可行。但插入也是需要开销的。比如数组 arrayX[N]方法一\ for each @x in arrayX { select id from table1 where id=@x)方法二\ for each @x in arrayX { @s += @x + "),(" } insert into tableTemp (id) values @s; select * from tableTemp where not exists (select 1 from table1 where id=tableTemp .id)效率上明显可以分析出差异。
for each @x in arrayX { select id from table1 where id=@x)方法二\
for each @x in arrayX { @s += @x + "),(" }
insert into tableTemp (id) values @s;
select * from tableTemp where not exists (select 1 from table1 where id=tableTemp .id)效率上明显可以分析出差异。
2、定义一个集合notinDBList
for(int i=0;i<id数组.size;i++){
if(!allIdList.contains(id数组[i])){
notinDBList.add(id数组[i]));
}
}
3、最后notinDBList里面存的就是你要的结果