列计算查询

SQL code

create table t(ta int ,n1 varchar(10),m1 varchar(10), n2 varchar(10),m2 varchar(10))insert into t select 1,'10','20','12','25' union  select 2,'22','20','50','30'unionselect 3,'55','700','100' ,'600' union select 4,'42','50','40' ,'48'unionselect 5,'85','80', '80','90'unionselect 6,'88','300','50','70' union  select 7,'90','70', '40','80' select * from t--drop table t-----------按照10为判断值  判断界限的应该是-10<=x<=10-----------X值为n2-n1,m2-m1的结果  计算结果在上面的范围内 用1表示 否则为0---------- 或者换其他的方式思考

SQL code

select * from twhere abs(n2-n1)<10 or ABS(m2-m1)<10/*1    10    20    12    254    42    50    40    485    85    80    80    90*/
[解决办法]
SQL code
select *,compute=(if (abs(n2-n1)<=10 and ABS(m2-m1)<=10)1else 0)from t
[解决办法]
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SQL code
select *,n2-n1 as su1,m2-m1 as su2,case when abs(n2-n1)<10 and ABS(m2-m1)<10 then 1 else 0 end as sigfrom t/*ta          n1          m1          n2          m2          su1         su2         sig----------- ----------- ----------- ----------- ----------- ----------- ----------- -----------1           10          20          12          25          2           5           12           22          20          50          30          28          10          03           55          700         100         600         45          -100        04           42          50          40          48          -2          -2          15           85          80          80          90          -5          10          06           88          300         50          70          -38         -230        07           90          70          40          80          -50         10          0(7 行受影响)*/