Question

我有以下表格：

表名::学生

studentid     studentname
1001          Charlie Sheen
1002          John Cryer

表名:: studentpayment1

paymentid  studentid    fee_month  fee_year  totalamount
  1234          1001     February    2012        $500
  4321          1002     January     2012       $1500

表名:: studentpayment2

 id          pid        fee_type        fee_amount
 1          1234        Monthly Fee           $500
 2          4321        Exam Fee             $1500

现在，我正在试图找出未支付2012年2月份“月费”的学生的姓名。

我尝试了以下代码，但它没有生成正确的信息。

$year="2012"; $month="February"; $fee_type="Monthly Fee";


SELECT DISTINCT studentid,
                studentname
FROM   student
WHERE  NOT EXISTS (SELECT   *
                   FROM     studentpayment1
                     JOIN   studentpayment2
                     ON     studentpayment1.paymentid = studentpayment2.pid
                   WHERE    fee_month = '$month'
                     AND    fee_type = '$fee_type'
                     AND    fee_year = '$year'
                     AND    student.studentid = studentpayment1.studentid)
LIMIT  $perPage

如何修复此查询？

Answer 1

这将为您提供所有在2012年2月没有支付任何费用，2012年2月没有任何费用或2012年2月没有月费的学生

SELECT * FROM student s 
   LEFT JOIN studentpayment1 sp1 
      ON s.studentid = sp1.studentid 
        AND ((sp1.fee_month = 'February' AND sp1.fee_year = '2012') OR sp1.fee_month is null)
   LEFT JOIN studentpayment2 sp2 
      ON sp1.paymentid = sp2.pid AND ( sp2.fee_type  = 'Monthly Fee' OR  sp2.fee_type is null)
   WHERE sp1.fee_month is null or sp2.fee_type is null

祝你好运

BTW这里有一些很好的架构建议：将选项卡上的所有主键更改为'id'并与你命名foriegn键一致：学生表的primay键应该只是studentpayment1中的'id'表应该有一个名为student_id的列。 studentpayment2表应该有一个指向studentpayment1的列，名为studentpayment1_id而不是'pid'。

Answer 2

您的变量不应该在单引号内：

"WHERE    fee_month = '".$month."' "

更好的方法是绑定变量。

关于绑定的讨论：How to bind SQL variables in Php?

Answer 3

我没有尝试过实时运行此查询，但这应该会帮助你。

select distinct studentid from student where studentid NOT EXISTS (select studentid from studentpayment1 where  fee_month = '$month'
                     AND    fee_type = '$fee_type'
                     AND    fee_year = '$year')

或

select distinct studentid from student where NOT EXISTS (select studentid from studentpayment1 where  fee_month = '$month'
                     AND    fee_type = '$fee_type'
                     AND    fee_year = '$year')

Answer 4

如果你的子查询是正确的（我没看过它），下面的查询应该可以正常工作

SELECT DISTINCT studentid,studentname FROM student WHERE
studentid NOT IN (SELECT studentid FROM studentpayment1 JOIN studentpayment2 ON studentpayment1.paymentid=studentpayment2.pid WHERE 
fee_month='$month' AND fee_type='$fee_type' AND fee_year='$year' AND student.studentid=studentpayment1.studentid ) LIMIT $perPage

Answer 5

看起来这会做的事情：

$year="2012"; $month="February"; $fee_type="Monthly Fee";

$students_with_debt = "SELECT name FROM student WHERE studentid NOT IN (SELECT studentid FROM studentpayment1 p1 INNER JOIN studentpayment2 p2 ON p1.paymentid = p2.pid WHERE fee_month = '$month' AND p1.fee_year = '$year' AND p2.fee_type = '$fee_type');"

Answer 6

请尝试使用LEFT JOIN：

SELECT student.studentid, student.studentname FROM student
LEFT JOIN studentpayment1 ON studentpayment1.studentid = student.studentid
  AND studentpayment1.fee_month = ?
  AND studentpayment1.fee_year  = ?
LEFT JOIN studentpayment2 ON studentpayment2.pid = studentpayment1.paymentid
  AND studentpayment2.fee_type = ?
WHERE studentpayment2.id IS NULL

确保将索引放在fee_month，fee_year和fee_type上，并替换所有？用你的变量。

Mysql选择不存在的地方

6 个答案: