我喜欢添加54个IF嵌套，我尝试通过在每8个之后将它们打散来添加＆选项，因为我这样做只允许多进行8个嵌套，其余的则不能工作。我有什么办法可以解决这个问题？
=IF(E9<231,"1",IF(E9<321,"2",IF(E9<411,"3",IF(E9<501,"4",IF(E9<591,"5",IF(E9<681,"6",IF(E9<771,"7",IF(E9<861,"8",)))))))))IF(E9<951,"9",IF(E9<1041,"10",IF(E9<1131,"11",IF(E9<1221,"12",IF(E9<1311,"13",IF(E9<1401,"14",IF(E9<1491,"15",IF(E9<1581,"16",))))))))IF(E9<1581,"16",IF(E9<1671,"17",IF(E9<1761,"18",IF(E9<1851,"19",IF(E9<1941,"20",IF(E9<2031,"21",IF(E9<2121,"22",)))))))IF(E9<2211,"23",IF(E9<2301,"24",IF(E9<2391,"25",IF(E9<2481,"26",IF(E9<2571,"27",IF(E9<2661,"28",IF(E9<2751,"29",IF(E9<2841,"30",)))))))) IF(E9<2931,"31", IF(E9<3021,"32", IF(E9<3111,"33",IF(E9<3201,"34", IF(E9<3291,"35", IF(E9<3381,"36", IF(E9<3471,"37", IF(E9<3561,"38",)))))))) IF(E9<3651,"39", IF(E9<3741,"40", IF(E9<3831,"41", IF(E9<3921,"42", IF(E9<4011,"43", IF(E9<4101,"44", IF(E9<4191,"45", IF(E9<4281,"46",)))))))) IF(E9<4371,"47", IF(E9<4461,"48", IF(E9<4551,"49", IF(E9<4641,"50", IF(E9<4731,"51", IF(E9<4821,"52", IF(E9<4911,"53", IF(E9<5001,"54",)))))))
答案 0 :(得分：6)
使用MATCH（）
=MATCH(E9,{0,231,321,411,501,...})
将...替换为其余阈值。
正如@SolarMike所说，您还可以将阈值放在一个范围内。然后，请参考该范围，而不是对数组进行“硬编码”。因此，如果您将阈值设置为A1：A55，则公式为：
=MATCH(E9,A1:A55)
答案 1 :(得分：3)
嵌套函数有时具有可以通过一些数学利用的模式。
=INT(MAX(E9-51,90)/90)
您没有提供E列中的价位大于5000的准备金，但可以轻松添加。