重新生成这串sql 解决division by zero的问题 select PROD_DIS_TYPE , case when PROD_DIS_TYPE = '1' then DECODE(sum(FIRST_BIZ), 0,0, ROUND(sum(case when CONTRACT_CODE = 'NHT-1904506' then FIRST_BIZ else 0 end) / sum(FIRST_BIZ), 2))::int else 0 end as FIRST_BIZ , case when PROD_DIS_TYPE = '2' then DECODE(sum(APPR_DESIGN_PROD), 0, 0, ROUND(sum(case when CONTRACT_CODE = 'NHT-1904506' then APPR_DESIGN_PROD else 0 end) / sum(APPR_DESIGN_PROD), 2))::int else 0 end as APPR_DESIGN_PROD from T_SPDI_CONT_SUB_PROJ where UNITERM_CODE = 'N1900620-04-008' and SUB_PROJ_CODE = 'N1900620-04-302' and PROD_DIS_TYPE = '1' group by PROD_DIS_TYPE

可以在除法运算前加上 NULLIF 函数，将分母为 0 的情况转换为 NULL，避免了 division by zero 的错误。修改后的 SQL 如下：

SELECT 
    PROD_DIS_TYPE, 
    CASE WHEN PROD_DIS_TYPE = '1' 
        THEN DECODE(SUM(FIRST_BIZ), 0, 0, ROUND(SUM(CASE WHEN CONTRACT_CODE = 'NHT-1904506' THEN FIRST_BIZ ELSE 0 END) / NULLIF(SUM(FIRST_BIZ), 0), 2))::INT 
        ELSE 0 
    END AS FIRST_BIZ, 
    CASE WHEN PROD_DIS_TYPE = '2' 
        THEN DECODE(SUM(APPR_DESIGN_PROD), 0, 0, ROUND(SUM(CASE WHEN CONTRACT_CODE = 'NHT-1904506' THEN APPR_DESIGN_PROD ELSE 0 END) / NULLIF(SUM(APPR_DESIGN_PROD), 0), 2))::INT 
        ELSE 0 
    END AS APPR_DESIGN_PROD 
FROM 
    T_SPDI_CONT_SUB_PROJ 
WHERE 
    UNITERM_CODE = 'N1900620-04-008' 
    AND SUB_PROJ_CODE = 'N1900620-04-302' 
    AND PROD_DIS_TYPE = '1' 
GROUP BY 
    PROD_DIS_TYPE