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多列外键引用

2014-11-23 62 views
1

我有这样的：多列外键引用

Table "schedules" 
---------------------------------------------------------------------- 
| sched_id | sched_name | sc_id1 | sc_id2 | sc_id3 | sc_id4 | sc_id5 | 
---------------------------------------------------------------------- 
| 1  | block 1 | 1 | 2 | 3 | 4 | 5 | 
---------------------------------------------------------------------- 
| 2  | block 2 | 1 | 2 | 3 | NULL | NULL | 
---------------------------------------------------------------------- 

Table "subject_current" 
--------------------------------------- 
|sc_id | sl_id | schoolyear | semister| 
--------------------------------------- 
| 1 | 5 | 2014-2015 | 1st | 
--------------------------------------- 
| 2 | 6 | 2014-2015 | 1st | 
--------------------------------------- 
| 3 | 7 | 2014-2015 | 1st | 
--------------------------------------- 
| 4 | 8 | 2014-2015 | 1st | 
--------------------------------------- 
| 5 | 9 | 2014-2015 | 1st | 
--------------------------------------- 

Table "subject_list" 
------------------------------------------------------------- 
|sl_id | subject_code | subject_description | subject_prereq| 
------------------------------------------------------------- 
| 5 |  math1 |   algebra  |  none  | 
------------------------------------------------------------- 
| 6 |  math2 | trigonometry  |  none  | 
------------------------------------------------------------- 
| 7 |  math3 |  Calculus  |  none  | 
------------------------------------------------------------- 
| 8 |  eng1 |  english 1  |  none  | 
------------------------------------------------------------- 
| 9 |  hum1 |  humanities  |  none  | 
-------------------------------------------------------------

列sc_id1 - sc_id5是一个外键引用到这是sc_id从表subject_current相同的密钥。并且列sl_id来自表subject_current参照sl_id来自表subject_list。

我的问题是，我怎么能检索使用仅从table时间表将数据从tablesubject_list的数据？我想实现的是这样的：

(This will echo on my page) 
-------------------------------------------------------------------------------- 
| sched_id | sched_name |      Subjects      | 
-------------------------------------------------------------------------------- 
| 1  | block 1 | algebra, tirgonometry, calculus, english1, humanities| 
-------------------------------------------------------------------------------- 
| 2  | block 2 |   algebra, tirgonometry, calculus    | 
--------------------------------------------------------------------------------

我搜索了关于LEFT JOIN但它有点难受，因为我只是在这一领域的初学者，有三个tables可能不得不JOIN。所以，如果你能提出一些建议，我很乐意欣赏。

编辑

我发现这一点，我几乎没有，但只显示表中的第一sc_idschedule_subject_currents：

<?php 

    echo "<table class='opensubtbl' cellspacing='0' cellpadding='5' ><tr>"; 

    echo "<th>Schedule ID</th>"; 
    echo "<th>Sched Name</th>"; 
    echo "<th>Subjects ID</th></tr>"; 

    //$sched = mysql_query("select * from schedules s, schedule_subject_currents h, subject_current c, subject_list l where s.sched_id = h.sched_id and h.sc_id = c.sc_id and c.sl_id = l.sl_id"); 
    //$sched = mysql_query("SELECT * FROM schedules"); 

    $sched = mysql_query(" 

     SELECT * 
     FROM schedules s 
     LEFT JOIN schedule_subject_currents ssc ON ssc.sched_id = s.sched_id 
     LEFT JOIN subject_current c ON c.sc_id = ssc.sc_id 
     LEFT JOIN subject_list l ON l.sl_id = c.sl_id 
     GROUP BY s.sched_id, sched_name 
     ORDER BY sched_name 

    "); 

    while($rows_s = mysql_fetch_assoc($sched)){ 
     $s_schedid = $rows_s['sched_id']; 
     $s_schedname = $rows_s['sched_name']; 
     $s_subdesc = $rows_s['subject_description']; 
     $s_scid = $rows_s['sc_id']; 

     echo "<tr>"; 
     echo "<td>$s_schedid</td>"; 
     echo "<td>$s_schedname</td>"; 

     echo "<td>$s_subdesc</td>"; 
     echo "</tr>"; 
    } 

    echo "</table>"; 

?>

来源

2014-11-23 Archie Zineg

+1

@F abioCardoso，感谢回复先生。但是sched_id不匹配sc_id。想象一下，你有一个有很多主题ID的时间表，并且主题ID保存来自另一个表的数据。 – 2014-11-23 18:01:10

+0

像'idn'的气味列，您需要通过创建其他表来对其进行标准化 – 2014-11-23 20:30:39

回答

2

避免存储相同类型的信息在不同的栏目中，否则你最终会遇到像你遇到的问题。相反，您应该将每条信息存储在自己的行中。

首先，通过引入另一个名为的表来标准化数据模式，例如， scedule_subject_currents，让你有

Table "schedules" 
------------------------- 
| sched_id | sched_name | 
------------------------- 
| 1  | block 1 | 
------------------------- 
| 2  | block 2 | 
-------------------------

和

Table "schedule_subject_currents" 
-------------------- 
| sched_id | sc_id | 
-------------------- 
| 1  | 1 | 
-------------------- 
| 1  | 2 | 
-------------------- 
| 1  | 3 | 
-------------------- 
| 1  | 4 | 
-------------------- 
| 1  | 5 | 
-------------------- 
| 2  | 1 | 
-------------------- 
| 2  | 2 | 
-------------------- 
| 2  | 3 | 
--------------------

然后你就可以使用查询

SELECT s.sched_id, s.sched_name, GROUP_CONCAT(subject_description) as Subjects 
FROM schedules s 
LEFT JOIN schedule_subject_currents ssc ON ssc.sched_id = s.sched_id 
LEFT JOIN subject_current c ON c.sc_id = ssc.sc_id 
LEFT JOIN subject_list l ON l.sl_id = c.sl_id 
GROUP BY s.sched_id, sched_name 
ORDER BY sched_name

来源

2014-11-23 20:18:50

+0

哇！这对我来说现在是有意义的。从今以后，这对我来说是一个很好的做法。但我对“主题”，“主题”，“sced_id”和“sced_name”感到困扰。 “sced_id”和“sced_name”可能是拼写错误。但是“主题”和/或“主题”，我不明白。你是从哪里做的？我很困惑。 – 2014-11-24 03:17:14

+0

我已经尝试过您查询并更改某些内容以使其正常工作，但它仅显示表** schedule_subject_currents **中的第一个'sc_id'。如果我删除了“GROUP BY”，则会显示所有的“sc_id”。但我希望它被分组，看起来像我的问题上的表格。 – 2014-11-24 04:08:40

+0

@ArchieZineg当你从这个答案中得到一个查询和模式为你工作时，一定要不接受另一个丑陋的非关系型答案并接受这个答案。 – philipxy 2014-11-24 09:30:00

1

尝试while嵌套查询：

<?php 

    echo "<table class='opensubtbl' cellspacing='0' cellpadding='5' ><tr>"; 

    echo "<th>Schedule ID</th>"; 
    echo "<th>Sched Name</th>"; 
    echo "<th>Subjects</th>"; 
    echo "<th></th></tr>"; 

    //$sched = mysql_query("select * from schedules s, schedule_subject_currents h, subject_current c, subject_list l where s.sched_id = h.sched_id and h.sc_id = c.sc_id and c.sl_id = l.sl_id"); 
    //$sched = mysql_query("SELECT * FROM schedules"); 
    //$sched = mysql_query("SELECT * FROM schedules s LEFT JOIN schedule_subject_currents ssc ON ssc.sched_id = s.sched_id LEFT JOIN subject_current c ON c.sc_id = ssc.sc_id LEFT JOIN subject_list l ON l.sl_id = c.sl_id ORDER BY sched_name "); 

    $sched = mysql_query("SELECT * FROM schedules"); 

    while($rows_s = mysql_fetch_assoc($sched)){ 
     $s_schedid = $rows_s['sched_id']; 
     $s_schedname = $rows_s['sched_name']; 

     echo "<tr>"; 
     echo "<td>$s_schedid</td>"; 
     echo "<td>$s_schedname</td>"; 
     echo "<td>"; 

     $schedsubcur = mysql_query("SELECT * FROM schedule_subject_currents WHERE sched_id='$s_schedid'"); 
     while($rows_ssc = mysql_fetch_assoc($schedsubcur)){ 
      $ssc_scid = $rows_ssc['sc_id']; 
      $schedsublist = mysql_query("SELECT * FROM subject_current WHERE sc_id='$ssc_scid'"); 
      while($rows_ssl = mysql_fetch_assoc($schedsublist)){ 
       $ssl_slid = $rows_ssl['sl_id']; 
       $ssublist = mysql_query("SELECT * FROM subject_list WHERE sl_id='$ssl_slid'"); 
       while($rows_ssubl = mysql_fetch_assoc($ssublist)){ 
        $ssubl_subdesc = $rows_ssubl['subject_description']; 
        echo $ssubl_subdesc."($ssc_scid), "; 
       } 
      } 
     } 

     echo "</td>"; 
     echo "</tr>"; 
    } 

    echo "</table>"; 

?>

来源

2014-11-24 04:48:45 AlexJaa

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