选择SQL查询以从多个表中获取数据并在一个表中查看

Question

出现的问题是，在雇主输入结帐时间数据之前自动结帐时间。

签入时间输入后，结帐栏将为空，但它会自动生成结帐时间，并在接下来的几天中提取前几天的数据并重复该时间。我只是怀疑，调用时会出现一些错误一个查询中来自3个表的数据：* tbl_student *时间* check_out

 Table 1 ( time ) id id_number1 date1 check_in ----------------------------------------------------- 30 10001 2018-02-04 22:07:59 31 10002 2018-02-04 22:08:09 32 10001 2018-02-05 08:21:10 Table 2 ( check_out) id id_number2 date2 check_out ----------------------------------------------------- 10 10001 2018-02-04 22:09:09 11 10002 2018-02-04 22:09:21 12 10001 2018-02-05 09:21:22 Table 3 ( tbl_student ) id id_number Password full_name ----------------------------------------------------- 2 10001 10001 Alma 3 10002 10002 Daniel 

淘汰

 <?php include ("connection.php"); $id = $_GET['employee-ID']; $query = "SELECT tbl_student.full_name,tbl_student.Id_Number, time.Id_Number1, time.date1 ,time.check_in, check_out.check_out FROM tbl_student, time, check_out WHERE check_out.Id_Number2 ='$id' AND time.Id_Number1 = '$id' AND tbl_student.Id_Number = '$id'"; $result = mysqli_query ($link, $query) or die("select Error ".mysqli_error($link)); $i = 1; while ($row = mysqli_fetch_array($result)) { ?> <tbody> <tr> <td><?php echo $i; ?></td> <td><?php echo $row['full_name']; ?></td> <td><?php echo $row['Id_Number']; ?></td> <td><?php echo $row['check_in']; ?></td> <td><?php echo $row['check_out']; ?></td> </tr> <?php $i++; } mysqli_close ($link); ?> <p> </p> 

错误

 <?php include ("connection.php"); $id = $_GET['employee-ID']; $query = "SELECT tbl_student.full_name, tbl_student.Id_Number, time.Id_Number1, time.date1, time.check_in,check_out.check_out, time.time_rn, check_out.check_out_rn FROM tbl_student ,(SELECT time.*, row_number() OVER ( partition by time.id_number1 order by time.date1) time_rn FROM time WHERE 1=1 ) time, (SELECT check_out.*, row_number() OVER (partition by check_out.id_number2 order by check_out.date2) check_out_rn from check_out where 1=1) check_out WHERE check_out.Id_Number2='$id' AND time.Id_Number1 ='$id' AND tbl_student.Id_Number = '$id' AND time_rn = check_out_rn (+)"; $result = mysqli_query ($link, $query) or die("select Error ".mysqli_error($link)); while ($row = mysqli_fetch_array($result)) { ?> 

Answer 1

好的，这就是问题。 如果运行查询，您会发现它只是在time和check_out表之间进行交叉 check_out或笛卡尔 check_out 。 您几乎从不想要这个。 它显示了check_in和check_out时间的所有可能组合，因为您没有告诉它check_in和check_out时间应该如何关联。

select full_name, id_number, date1, check_in, check_out
from tbl_student, time, check_out
where id_number = 10001 and id_number1 = 10001 and id_number2 = 10001;

FULL_NAME  ID_NUMBER DATE1     CHECK_IN CHECK_OUT
--------- ---------- --------- -------- ---------
Alma           10001 04-FEB-18 22:07:59 22:09:09 
Alma           10001 04-FEB-18 22:07:59 09:21:22 
Alma           10001 05-FEB-18 08:21:10 22:09:09 
Alma           10001 05-FEB-18 08:21:10 09:21:22 

在这种情况下，我假设您希望它显示与该用户的check_in时间顺序相匹配的check_out时间 。 因此，用户的第三次签到应与他们的第三次签出相匹配。 为此，我将向每个表添加“由id_number分区的行号”，以便我们将它们匹配。

SELECT tbl_student.full_name, tbl_student.Id_Number, 
    time.Id_Number1, time.date1, time.check_in,
    check_out.check_out,
        time.time_rn, -- for testing
        check_out.check_out_rn -- for testing
  FROM tbl_student, 
    (select time.*,
        row_number() over (partition by time.id_number1 order by time.date1) time_rn
        from time where 1=1) time, 
    (select check_out.*,
        row_number() over (partition by check_out.id_number2 order by check_out.date2) check_out_rn
        from check_out where 1=1) check_out 
  WHERE check_out.Id_Number2=10001 AND time.Id_Number1 =10001 AND tbl_student.Id_Number = 10001
    and time_rn = check_out_rn (+)
  ;

输出：

FULL_NAME  ID_NUMBER ID_NUMBER1 DATE1     CHECK_IN CHECK_OUT    TIME_RN CHECK_OUT_RN
--------- ---------- ---------- --------- -------- --------- ---------- ------------
Alma           10001      10001 04-FEB-18 22:07:59 22:09:09           1            1
Alma           10001      10001 05-FEB-18 08:21:10 09:21:22           2            2

---

顺便说一下，这有点尴尬。 如果可以更改架构，建议您删除check_out表，然后执行以下操作：

create table time (id number, 
                   id_number1 number, 
                   date1 date, 
                   check_in varchar2(8), 
                   check_out varchar2(8));

这样一来，您就可以知道何时尚未有人结帐，因为time.check_out为null。

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编辑：看来您使用的是MySQL 8.0之前的版本，该版本不支持窗口函数。 试试这个吧。 在某些方面，它是一个更简单的查询。

SELECT tbl_student.full_name, tbl_student.Id_Number, 
time.Id_Number1, time.date1, time.check_in,
(select min(check_out) 
  from check_out
  where Id_Number2=time.Id_Number1
    and check_out.date2 >= time.date1
    and check_out.check_out >= time.check_in) check_out
FROM tbl_student, time
WHERE time.Id_Number1 =10001 AND tbl_student.Id_Number = 10001;