只要name(姓名)+任意一个电话相同的情况下，就提示该学员已存在

[fkueyga6]

xy表：
id name(姓名) mother_mobile(妈妈电话) father_mobile(爸爸电话) home_tel(家庭电话)

我现在是想做个判断，只要name(姓名)+任意一个电话相同的情况下，就提示该学员已存在
下面是我写的SQL语句，好复杂，而且我不知道对不对，有没有更简单的？
xy.name='".$arr_xy_info['name']."'
and (
((xy.mother_mobile='".$arr_xy_info['mother_mobile']."' or xy.father_mobile='".$arr_xy_info['mother_mobile']."' or xy.home_tel='".$arr_xy_info['mother_mobile']."') and xy.father_mobile<>'' and xy.home_tel<>'' and xy.mother_mobile<>0)
or ((xy.mother_mobile='".$arr_xy_info['father_mobile']."' or xy.father_mobile='".$arr_xy_info['father_mobile']."' or xy.home_tel='".$arr_xy_info['father_mobile']."') and xy.father_mobile<>'' and xy.home_tel<>'' and xy.mother_mobile<>0)
or ((xy.mother_mobile='".$arr_xy_info['home_tel']."' or xy.father_mobile='".$arr_xy_info['home_tel']."' or xy.home_tel='".$arr_xy_info['home_tel']."') and xy.father_mobile<>'' and xy.home_tel<>'' and xy.mother_mobile<>0)
)

[蜗牛慢慢向上爬]

个人觉得用程序的写的话简单些

[hchyyooh]

同楼上，用程序判断吧

[tianboo]

不考虑效率的前提下： select id from ( select id,name,concat(mother_mobile,father_mobile,home_tel) `phones` from xy ) `a` where name='test' and phones likes('%181822002200%') limit 0,1;

[tianboo]

引用 3 楼 tianboo 的回复:

不考虑效率的前提下： select id from ( select id,name,concat(mother_mobile,father_mobile,home_tel) `phones` from xy ) `a` where name='test' and phones likes('%181822002200%') limit 0,1;

更正一下.like 写成 likes

[傲雪星枫]

$sqlstr = "select id from xy where name='".$arr_xy_info['name']." and id in(select id from xy where mother_mobile='".$arr_xy_info['mother_mobile']."' or father_mobile='".$arr_xy_info['father_mobile']."' or home_tel='".$arr_xy_info['home_tel']."')";