Oracle APEX PL/SQL 程序调试？

Question

我正在使用 PL/SQL 程序将之前生成的 Table 插入另一个现有 table:

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;

但是不行。没有错误消息或其他程序失败的迹象，但它没有在最终 table.

中显示任何结果

第一个问题，有没有迹象表明我做错了什么？

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

Answer 1

我认为您不能那样引用 seq_id。

使用APEX_COLLECTION.DELETE_MEMBER去掉seq_id1.

例如

BEGIN

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

END;

然后照常继续。

编辑 - 失败的原因是错误的数据类型错误。如果您检查调试器，您应该能够看到这一点。

Answer 2

在插入数据库之前，必须为每个成员属性更新顶点集合。

第一个 Apex 流程： 提交时和计算前

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;

第二个 Apex 流程： 提交后

declare
    v_msg varchar2(4000);
begin
    for x in (select * from apex_collections where collection_name = 'COLLECTION' and seq_id != 1 order by seq_id) 
    loop            
        begin
            INSERT INTO REI_LABOUR_COST (DEALER_NUMBER, DEALER_STATUS, BILLING_PARTNER, BSI_GW, YEAR, QUANTITY_FRU, LABOUR_EUR, LABOUR_LOCAL, REQUESTED_RATE) 
            values(x.c001, x.c002, x.c003, x.c004, x.c005, x.c006, x.c007, x.c008, x.c009);
        exception when others then
            v_msg := ''||sqlerrm;
            wwv_flow.debug('ERR_1: '||v_msg);                   
        end;
    end loop;   
exception when others then
    v_msg := ''||sqlerrm;
    wwv_flow.debug('ERR_2: '||v_msg);   
end;

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