Oracle ORA-01873: the leading precision of the interval is too small 解决办法
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Oracle 执行 对时间间隔进行操作的 sql 语句时,报错 ORA-01873: the leading precision of the interval is too small
select a.* from (select case when num_to_date(nvl(VALIDITY_UNTIL_DATE,0))< sysdate THEN 2 when num_to_date(nvl(VALIDITY_UNTIL_DATE,0)) - interval '180' day < sysdate then 1 else 0 end FROM YH4_FIRE_LICENSE WHERE ENTERPRISE_ID =1 order by LICENSE_ID desc ) a where rownum = 1
这个是因为 day 时间间隔默认是只支持 -99 ~ 99 的范围,这个时候只要在 把 interval '-183' day 改成 interval '-183'day(4) 即可,把支持时间间隔范围改成 -9999 ~ 9999 四位。
select a.* from (select case when num_to_date(nvl(VALIDITY_UNTIL_DATE,0))< sysdate THEN 2 when num_to_date(nvl(VALIDITY_UNTIL_DATE,0)) - interval '9999' day(4) < sysdate then 1 else 0 end FROM YH4_FIRE_LICENSE WHERE ENTERPRISE_ID =1 order by LICENSE_ID desc ) a where rownum = 1
对 月 和 年 的时间间隔操作,应该类似,后续有时间在测试更新
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