MySql Select数据在PHP中是多对多关系

Question

• 条目（表）

   ent_seq entry 100010 1 100020 2 100030 3 100040 4 

• K_ELE（表格）

   KID entry keb 1 1 食べる2 2 日本3 3 先生 

• R_ELE（表）

   RID entry reb 1 1 たべる2 2 にほん3 2 にっぽん4 3 せんせい5 4 ビール 

• 感（表）

   SID entry 1 1 2 2 3 3 4 4 

• GLOSS_ENG（SENSE表的子表）

   GID SID mean 1 1 eat 2 2 Japan 3 2 Japanese 4 3 theacher 5 4 beer 

• UPDATE（SENSE表的子表）

    UPID SID up_date 1 1 2012.12.12 17:31 2 2 2012.12.12 17:31 3 2 2012.12.12 17:31 4 3 2012.12.12 17:31 5 4 2012.12.12 17:31 

我喜欢显示这张桌子。

   ent_seq  keb     reb            mean             update

   100010   食べる   たべる           eat             2012.12.12 17:31
   100020   日本    にほん; にっぽん   Japan; Japanese 2012.12.12 17:31
   100030   先生    せんせい          theacher        2012.12.12 17:31
   100040           ビール   beer                    2012.12.12 17:31

这个数据库很大195,000行请帮助我！

我的简单代码KKK

 <?php include('con_db.php');
$query = mysql_query("SELECT * FROM entry");
$number=mysql_num_rows($query);
for ($count=1; $count<= 20 ; $count++)
{ $entry = RAND(1,$number);

$query1 = "SELECT   
        entry.ent_seq, k_ele.keb, r_ele.reb, s_gloss_eng.gloss
        FROM entry, k_ele, r_ele, sense, s_gloss_eng
        WHERE entry.entry_pkey = $entry
        AND   sense.sense_pkey = s_gloss_eng.sense_pkey
        AND   entry.entry_pkey = sense.entry_pkey
        AND   entry.entry_pkey = r_ele.entry_pkey
        AND   entry.entry_pkey = k_ele.entry_pkey
        ORDER BY entry.ent_seq
        ";
$view2 = mysql_query($query1);
$row = mysql_fetch_assoc($view2);

    $ent_seq=$row['ent_seq'];
    $keb=$row['keb'];
    $reb=$row['reb'];
    $mean_eng=$row['mean'];
    $update=$row['update'];
?>

      <table width="614" border="1">
   <tr>
     <td>ent_seq</td>
     <td>keb</td>
     <td>reb </td>
     <td>mean</td>
     <td>update</td>
  </tr>
  <tr>
     <td>$ent_seq</td>
     <td>$keb</td>
     <td>$reb</td>
     <td>$mean_eng</td>
     <td>$update</td>
  </tr>
  </table>
 <?php
 }
 ?>

Answer 1

试试这个查询-

SELECT
  e.ent_seq,
  e.entry,
  ke.keb,
  re.reb,
  s.mean,
  s.up_date
FROM entry e
LEFT JOIN (SELECT entry, GROUP_CONCAT(keb) keb FROM k_ele GROUP BY entry) ke
  ON ke.entry = e.entry
LEFT JOIN (SELECT entry, GROUP_CONCAT(reb) reb FROM r_ele GROUP BY entry) re
  ON re.entry = e.entry
LEFT JOIN (
  SELECT s.entry, mean, up_date FROM sense s
    LEFT JOIN (SELECT sid, GROUP_CONCAT(mean) mean FROM gloss_eng
               GROUP BY sid
              ) ge
      ON ge.sid = s.sid
    LEFT JOIN (SELECT sid, GROUP_CONCAT(DISTINCT up_date) up_date FROM `update`
               GROUP BY sid
              ) u
      ON u.sid = s.sid
  GROUP BY s.sid
  ) s
  ON s.entry = e.entry

Answer 2

SELECT e.ent_seq, 
    GROUP_CONCAT(DISTINCT kei.keb SEPARATOR'; ')AS kebs, 
    GROUP_CONCAT(DISTINCT rei.reb SEPARATOR '; ') AS rebs,
    GROUP_CONCAT(DISTINCT sei.mean SEPARATOR '; ') AS means

FROM entry e

LEFT JOIN k_ele kei ON e.entry = kei.entry
LEFT JOIN r_ele rei ON e.entry = rei.entry
LEFT JOIN sense se ON e.entry = se.entry 
LEFT JOIN s_gloss_eng sei ON se.SID = sei.SID

WHERE e.ent_seq IN (SELECT jt5.jlpt5_seq FROM jlptn5 jt5
                   WHERE jt5.jlpt5_null = 0
                   ORDER BY RAND())
GROUP BY e.entry    
ORDER BY RAND()
LIMIT 20