问题描述
我知道我在问题标题中没有明确说明。让我解释一下。
说我有一个表SOURCE_TABLE
,其中有1列看起来像这样:
Filter
------------------|
Name='John'
Surname = 'Smith'
Age = '25'
我想将此表用作过滤器。如下所示:
SELECT * FROM TARGET_TABLE WHERE (SELECT FILTER FROM SOURCE_TABLE)
我听说也许评估功能可以帮助我,但老实说,我不明白该怎么做。
您知道有什么方法可以将列用作我的过滤器源吗?
Edit1:
DECLARE
my_filter VARCHAR2(100);
my_query VARCHAR2(500);
BEGIN
my_query := 'SELECT FILTER FROM SOURCE_TABLE WHERE ROWNUM=1';
EXECUTE IMMEDIATE my_query INTO my_filter;
EXECUTE IMMEDIATE 'SELECT * FROM TARGET_TABLE WHERE '|| my_filter;
END;
@Sujitmohanty30我在学习EXECUTE IMMEDIATE
之后提出了上述建议。但是我偶然发现了一个问题。关于最终结果,这应该是动态的,我希望最后看到选择查询的结果。
解决方法
假设我们有以下两个表格:
create table TARGET_TABLE(name varchar2(30),surname varchar2(30));
insert into TARGET_TABLE(name,surname) values ('John','Doe');
insert into TARGET_TABLE(name,surname) values ('Ann','Smith');
insert into TARGET_TABLE(name,surname) values ('Steven','Feuerstein');
insert into TARGET_TABLE(name,'King');
commit;
create table SOURCE_TABLE(filter) as
select q'[name='John']' from dual union all
select q'[name='Ann']' from dual union all
select q'[name='Steven' and surname='King']' from dual union all
select q'[surname='Feuerstein']' from dual;
然后,您可以使用xmltable和dbms_xmlgen.getXMLtype动态获取此类行:
https://dbfiddle.uk/?rdbms=oracle_18&fiddle=72abdf18b149cf30882cb4e1736c9c33
select *
from SOURCE_TABLE,xmltable(
'/ROWSET/ROW'
passing dbms_xmlgen.getXMLtype('select * from TARGET_TABLE where '|| SOURCE_TABLE.filter)
columns
name varchar2(30) path 'NAME',surname varchar2(30) path 'SURNAME'
) x;
结果:
FILTER NAME SURNAME
-------------------------------- ------------------------------ ------------------------------
name='John' John Doe
name='Ann' Ann Smith
name='Steven' and surname='King' Steven King
surname='Feuerstein' Steven Feuerstein