位数 <>b.小数位数
or
a.允许空 <>b.允许空
or
a.默认值 <>b.默认值
or
a.字段说明 <>b.字段说明
order
by
isnull
(a.表名1,b.表名2),
isnull
(a.序号,b.序号)
--isnull(a.字段名,b.字段名)
go
|
"Execution Results"
PS: Please use the system Administrator (sysadmin) role account for SQL execution above. Other roles I have not tried, there is time to try. When I used an account test that mapped only the library "owner" permission, I reported the following error: The EXECUTE permission on object ' p_compdb ' (Database ' master ', schema ' dbo ') was rejected.
When writing, this article uses the SQL SERVER 2005 version and does not guarantee that other versions will work correctly. If the other version has problems, you can leave a message, study together. The debug environment details are as follows:
Microsoft SQL Server Management Studio 9.00.1399.00
Microsoft Analysis Services Client Tools 2005.090.1399.00
Microsoft Data Access Component (MDAC) 6.1.7600.16385 (win7_rtm.090713-1255)
Microsoft MSXML 3.0 4.0 6.0
Microsoft Internet Explorer 8.0.7600.16385
Microsoft. NET Framework 2.0.50727.4984
Operating system 6.1.7600
Two databases compare and contrast view stored procedures and table structure differences