ID_1, ID_2, ID_3等列的存在意味着你的表不是第一范式（1 NF），这是数据建模/设计的第一个最基本的步骤。不应该有这样的重复组。正确的解决方法是将这些ID分解到它们自己的表中：

table_1                           table1_table2_xref    table_2
-------                           -----------           -------
{table_1_pk},                     {table_1_pk},         {id}
scalar attributes of table_1...   {id        }          table_2 attrs...

一旦这是固定的，你然后简单地做：

SELECT *
  FROM table_1 t1
  LEFT JOIN table1_table2_xref x ON x.table_1_pk = t1.table_1_pk
  LEFT JOIN table_2 t2 ON x.id = t2.id

这将使您无论有多少ID都能很好地执行，极大地简化了代码，并使您的设计具有弹性和可扩展性（您可以在不更改代码的情况下添加ID）。

如果你只希望每个内部ID有一个值，那么你可以使用条件聚合来透视：

SELECT t1.id,
       MAX(CASE WHEN T2.id = T1.id_1 THEN T2.key   END) AS key1,
       MAX(CASE WHEN T2.id = T1.id_1 THEN T2.value END) AS value1,
       MAX(CASE WHEN T2.id = T1.id_2 THEN T2.key   END) AS key2,
       MAX(CASE WHEN T2.id = T1.id_2 THEN T2.value END) AS value2,
       MAX(CASE WHEN T2.id = T1.id_3 THEN T2.key   END) AS key3,
       MAX(CASE WHEN T2.id = T1.id_3 THEN T2.value END) AS value3
FROM   TABLE_1 T1
       LEFT OUTER JOIN TABLE_2 T2
         ON T2.ID IN (T1.ID_1, T1.ID_2, T1.ID_3)
GROUP BY t1.id;

其中，对于示例数据：

CREATE TABLE table_1 ( id, id_1, id_2, id_3 ) AS
SELECT 1, 1, 2, 3 FROM DUAL UNION ALL
SELECT 2, 4, 5, 6 FROM DUAL;

CREATE TABLE table_2 (id, key, value) AS
SELECT LEVEL, 'key' || level, CHR(64+LEVEL) FROM DUAL CONNECT BY LEVEL <= 6;

输出：
| ID|按键1|数值1|按键2|价值2|按键3|价值3|
| --------------|--------------|--------------|--------------|--------------|--------------|--------------|
| 1|键1|A|键2|B|键3|C|
| 第二章|按键4|D|键5|E级|键6|F|
fiddle