Question

我的查询中有一个错误，但找不到原因。我想根据条件找到相同的电话号码。

我尝试了if条件。

SELECT count(t.id) as sayi from student as t 
    inner join student s on s.id!= t.id                                  
    where case 
          when one_or_two=$one_or_two and one_or_two = 1 then 
                s.first_phone = t.first_phone 
          when one_or_two=$one_or_two and one_or_two = 2 then 
                s.sec_phone = t.sec_phone
          else 1=1
          end
    and t.id=4327

如果$ one_or_two = 1

此条件应正常运行s.first_phone = t.first_phone

如果$ one_or_two = 2

此条件应正常运行s.sec_phone = t.sec_phone

Answer 1

您的状况

when one_or_two = $one_or_two and one_or_two = 1

等价于

 one_or_two = 1

因此您的代码应减少为

  SELECT count(t.id) as sayi 
  from student as t 
  inner join student s on s.id!= t.id                                  
  where case 
            when one_or_two = 1 
              then 
                  s.first_phone = t.first_phone 
            when one_or_two = 2 then 
                  s.sec_phone = t.sec_phone
            else 1=1
            end

      and t.id=4327

或

  SELECT count(t.id) as sayi 
  from student as t 
  inner join student s on s.id!= t.id                                  
  where case 
            when one_or_two = 1 AND   s.first_phone = t.first_phone  THEN 1 else 0 
            when one_or_two = 2 AND  s.sec_phone = t.sec_phone then 1 else 0
            else 0
            end
  and t.id=4327

在Mysql所在的地方用例

1 个答案: