eat*_*ode 9 sql oracle where-in xmltable
我正在处理我正在处理的项目中的问题,我不能给你实际的代码,但我已经创建了一个可执行的示例代码,如下所示
这里temp和temp_id两个表
temp table包含以逗号分隔的id列表 VARCHAR2temp_id 表包含实际的ID NUMBER我想要搜索行temp_id通过获取表ids从逗号分隔的ID列表temp表
//DDLs to create table
CREATE TABLE temp(ids VARCHAR2(4000));
CREATE TABLE temp_id(data_id NUMBER);
//DMLs to populate test data
INSERT INTO temp VALUES('1, 2, 3');
INSERT INTO temp_id VALUES(1);
INSERT INTO temp_id VALUES(2);
INSERT INTO temp_id VALUES(3);
INSERT INTO temp_id VALUES(4);
INSERT INTO temp_id VALUES(5);
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此查询不起作用
SELECT * FROM temp_id WHERE data_id IN (SELECT to_number(COLUMN_VALUE) FROM XMLTABLE(SELECT ids FROM temp));
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工作查询
SELECT * FROM temp_id WHERE data_id IN (SELECT to_number(COLUMN_VALUE) FROM XMLTABLE('1, 2, 3'));
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上面两个查询之间的差异是我temp在第一个查询中使用表中的列,并varchar2在第二个查询中直接引用.没有理由不起作用?我错过了什么吗?我认为可能存在一些数据类型不匹配但无法弄清楚.
您的要求称为Varying IN-lists.请参阅WHERE子句中的Varying IN值列表
原因: IN ('1, 2, 3')是不一样的IN (1, 2, 3)ORIN('1', '2', '3')
因此,
SELECT*FROM temp_id WHERE data_id IN(SELECT ids FROM temp);
和...一样
SELECT*FROM temp_id WHERE data_id IN('1,2,3');
会引发错误 ORA-01722: invalid number -
SQL> SELECT * FROM temp_id WHERE data_id IN('1, 2, 3');
SELECT * FROM temp_id WHERE data_id IN('1, 2, 3')
*
ERROR at line 1:
ORA-01722: invalid number
SQL> SELECT * FROM temp_id WHERE data_id IN(SELECT ids FROM temp);
SELECT * FROM temp_id WHERE data_id IN(SELECT ids FROM temp)
*
ERROR at line 1:
ORA-01722: invalid number
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与...不一样
SELECT*FROM temp_id WHERE data_id IN(1,2,3);
哪个会给你正确的输出 -
SQL> SELECT * FROM temp_id WHERE data_id IN(1, 2, 3);
DATA_ID
----------
1
2
3
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方案:
根据您的要求,您可以像这样实现 -
SQL> SELECT * FROM temp;
IDS
--------------------------------------------------------------
1, 2, 3
SQL> SELECT * FROM temp_id;
DATA_ID
----------
1
2
3
4
5
SQL> WITH data AS
2 (SELECT to_number(trim(regexp_substr(ids, '[^,]+', 1, LEVEL))) ids
3 FROM temp
4 CONNECT BY instr(ids, ',', 1, LEVEL - 1) > 0
5 )
6 SELECT * FROM temp_id WHERE data_id IN
7 (SELECT ids FROM data
8 )
9 /
DATA_ID
----------
1
2
3
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或者,您可以创建自己的TABLE函数或Pipelined函数来实现此目的.您的目标应该是将逗号分隔的IN列表拆分为多行.你是怎么做的取决于你!
工作演示
我们以模式中的标准EMP表为例SCOTT.
我有一个字符串中的作业列表,我想计算这些工作的员工:
SQL> SET serveroutput ON
SQL> DECLARE
2 str VARCHAR2(100);
3 cnt NUMBER;
4 BEGIN
5 str := q'[CLERK,SALESMAN,ANALYST]';
6 SELECT COUNT(*) INTO cnt FROM emp WHERE JOB IN (str);
7 dbms_output.put_line('The total count is '||cnt);
8 END;
9 /
The total count is 0
PL/SQL procedure successfully completed.
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哦! 发生了什么?标准emp表应该给出输出10.原因是变化的IN列表.
让我们看看正确的方法:
SQL> SET serveroutput ON
SQL> DECLARE
2 str VARCHAR2(100);
3 cnt NUMBER;
4 BEGIN
5 str := q'[CLERK,SALESMAN,ANALYST]';
6 SELECT COUNT(*)
7 INTO cnt
8 FROM emp
9 WHERE job IN
10 (SELECT trim(regexp_substr(str, '[^,]+', 1, LEVEL))
11 FROM dual
12 CONNECT BY instr(str, ',', 1, LEVEL - 1) > 0
13 );
14 dbms_output.put_line('The total count is '||cnt);
15 END;
16 /
The total count is 10
PL/SQL procedure successfully completed.
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