Question

我正在使用PL / SQL过程将之前生成的一个表插入到另一个现有表中：

INSERT INTO REI_LABOUR_COST (DEALER_NUMBER, DEALER_STATUS, BILLING_PARTNER, BSI_GW, YEAR, QUANTITY_FRU, LABOUR_EUR, LABOUR_LOCAL, REQUESTED_RATE)
select c001 as DEALER_NUMBER
     , c002 as DEALER_STATUS
     , c003 as BILLING_PARTNER
     , c004 as BSI_GW
     , c005 as YEAR
     , c006 as QUANTITY_FRU
     , c007 as LABOUR_EUR
     , c008 as LABOUR_LOCAL
     , c009 as REQUESTED_RATE
from apex_collections col
where collection_name = 'COLLECTION' and seq_id != 1
order by seq_id;

但它不起作用。没有错误消息或程序失败的其他迹象，但它没有在最终表格中显示任何结果。

第一个问题，有什么迹象表明我做错了吗？

第二个问题，如果命令失败，我会在哪里看到错误消息？我可以在某个地方启用它吗？

Answer 1

在插入数据库之前，必须为每个成员属性更新apex集合。

First Apex流程： On Submit and Computation

 ps aux --sort=-%mem | awk 'NR<=10{print $0}'
USER       PID %CPU %MEM    VSZ   RSS TTY      STAT START   TIME COMMAND
ubuntu   27463  3.9 33.8 1045284 343428 ?      Sl   13:11   0:05 sidekiq 4.1.2 Quickmilk [1 of 1 busy]
ubuntu   27466  1.5 31.3 841340 317708 ?       Sl   13:11   0:02 Passenger RubyApp: /var/www/dairylac/Quickmilk/current/public (staging)
ubuntu   27370  2.8 18.6 492812 189012 ?       Sl   13:11   0:04 Passenger AppPreloader: /var/www/dairylac/Quickmilk/current
postgres 27476  0.2  4.0 192540 41344 ?        Ss   13:11   0:00 postgres: enbake quickmilk_dev_one 127.0.0.1(33998) idle
postgres 27915 21.0  3.5 189828 36296 ?        Ss   13:13   0:03 postgres: enbake quickmilk_dev_one 127.0.0.1(34040) idle in transaction
postgres  1489  0.0  3.4 185752 34680 ?        Ss   Aug29   0:07 postgres: writer process   
ubuntu   27444  0.0  1.5  42764 15228 ?        Ssl  13:11   0:00 /var/www/dairylac/Quickmilk/shared/bundle/ruby/2.1.0/gems/appsignal-1.2.5/ext/appsignal-agent
root      1078  0.0  0.9 278696 10040 ?        S<sl Aug29   0:01 /usr/lib/snapd/snapd
redis     1069  0.0  0.8  39236  8976 ?        Ssl  Aug29   0:42 /usr/local/bin/redis-server 127.0.0.1:6379

第二个Apex流程：提交后

declare
    y pls_integer := 0;
    v_msg varchar2(4000);
begin
    if not apex_collection.collection_exists(p_collection_name=>'COLLECTION') then
        wwv_flow.debug('No Apex collection found!');
    else
        for x in (select * from apex_collections where collection_name = 'COLLECTION' and seq_id != 1 order by seq_id) 
        loop
           y := y+1;
            apex_collection.update_member_attribute (p_collection_name=> 'COLLECTION', p_seq=> x.seq_id,p_attr_number =>1,p_attr_value=>wwv_flow.g_f01(y));     
            apex_collection.update_member_attribute (p_collection_name=> 'COLLECTION', p_seq=> x.seq_id,p_attr_number =>2,p_attr_value=>wwv_flow.g_f02(y)); 
            apex_collection.update_member_attribute (p_collection_name=> 'COLLECTION', p_seq=> x.seq_id,p_attr_number =>3,p_attr_value=>wwv_flow.g_f03(y)); 
            apex_collection.update_member_attribute (p_collection_name=> 'COLLECTION', p_seq=> x.seq_id,p_attr_number =>4,p_attr_value=>wwv_flow.g_f04(y)); 
            apex_collection.update_member_attribute (p_collection_name=> 'COLLECTION', p_seq=> x.seq_id,p_attr_number =>5,p_attr_value=>wwv_flow.g_f05(y)); 
            apex_collection.update_member_attribute (p_collection_name=> 'COLLECTION', p_seq=> x.seq_id,p_attr_number =>6,p_attr_value=>wwv_flow.g_f06(y)); 
            apex_collection.update_member_attribute (p_collection_name=> 'COLLECTION', p_seq=> x.seq_id,p_attr_number =>7,p_attr_value=>wwv_flow.g_f07(y)); 
            apex_collection.update_member_attribute (p_collection_name=> 'COLLECTION', p_seq=> x.seq_id,p_attr_number =>8,p_attr_value=>wwv_flow.g_f08(y)); 
            apex_collection.update_member_attribute (p_collection_name=> 'COLLECTION', p_seq=> x.seq_id,p_attr_number =>9,p_attr_value=>wwv_flow.g_f09(y)); 
        end loop;   
    end if;
exception when others then
    v_msg := ''||sqlerrm;
    wwv_flow.debug('ERR: '||v_msg); 
end;

Answer 2

我认为你不能像那样引用seq_id。

使用APEX_COLLECTION.DELETE_MEMBER摆脱seq_id 1.

E.g。

BEGIN

 APEX_COLLECTION.DELETE_MEMBER(
 p_collection_name => 'COLLECTION',
 p_seq => '1'); 

END;

然后继续正常。

编辑 - 它失败的原因是错误的数据类型错误。如果你检查调试器，你应该能看到这个。

Oracle APEX PL / SQL过程调试？

2 个答案: