如何在表数据库Oracle中插入100万个随机行

时间:2012-07-31 10:45:30

标签: sql oracle

我无法弄清楚如何在表格中放入大量数据。数据不得重复

建议,可能还有其他方法吗?

create table COUNTRIES (
  COUNTRY_ID   VARCHAR2(7),
  COUNTRY_NAME VARCHAR2(40),
  constraint COUNTRY_C_ID_PK primary key (COUNTRY_ID)
);


Begin
For IDS in 1..1000000
Loop
INSERT INTO "SYSTEM"."COUNTRIES" (COUNTRY_ID, COUNTRY_NAME) VALUES (dbms_random.string('L', 7), dbms_random.string('L', 15));
Commit;
End loop;
End; 

2 个答案:

答案 0 :(得分:27)

如果您只想要大量数据而不关心内容的随机性,

 insert into countries select rownum, 'Name'||rownum from dual
   connect by rownum<=1000000;

应该这样做。

答案 1 :(得分:5)

如果你有一个非常具体的随机定义,并且不能允许重复,那么异常处理可以帮助避免重复。

这种方法会很慢。如果您需要多次执行此操作,或者需要执行大量数据,您可能需要放宽对“随机”的定义,并使用像Erich这样的解决方案。

--Create temporary unique constraint.  (Assuming you want each column to be unique?)
alter table countries add constraint countries_name_uq unique (country_name);

--Insert random data until it worked 1 million times.
Declare
    rows_inserted number := 0;
Begin
    Loop
        Begin
            INSERT INTO COUNTRIES(COUNTRY_ID, COUNTRY_NAME)
            VALUES(dbms_random.string('L', 7), dbms_random.string('L', 15));
            --Only increment counter when no duplicate exception
            rows_inserted := rows_inserted + 1;
        Exception When DUP_VAL_ON_INDEX Then Null;
        End;
        exit when rows_inserted = 1000000;
    End loop;
    commit;
End;
/

--Drop the temporary constraint
alter table countries drop constraint countries_name_uq;

--Double-check the count of distinct rows
select count(*) from
(
    select distinct country_id, country_name from countries
);

Result
------
1000000