Sql 我们可以在select子句中使用关系运算符吗

我想制作一列，其中我的金额小于某个值。当我尝试此操作时，查询将运行

select LOSA_APP.app_ref_no AS "App.Ref.No.",
   LOSA_EXP_SUMM_Z.group_exp AS "Group Exposure Amount",
   LOSA_EXP_SUMM_Z.group_exp - 25000 AS "Less than",
   columns AS "some Name"
   columns AS "some Name"
from 
     losa_app LOSA_APP
INNER JOIN
    code_branch CODE_BRANCH
ON
    LOSA_APP.attend_branch = CODE_BRANCH.branch_id
....
where 
    LOSA_APP.app_status in ('A','R')
and
    ....
 or 
 (
    '&&aplication' = 'Enhancement' 
    and 
    (
        nvl(LOSA_APP.review_type, '-') IN ('Enhancement', 'Additional')  and nvl(LOSA_APP.review_freq, '-') IN ('Enhancement', 'New')
    )
);

但是，当我将行

LOSA\u EXP\u sum\u Z.group\u EXP-25000更改为“小于”

时，将

LOSA\u EXP\u sum\u Z.group\u EXP<25000更改为“小于”

，那么我得到的错误是

ORA-00923: FROM keyword not found where expected
00923. 00000 -  "FROM keyword not found where expected"
*Cause:    
*Action:
Error at Line: 3 Column: 34

第3行是

LOSA\u EXP\u SUMM\u Z.group\u EXP<25000，表示“小于”

。当我们使用

减号
并在
关系运算符的情况下给出错误时，为什么会出现这种情况

谢谢
如果要在
选择中使用关系运算符
，则需要在
案例
语句中使用它，如下所示：

select LOSA_APP.app_ref_no AS "App.Ref.No.",
   LOSA_EXP_SUMM_Z.group_exp AS "Group Exposure Amount",
   CASE 
        WHEN LOSA_EXP_SUMM_Z.group_exp < 25000 
        then 'true'
        else 'false' 
     end AS "Less than",
   columns AS "some Name"
   columns AS "some Name"
from losa_app LOSA_APP
INNER JOIN code_branch CODE_BRANCH
    ONLOSA_APP.attend_branch = CODE_BRANCH.branch_id
....
where  LOSA_APP.app_status in ('A','R')
and
    ....
 or 
 (
    '&&aplication' = 'Enhancement' 
    and 
    (
        nvl(LOSA_APP.review_type, '-') IN ('Enhancement', 'Additional')  and nvl(LOSA_APP.review_freq, '-') IN ('Enhancement', 'New')
    )
);

select LOSA_APP.app_ref_no AS "App.Ref.No.",
   LOSA_EXP_SUMM_Z.group_exp AS "Group Exposure Amount",
   CASE WHEN LOSA_EXP_SUMM_Z.group_exp < 25000 
            then LOSA_EXP_SUMM_Z.group_exp - 25000
        else LOSA_EXP_SUMM_Z.group_exp end AS "Less than",
   columns AS "some Name"
   columns AS "some Name"
from losa_app LOSA_APP
INNER JOIN code_branch CODE_BRANCH
    ONLOSA_APP.attend_branch = CODE_BRANCH.branch_id
....
where  LOSA_APP.app_status in ('A','R')
and
    ....
 or 
 (
    '&&aplication' = 'Enhancement' 
    and 
    (
        nvl(LOSA_APP.review_type, '-') IN ('Enhancement', 'Additional')  and nvl(LOSA_APP.review_freq, '-') IN ('Enhancement', 'New')
    )
);