如何生成的Oracle架构（编写脚本）的整个DDL？如何生成的Oracle架构（编写脚本）的整个DD

谁能告诉我怎么可以生成DDL所有表，视图，索引，包，过程，函数，触发器，类型，序列，同义词，赠款等Oracle模式里面？ 理想情况下，我想行过副本，但是不那么重要。

我想这样做就不能手动每次某种形式的计划作业，并让排除了使用在SQL Developer中的向导。

理想的情况是，因为我将在具有赠款和同义词彼此几种模式运行这个，我想有一种方法做一个查找/输出更换这样的模式名称匹配任何我的新的模式的名字会成为。

谢谢！

Answer 1:

您可以后台架构出通过SQL * Plus和DBMS_METADATA包中的文件。 然后，通过SED另一个替代方案名称。 这适用于甲骨文10及更高版本。

sqlplus<<EOF
set long 100000
set head off
set echo off
set pagesize 0
set verify off
set feedback off
spool schema.out

select dbms_metadata.get_ddl(object_type, object_name, owner)
from
(
    --Convert DBA_OBJECTS.OBJECT_TYPE to DBMS_METADATA object type:
    select
        owner,
        --Java object names may need to be converted with DBMS_JAVA.LONGNAME.
        --That code is not included since many database don't have Java installed.
        object_name,
        decode(object_type,
            'DATABASE LINK',      'DB_LINK',
            'JOB',                'PROCOBJ',
            'RULE SET',           'PROCOBJ',
            'RULE',               'PROCOBJ',
            'EVALUATION CONTEXT', 'PROCOBJ',
            'CREDENTIAL',         'PROCOBJ',
            'CHAIN',              'PROCOBJ',
            'PROGRAM',            'PROCOBJ',
            'PACKAGE',            'PACKAGE_SPEC',
            'PACKAGE BODY',       'PACKAGE_BODY',
            'TYPE',               'TYPE_SPEC',
            'TYPE BODY',          'TYPE_BODY',
            'MATERIALIZED VIEW',  'MATERIALIZED_VIEW',
            'QUEUE',              'AQ_QUEUE',
            'JAVA CLASS',         'JAVA_CLASS',
            'JAVA TYPE',          'JAVA_TYPE',
            'JAVA SOURCE',        'JAVA_SOURCE',
            'JAVA RESOURCE',      'JAVA_RESOURCE',
            'XML SCHEMA',         'XMLSCHEMA',
            object_type
        ) object_type
    from dba_objects 
    where owner in ('OWNER1')
        --These objects are included with other object types.
        and object_type not in ('INDEX PARTITION','INDEX SUBPARTITION',
           'LOB','LOB PARTITION','TABLE PARTITION','TABLE SUBPARTITION')
        --Ignore system-generated types that support collection processing.
        and not (object_type = 'TYPE' and object_name like 'SYS_PLSQL_%')
        --Exclude nested tables, their DDL is part of their parent table.
        and (owner, object_name) not in (select owner, table_name from dba_nested_tables)
        --Exclude overflow segments, their DDL is part of their parent table.
        and (owner, object_name) not in (select owner, table_name from dba_tables where iot_type = 'IOT_OVERFLOW')
)
order by owner, object_type, object_name;

spool off
quit
EOF

cat schema.out|sed 's/OWNER1/MYOWNER/g'>schema.out.change.sql

把一切都放在一个脚本，并通过cron（调度器）运行它。 当使用先进的功能，导出对象可能会非常棘手。 如果你需要一些例外添加到上面的代码，请不要惊讶。

Answer 2:

如果你想为每个对象单独生成DDL，

查询是：

--generate DDL针对所有用户对象

--1。 所有表

SELECT DBMS_METADATA.GET_DDL('TABLE', TABLE_NAME) FROM USER_TABLES;

--2。 所有索引

SELECT DBMS_METADATA.GET_DDL('INDEX', INDEX_NAME) FROM USER_INDEXES WHERE INDEX_TYPE ='NORMAL';

--3。 所有视图

SELECT DBMS_METADATA.GET_DDL('VIEW', VIEW_NAME) FROM USER_VIEWS;

要么

SELECT TEXT FROM USER_VIEWS

--4。 FOR ALL MATERILIZED VIEWS

SELECT QUERY FROM USER_MVIEWS

--5。 FOR ALL功能

SELECT DBMS_METADATA.GET_DDL('FUNCTION', OBJECT_NAME) FROM USER_PROCEDURES WHERE OBJECT_TYPE = 'FUNCTION'

================================================== =============================================

像LOB一些OBJECT_TYPE功能GET_DDL支持犯规，实体化视图，表分区

因此，对于生成DDL将综合查询：

SELECT OBJECT_TYPE, OBJECT_NAME,DBMS_METADATA.GET_DDL(OBJECT_TYPE, OBJECT_NAME, OWNER)
  FROM ALL_OBJECTS 
  WHERE (OWNER = 'XYZ') AND OBJECT_TYPE NOT IN('LOB','MATERIALIZED VIEW', 'TABLE PARTITION') ORDER BY OBJECT_TYPE, OBJECT_NAME;

Answer 3:

没有与对象，如PACKAGE_BODY一个问题：

SELECT DBMS_METADATA.get_ddl(object_Type, object_name, owner) FROM ALL_OBJECTS WHERE OWNER = 'WEBSERVICE';


ORA-31600 invalid input value PACKAGE BODY parameter OBJECT_TYPE in function GET_DDL
ORA-06512: на  "SYS.DBMS_METADATA", line 4018
ORA-06512: на  "SYS.DBMS_METADATA", line 5843
ORA-06512: на  line 1
31600. 00000 -  "invalid input value %s for parameter %s in function %s"
*Cause:    A NULL or invalid value was supplied for the parameter.
*Action:   Correct the input value and try the call again.



SELECT DBMS_METADATA.GET_DDL(REPLACE(object_type,' ','_'), object_name, owner)
  FROM all_OBJECTS 
  WHERE (OWNER = 'OWNER1');

Answer 4:

包装用程序GET_DDL将返回两种规格，车身，所以这将是更好的改变对ALL_OBJECTS查询，以便包体上没有选择返回。

到目前为止，我改变了查询到这一点：

SELECT DBMS_METADATA.GET_DDL(REPLACE(object_type, ' ', '_'), object_name, owner)
FROM all_OBJECTS
WHERE (OWNER = 'OWNER1')
and object_type not like '%PARTITION'
and object_type not like '%BODY'
order by object_type, object_name;

虽然可能会根据你所得到的对象类型需要其他的变化...

Answer 5:

首先导出模式元数据：

expdp dumpfile=filename logfile=logname directory=dir_name schemas=schema_name

然后使用进口sqlfile选项（它不会导入数据将只写模式DDL该文件）

impdp dumpfile=filename logfile=logname directory=dir_name sqlfile=ddl.sql

Answer 6:

要生成，即用户的整个模式的DDL脚本，您可以使用DBMS_METADATA.GET_DDL。

在执行SQL *以下脚本加上通过创建蒂姆·霍尔 ：

提示时提供的用户名 。

set long 20000 longchunksize 20000 pagesize 0 linesize 1000 feedback off verify off trimspool on
column ddl format a1000

begin
   dbms_metadata.set_transform_param (dbms_metadata.session_transform, 'SQLTERMINATOR', true);
   dbms_metadata.set_transform_param (dbms_metadata.session_transform, 'PRETTY', true);
end;
/

variable v_username VARCHAR2(30);

exec:v_username := upper('&1');

select dbms_metadata.get_ddl('USER', u.username) AS ddl
from   dba_users u
where  u.username = :v_username
union all
select dbms_metadata.get_granted_ddl('TABLESPACE_QUOTA', tq.username) AS ddl
from   dba_ts_quotas tq
where  tq.username = :v_username
and    rownum = 1
union all
select dbms_metadata.get_granted_ddl('ROLE_GRANT', rp.grantee) AS ddl
from   dba_role_privs rp
where  rp.grantee = :v_username
and    rownum = 1
union all
select dbms_metadata.get_granted_ddl('SYSTEM_GRANT', sp.grantee) AS ddl
from   dba_sys_privs sp
where  sp.grantee = :v_username
and    rownum = 1
union all
select dbms_metadata.get_granted_ddl('OBJECT_GRANT', tp.grantee) AS ddl
from   dba_tab_privs tp
where  tp.grantee = :v_username
and    rownum = 1
union all
select dbms_metadata.get_granted_ddl('DEFAULT_ROLE', rp.grantee) AS ddl
from   dba_role_privs rp
where  rp.grantee = :v_username
and    rp.default_role = 'YES'
and    rownum = 1
union all
select to_clob('/* Start profile creation script in case they are missing') AS ddl
from   dba_users u
where  u.username = :v_username
and    u.profile <> 'DEFAULT'
and    rownum = 1
union all
select dbms_metadata.get_ddl('PROFILE', u.profile) AS ddl
from   dba_users u
where  u.username = :v_username
and    u.profile <> 'DEFAULT'
union all
select to_clob('End profile creation script */') AS ddl
from   dba_users u
where  u.username = :v_username
and    u.profile <> 'DEFAULT'
and    rownum = 1
/

set linesize 80 pagesize 14 feedback on trimspool on verify on

Answer 7:

这个查询的输出是非常干净的（原来这里 ）

clear screen
accept uname prompt 'Enter User Name : '
accept outfile prompt  ' Output filename : '

spool &&outfile..gen

SET LONG 20000 LONGCHUNKSIZE 20000 PAGESIZE 0 LINESIZE 1000 FEEDBACK OFF VERIFY OFF TRIMSPOOL ON

BEGIN
   DBMS_METADATA.set_transform_param (DBMS_METADATA.session_transform, 'SQLTERMINATOR', true);
   DBMS_METADATA.set_transform_param (DBMS_METADATA.session_transform, 'PRETTY', true);
END;
/

SELECT dbms_metadata.get_ddl('USER','&&uname') FROM dual;
SELECT DBMS_METADATA.GET_GRANTED_DDL('SYSTEM_GRANT','&&uname') from dual;
SELECT DBMS_METADATA.GET_GRANTED_DDL('ROLE_GRANT','&&uname') from dual;
SELECT DBMS_METADATA.GET_GRANTED_DDL('OBJECT_GRANT','&&uname') from dual;

spool off

文章来源: How to generate entire DDL of an Oracle schema (scriptable)?