postgresql sql 总结

天成1

[size=x-small]

//更新过期的
update  
capture  set  flag=1    where   (extract (epoch from  now())::bigint-extract (epoch from capture_time )::bigint)>time_val
--系统表
pg_stat_activity 数据库活动进程视图
Pg_database 系统数据库字典
Pg_stat_database系统统计数据库字典视图
Pg_stat_sys_tables 系统字典表
Pg_stat_sys_indexes系统字典表索引
Pg_stat_user_tables用户表
Pg_stat_user_indexes 用户表索引
pg_stat_all_tables 数据库所有的表（包括系统表和用户表）
pg_stat_all_indexs数据库所有的表的索引（包括系 统表和用户表的索引）
select datname,procpid,query_start,current_query,waiting,client_addr ,usename from pg_stat_activity  ;
--查询更新语句
select datname,procpid,query_start,current_query,waiting,client_addr ,usename from pg_stat_activity  where  and current_query like '%update%';  
ps aux|grep postgres|grep 14190
--取消查询语句
select pg_cancel_backend(14190);
--kill 各种sql
pg_terminate_backend(pid int),

psql -h .. -p .. -d db_name -U username  -c "copy public.test_1 to stdout" > test_1.out
--
除0
NULLIF 等价＝case when
select   count(nullif(answer_time_of_date,'0')) as succ_calls  from  client_cdr
select     count(case  when  answer_time_of_date  ='0' then null else answer_time_of_date end) as succ_calls  from  client_cdr
select    count(case  when  release_cause_from_protocol_stack  ~* '^486*' then release_cause_from_protocol_stack else null end) as busy_calls  from  client_cdr
exchange=# select  3/NULLIF(0,0);
SELECT COALESCE(SUM(amount), 0) / NULLIF(SUM(current_balance), 0) as average_price FROM client_payment
--防止除0  
select  3/0
select   3/NULLIF(0,0)
SELECT COALESCE(SUM(cost), 0) / NULLIF(SUM(amount), 0) as average_price FROM sales

SELECT *, first_name||' '||last_name AS full_name FROM test_table;
--取出几个列中最大的值
select        LEAST(current_balance,amount)   from    client_payment  where   client_payment_id=1
--取出几个列中最大的值
select        greatest(current_balance,amount)   from    client_payment  where   client_payment_id=1
--取出几个列中不为空的值
select             COALESCE(current_balance,amount)     from    client_payment  where   client_payment_id=1
--取出几个列中空的值
select             NULLIF(current_balance,amount)     from    client_payment  where   client_payment_id=1

在PostgreSQL中实现行号
2010-05-28 09:30
建立一个序列，每次查询果将其当前值设置为0，然后用nextval()函数取值就可以
create sequence test_row;
select setval('test_row',1);
select nextval('test_row')-1 as rowno ,table.* from table;

:%s/,\n/\r/g  --去掉最后一个,

:%s/^1//g--去掉第一个1


--PostgreSQL中如何删除重复记录
--ostgreSQL中删除重复记录其实很简单，不论有多少行重复，只要在要删除重复记录的表中table加一列rownum字段(id为table表中的主键)，类型设置为serial类型即可，然后执行sql
--删除原理   添加一个自增列,  按照 重复的 字段分组，取出rownum最大的一列,这一列就是去掉重复后的列
insert   into version    values(3,'22','333')
--添加自增字段
ALTER TABLE "version" ADD COLUMN rownum   serial;
--去掉重复后的结果
select   *   from    version   where  rownum  in  (   select    max(rownum)   from    version   group by  table_name )
--删除重复
delete   from    version   where  rownum   not  in  ( select    max(rownum)   from    version   group by  table_name )

--知道有什么用户在访问，访 问什么库以及访问的进程
select datid,datname,procpid,usename, xact_start from pg_stat_activity;
--kill 查询  ill  Select 查询，而updae,delete DML不生效
select pg_cancel_backend(线程id);

--查看有哪些SQL正在执行
select datname,procpid,query_start,current_query,waiting,client_addr from pg_stat_activity where waiting='t';


--最近15分钟的acd ,asr ,pdd,ca
--15 min
select   
sum(call_count_asr::integer*asr::real)/sum(call_count_asr::integer) as asr,
(sum(acd::real)/sum(call_count::integer))/60 as acd,
(sum(pdd::real) / sum(call_count::integer)) as pdd, sum(ca::integer) as ca
into  r     from  host_info
where ip=host_ip and   time::bigint between extract(epoch from now())::bigint-(60*15)      and  extract(epoch from now())::bigint;
select extract (epoch from  now())::bigint  //当前时间对应的时间戳
select  now()//现在时间
update host_info  set time=(select  extract(epoch from '2010-12-21     01:52:10.849097+00'::timestamp with time zone )::bigint)
--累加sql
select  invoice_id,  client_id ,
(select sum(total_amount) as past_due from invoice as inner_invoice where client_id = invoice.client_id and inner_invoice.invoice_end <= invoice.invoice_end)  as total from invoice;
update host_info  set time=(select  extract(epoch from '2010-12-20 11:22:10.849097+00'::timestamp with time zone )::bigint)
----------备份cdr
class4_pr=# copy (select *  from  client_cdr   limit  1)    to  '/tmp/exports/cdr.csv'   csv  HEADER;
--导入cdr
class4_pr=# copy client_cdr  from '/tmp/exports/cdr.csv'  csv  HEADER ;


---------------------------------------------------------------重建触发器
psql carry_sst  < /home/test/record.sql

SELECT *
FROM prefixes
WHERE prefix @> '0123456789'
ORDER BY length(prefix) DESC
LIMIT 1;

EXPLAIN ANALYZE
touch /var/lock/subsys/local
svnserve -d -r /shanmin/svn


----------------------------------------------------------------------
COPY----------- 用法-
-----------------------------------------------------------------------
COPY  error_message_info  FROM   '/tmp/message.csv'  CSV HEADER ;//导入
COPY  (select * from   error_message_info  limit 2)  TO   '/tmp/message1.csv'  CSV HEADER ;//导出


---case when  用法
case when call_duration='' then '0' ELSE call_duration end ,
select    call_count,  case   when call_count='0'  then '1' ELSE   call_count  end
from  host_info  where    call_count='0'



---v---------------v------------------各种时间转换---v------------------v---------------

SELECT   to_timestamp(start_time_of_date::bigint) as time  from  cdr
--时间处理
select     extract(epoch from      (logout_time -login_time ))  from  acd_login_log

SELECT EXTRACT(EPOCH FROM INTERVAL '5 days 3 hours');


实现按月按年，按日统计
把数据库里的数据按每一天统计当天的点击量，当然是用 group by，可惜问题是时间字段存储为bigint型，看来还得先转成timestamp型的，于是翻了手册，写了如下SQL搞定！
select to_char(to_timestamp(accesstime), ‘YYYY-MM-DD’) as d, count(validclick) as c from dbtm_ad_list group by d order by c desc
其中to_timestamp()函数将int型数据转换为timestamp类型的值，格式为”YYYY-MM-DD”， 再将其值用to_char()函数转换为文本类型，便可以用group by分组了。

select        ( current_timestamp(0) - interval '24 hour')
--将微秒改为时间戳
SELECT  to_timestamp(substring(answer_time_of_date from 1 for 10) ::bigint)from  client_cdr   
//更新当前时间
update  host_info  set     time=(select  EXTRACT(EPOCH from current_timestamp(0)))::text
select     *  from current_timestamp(0)    =="2010-11-10 10:40:46+08"
select  EXTRACT(EPOCH from current_timestamp(0))  (1289356894)
--将 1289356894转换为  2010-11-10 10:40:46+08
SELECT   to_timestamp(time::bigint)    from  host_info   
-- Column: create_time  给字段设置默认值为当前时间
-- ALTER TABLE account_balance DROP COLUMN create_time;
ALTER TABLE account_balance ADD COLUMN create_time timestamp with time zone;
ALTER TABLE account_balance ALTER COLUMN create_time SET STORAGE PLAIN;
ALTER TABLE account_balance ALTER COLUMN create_time SET NOT NULL;
ALTER TABLE account_balance ALTER COLUMN create_time SET DEFAULT ('now'::text)::timestamp(0) with time zone;
COMMENT ON COLUMN account_balance.create_time IS '改动时间';

---v---------------v------------------各种时间转换---v------------------v---------------


--导出数据库表结构
[iyunv@class4 tmp]# /usr/local/pgsql/bin/pg_dump  -s  class4v3_test  >  tmp.sql
--导入数据库表结构
[iyunv@class4 tmp]# psql class4v3   < tmp.sql
--sql  
CASE account_payment.payment_method
WHEN 0 THEN '系统充值卡 '::text
WHEN 1 THEN '快钱'::text
WHEN 2 THEN '系统管理员(手工充值)'::text
WHEN 3 THEN '易宝'::text
ELSE 'other'::text
END AS payment_method1,
--pg_dump 导出sql数据
.   
利用select into语句创建一个临时表，然后使用pg_dump导出临时表。  
select * into test_tbl from some_tbl where some_field > some_value;  
pg_dump -d db_name -t test_tbl > /tmp/test.sql  
/tmp/test.sql里面包含的，就是要求的特定记录了。  



--模糊查询
$condition="'%".$key."%'";
select resource.alias,resource.resource_id , resource.name ,cps_limit,capacity,ingress,egress,active, a.ip_cnt , client_id
from  resource
left join (select count(*)as  ip_cnt,resource_id from resource_ip    group by  resource_id) a   on  a.resource_id=resource.resource_id
where resource.name   like $condition
or resource.alias like $condition
or ( select count(*)>0 from resource_ip where resource_ip.resource_id =resource.resource_id and resource_ip.ip::varchar like $condition )
order by resource.resource_id  limit '$pageSize' offset '$offset'

----前缀匹配的用法  只取
select into id_rate rate_id from rate
where rate_table_id = id_rate_table and code @> NEW.origination_destination_number::prefix_range and current_timestamp between effective_date and end_date order by code desc limit 1;
$code_where="  and  origination_destination_number::prefix_range <@ '$code'";
select  * from  cdr  where origination_destination_number::prefix_range <@ '$code'";

select trunk.trunk_ip||':'||trunk.port address, trunk.trunk_id from
prefix,trunk where trunk.trunk_id = prefix.trunk_id and prefix_code @>'%
s'  order by length (prefix_code::text) desc limit 1

--in的替代解决办法
SELECT　*
FROM　tab
WHERE　col1　IN　(SELECT　col2　FROM　TAB2)

select  * from client  where client_id::text in (select     client_id  from   class4_view_client_cost  )
改为：　
select  * from client  where  exists(select client_id from class4_view_client_cost
where class4_view_client_cost.client_id::integer=client.client_id)
SELECT　*
FROM　tab
WHERE　EXISTS　(SELECT　col2　FROM　TAB2　WHERE　col1　=　col2)


--建立索引
-- Index: code_index_code
-- DROP INDEX code_index_code;
CREATE INDEX code_index_code
ON code
USING btree
(code);

--索引测试
--索引前执行   4452 ms
select * from  class4_view_client_cost  
--索引后执行 14ms


--prefix 特有的索引
-- Index: idx_code_prefix
-- DROP INDEX idx_code_prefix;
CREATE INDEX idx_code_prefix
ON code
USING gist
(code gist_prefix_range_ops);



--导出sql

$copy_sql = "COPY ($download_sql)  TO   '$copy_file'  DELIMITER '$DELIMITER'  CSV $HEADER "; //daochu
--导入sql

cmd:='COPY ' || table_name || '(ani,dnis,action_ani,action_dnis,ani_method,dnis_method)  FROM  '''||filecsv ||''' CSV HEADER';

--删除重复行

id name score
1 aaa 30
2 aaa 30
3 aaa 36
4 bbb 30
5 bbb 30
6 aaa 30
7 bbb 30
8 ccc 24
9 ccc 51

保留同NAME中ID最小.
delete tb from tb t where id not in (select min(id) from tb where name = t.name)
保留同NAME中ID最大.
delete tb from tb t where id not in (select max(id) from tb where name = t.name)




相关方法见下:
--如何按字段删除重复记录
一张表里面以两个字段为唯一字段，当几条记录的这两个字段完全相同时，需要删除重复项，如下表
a b c d
1 2 3 4
1 5 3 5
1 2 7 9
以a、b为唯一字段，第一条和第三条的a、b完全相同，所以，需要删除第一条记录1 2 3 4 或者第三条记录1 2 7 9
即如下结果：
a b c d
1 2 3 4
1 5 3 5
或
a b c d
1 5 3 5
1 2 7 9

CREATE TABLE Tb1(id int, [a] varchar(255), varchar(255), [c] varchar(255), [d] varchar(255))
INSERT Tb1(id, [a], , [c], [d])
SELECT 1, '1','2','3','4'
UNION ALL SELECT 2, '1','5','3','5'
UNION ALL SELECT 3, '1','2','7','9'
UNION ALL SELECT 4, '1','4','7','6'
delete Tb1 where [id] not in (select max([id]) from Tb1 group by a,b )
select * from tb1
drop table tb1
如果要同时删除第一和第三行
即如下结果：
a b c d
1 5 3 5
语句如下：
delete m from tb t
inner join
(
select a ,b
from tb
group by a , b
having count(*)>1
)n
on m.a = n.a and m.b = n.b
或
delete * from tb as m,
(
select a ,b
from tb
group by a , b
having count(*)>1
)n
where m.a = n.a and m.b = n.b




------------------------------------------------------------------------------------
在几千条记录里,存在着些相同的记录,如何能用SQL语句,删除掉重复的呢?谢谢!
1、查找表中多余的重复记录，重复记录是根据单个字段（peopleId）来判断
select * from people
where peopleId in (select peopleId from people group by peopleId having count(peopleId) > 1)
2、删除表中多余的重复记录，重复记录是根据单个字段（peopleId）来判断，只留有rowid最小的记录
delete from people
where peopleId in (select peopleId from people group by peopleId having count(peopleId) > 1)
and rowid not in (select min(rowid) from people group by peopleId having count(peopleId )>1)
3、查找表中多余的重复记录（多个字段）
select * from vitae a
where (a.peopleId,a.seq) in (select peopleId,seq from vitae group by peopleId,seq having count(*) > 1)
4、删除表中多余的重复记录（多个字段），只留有rowid最小的记录
delete from vitae a
where (a.peopleId,a.seq) in (select peopleId,seq from vitae group by peopleId,seq having count(*) > 1)
and rowid not in (select min(rowid) from vitae group by peopleId,seq having count(*)>1)
5、查找表中多余的重复记录（多个字段），不包含rowid最小的记录
select * from vitae a
where (a.peopleId,a.seq) in (select peopleId,seq from vitae group by peopleId,seq having count(*) > 1)
and rowid not in (select min(rowid) from vitae group by peopleId,seq having count(*)>1)
比方说在A表中存在一个字段“name”，而且不同记录之间的“name”值有可能会相同，
现在就是需要查询出在该表中的各记录之间，“name”值存在重复的项；
Select Name,Count(*) From A Group By Name Having Count(*) > 1
如果还查性别也相同大则如下:
Select Name,sex,Count(*) From A Group By Name,sex Having Count(*) > 1
------------------------------------------------------------------------------------------------
declare @max integer,@id integer
declare cur_rows cursor local for select 主字段,count(*) from 表名 group by 主字段 having count(*) >； 1
open cur_rows
fetch cur_rows into @id,@max
while @@fetch_status=0
begin
select @max = @max -1
set rowcount @max
delete from 表名 where 主字段 = @id
fetch cur_rows into @id,@max
end
close cur_rows
set rowcount 0
方法二
　　有两个意义上的重复记录，一是完全重复的记录，也即所有字段均重复的记录，二是部分关键字段重复的记录，比如Name字段重复，而其他字段不一定重复或都重复可以忽略。
　　1、对于第一种重复，比较容易解决，使用
select distinct * from tableName
　　就可以得到无重复记录的结果集。
　　如果该表需要删除重复的记录（重复记录保留1条），可以按以下方法删除
select distinct * into #Tmp from tableName
drop table tableName
select * into tableName from #Tmp
drop table #Tmp
　　发生这种重复的原因是表设计不周产生的，增加唯一索引列即可解决。
　　2、这类重复问题通常要求保留重复记录中的第一条记录，操作方法如下
　　假设有重复的字段为Name,Address，要求得到这两个字段唯一的结果集
select identity(int,1,1) as autoID, * into #Tmp from tableName
select min(autoID) as autoID into #Tmp2 from #Tmp group by Name,autoID
select * from #Tmp where autoID in(select autoID from #tmp2)
　　最后一个select即得到了Name，Address不重复的结果集（但多了一个autoID字段，实际写时可以写在select子句中省去此列）
select * from tablename where id in (
select id from tablename
group by id
having count(id) > 1)


delete a
from 表 a where exists(select * from 表 where name=a.name and id>a.id)




--  级联删除  Foreign Key: class4_fkey_code_code_deck_id
-- ALTER TABLE code DROP CONSTRAINT class4_fkey_code_code_deck_id;
ALTER TABLE code
ADD CONSTRAINT class4_fkey_code_code_deck_id FOREIGN KEY (code_deck_id)
REFERENCES code_deck (code_deck_id) MATCH SIMPLE
ON UPDATE NO ACTION ON DELETE CASCADE;
--创建序列
CREATE SEQUENCE invocie_item_id_seq
INCREMENT 1
MINVALUE 1
MAXVALUE 9223372036854775807
START 137
CACHE 1;

--设置序列
ALTER TABLE invoice_item ALTER COLUMN id SET DEFAULT nextval('invocie_item_id_seq'::regclass);






select * from crosstab
('select code,rate_id,resource_id from resource join rate on
rate.rate_table_id=resource.rate_table_id
where resource_id in (448,449,450) order by 1,2')
AS               
ct(code prefix_range, res_id_1 integer, res_id_2 integer, res_id_3
integer, res_id_4 integer, res_id_5 integer, res_id_6 integer, res_id_7
integer, res_id_8
integer)   


select code,resource_id from resource
join rate on  rate.rate_table_id=resource.rate_table_id
where resource_id in (448,449,450) order by 1,2;