Question

我有一个包含三个数据库表的SQL语句。我正在比较一些ID和描述，现在需要比较结果FIRSTNAME，MIDDLENAME和SURNAME。

SELECT UGOVARATELJ_OSIGURANJA AS ULJEZ FROM INS_RAZNO.TEST_DOWNJONES 
WHERE UGOVARATELJ_OSIGURANJA IN 
(SELECT  FIRSTNAME||' '||MIDDLENAME||' '||SURNAME 
AS IME_PREZIME 
FROM (SELECT FIRSTNAME, MIDDLENAME, SURNAME 
FROM INS_RAZNO.AML_CTF
WHERE ID IN (SELECT ID FROM INS_RAZNO.AML_CTF_DESCRIPTIONS 
WHERE DESCRIPTION1='1' OR DESCRIPTION1='2')));

变量UGOVARATELJ_OSIGURANJA实际上是客户端名称，格式为NAME MIDDNAME SURNAME，名称的每个部分之间都有空格。我的想法是使用REMOVE()删除那些空格，但是当我将其与第二张表进行比较时，没有正确的结果。

我试图以此测试我的陈述：这仅是测试声明

SELECT REPLACE(UGOVARATELJ_OSIGURANJA, ' ','') AS ULJEZ FROM 
INS_RAZNO.TEST_DOWNJONES WHERE UGOVARATELJ_OSIGURANJA LIKE '%IVAN F%'

请注意，我必须在name和middname LIKE %IVAN F%之间设置空格，正确的方式应该是％IVANF％。

任何想法如何解决这个问题？预先感谢

编辑：我正在执行此替换（修剪），因为列表中的某些人没有MIDDLENAME

编辑：这是具有MIDDLENAME的名称的有效声明。当某人由于空格过多而没有MIDDLENAME时，该语句将无法正常工作。

Answer 1

尝试关注

将空格'％IVAN F％'替换为'％'

SELECT REPLACE(UGOVARATELJ_OSIGURANJA, ' ','') AS ULJEZ 
FROM INS_RAZNO.TEST_DOWNJONES 
WHERE REPLACE(UGOVARATELJ_OSIGURANJA, ' ','') LIKE '%IVAN%F%'

Answer 2

这是具有MIDDLENAME的名称的有效声明。当某人由于空格过多而没有MIDDLENAME时，该语句将无法正常工作。

您可以设置额外的空间为条件：

SELECT UGOVARATELJ_OSIGURANJA AS ULJEZ FROM INS_RAZNO.TEST_DOWNJONES 
WHERE UGOVARATELJ_OSIGURANJA IN 
(SELECT  FIRSTNAME||
   -- only add a space before middlename if it has a value
   CASE WHEN MIDDLENAME IS NOT NULL THEN ' ' END
   ||MIDDLENAME||' '||SURNAME 
AS IME_PREZIME 
FROM (SELECT FIRSTNAME, MIDDLENAME, SURNAME 
FROM INS_RAZNO.AML_CTF
WHERE ID IN (SELECT ID FROM INS_RAZNO.AML_CTF_DESCRIPTIONS 
WHERE DESCRIPTION1='1' OR DESCRIPTION1='2')));

这将从消除空格中消除错误匹配的可能性。

您可以通过删除一个不必要的子查询级别来稍微简化一下：

SELECT UGOVARATELJ_OSIGURANJA AS ULJEZ
FROM INS_RAZNO.TEST_DOWNJONES 
WHERE UGOVARATELJ_OSIGURANJA IN (
  SELECT  FIRSTNAME||
    CASE WHEN MIDDLENAME IS NOT NULL THEN ' ' END||MIDDLENAME
    ||' '||SURNAME
  FROM INS_RAZNO.AML_CTF
  WHERE ID IN (
    SELECT ID
    FROM INS_RAZNO.AML_CTF_DESCRIPTIONS 
    WHERE DESCRIPTION1='1' OR DESCRIPTION1='2'
  )
);

或将其更改为使用连接-我认为这是等效的：

SELECT TD.UGOVARATELJ_OSIGURANJA AS ULJEZ
FROM INS_RAZNO.AML_CTF_DESCRIPTIONS ACD
JOIN INS_RAZNO.AML_CTF AC ON AC.ID = ACD.ID
JOIN INS_RAZNO.TEST_DOWNJONES TD
ON TD.UGOVARATELJ_OSIGURANJ = AC.FIRSTNAME
  ||CASE WHEN AC.MIDDLENAME IS NOT NULL THEN ' ' END||AC.MIDDLENAME
  ||' '||AC.SURNAME
WHERE ACD.DESCRIPTION1 IN ('1', '2');

还要检查DESCRIPTION1的数据类型-如果实际上是数字，则应使用ACD.DESCRIPTION1 IN (1, 2)而不是强制在数字和字符串之间进行不必要的转换。

如何在SQL语句中比较修剪后的变量？

2 个答案: