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我有一个结果集，需要按原样显示在屏幕上。但问题是在使用 echo 命令显示结果集的每一行时，顺序正在发生变化。谁能说出为什么会发生这种情况并为我提供一种方法克服它。这是我的实际和打印输出。

实际结果集： 

JAIKE-ILENE-WACKI-MAZIE-REGLE-SBJ-KMMU  
LVZ-HARTY-MUGZY-STW 
MAZIE-SIXIE-SBJ-KMMU  
PXT-LOUIE-GATBY-RAZER-BUZIE-JAIKE-ILENE-WACKI-MAZIE  
SWANN-GATBY-RAZER-BUZIE-JAIKE-ILENE-WACKI-MAZIE

输出： 

 JAIKE-SBJ-ILENE-KMMU-WACKI-MAZIE-REGLE
 MUGZY-STW-LVZ-HARTY
 SBJ-KMMU-MAZIE-SIXIE
 ILENE-GATBY-WACKI-RAZER-MAZIE-BUZIE-PXT-JAIKE-LOUIE
 WACKI-RAZER-MAZIE-BUZIE-JAIKE-SWANN-ILENE-GATBY

这是我的代码 

$sql3="SELECT GROUP_CONCAT(l.fix_ident SEPARATOR '-') AS fix_seq,l.airport_ident,x.star_ident,x.transition_ident,
                 x.fix_ident from corept.std_star_leg l
                 JOIN
                    (SELECT DISTINCT c.airport_ident,c.star_ident,c.transition_ident,c.fix_ident
                     FROM corept.std_star_leg c
                     INNER JOIN
                          (SELECT star_ident,transition_ident,max(sequence_num) seq,route_type
                           FROM corept.std_star_leg
                           WHERE data_supplier='J'
                           AND airport_ident='KMMU'
                           GROUP BY star_ident,
                           transition_ident)b ON c.sequence_num=b.seq
                           AND c.star_ident=b.star_ident AND c.transition_ident=b.transition_ident
                           LEFT JOIN
                               (SELECT name,trans
                                FROM skyplan_deploy.deploy_stars
                                WHERE apt='KMMU'
                                AND name!=trans) d
                                ON d.name=c.star_ident
                                AND d.trans=c.fix_ident
                                WHERE c.data_supplier='J'
                                AND c.airport_ident='KMMU' AND d.name is null)x
                                where l.airport_ident='KMMU' and l.transition_ident=x.transition_ident
                                and l.star_ident=x.star_ident and l.data_supplier='J'
                                group by x.star_ident,x.transition_ident
                                order by l.star_ident,x.transition_ident,l.sequence_num";
  $res3=mysqli_query($mysqli,$sql3);  
if($res3)
  {
    while($newArray3=mysqli_fetch_array($res3,MYSQLI_ASSOC))
    {
    $apt=$newArray3['airport_ident'];
    $star_ident=$newArray3['star_ident'];
    $trans_ident=$newArray3['transition_ident'];
    $fix_ident=$newArray3['fix_ident'];
    $fix_seq=$newArray3['fix_seq'];
    echo $apt.",".$star_ident.",".$trans_ident.",".$fix_ident.",COREPT,".$fix_seq;
    echo "<br>";
    }
  }
 else
 {
   printf("ERROR:%s\n",mysqli_error($mysqli));
 }
4

1 回答 1

0

您的查询看起来过于复杂。它似乎是通过(grouped on and )找到分组的最大std_star_leg记录，不包括那些已经有匹配的 non-self-referencing的记录，然后返回将所有匹配值连接成字符串的结果再次分组？sequence_numstart_identtransition_identdeploy_starfix_ident

如果是这样，以下大大简化的查询应该会达到相同的结果：

SELECT   GROUP_CONCAT(fix_ident SEPARATOR '-') AS fix_seq,
         airport_ident,
         star_ident,
         transition_ident
FROM     corept.std_star_leg l NATURAL JOIN (
           SELECT   star_ident, transition_ident,
                    data_supplier, airport_ident,
                    MAX(sequence_num) sequence_num
           FROM     corept.std_star_leg
           WHERE    data_supplier = 'J'
                AND airport_ident = 'KMMU'
           GROUP BY star_ident, transition_ident
         ) b
WHERE    NOT EXISTS (
           SELECT NULL
           FROM   skyplan_deploy.deploy_stars d
           WHERE  d.name != d.trans
              AND d.name  = l.star_ident
              AND d.trans = l.fix_ident
              AND d.apt   = l.airport_ident
         )
GROUP BY star_ident, transition_ident

请注意，虽然您之前x.fix_ident在最外层的选择列表中进行了选择，但我省略了该列，因为它的值将由服务器从fix_seq.

现在，关于您的问题（这似乎fix_ident与值在GROUP_CONCAT()字符串中出现的顺序有关fix_seq——尽管从您的问题中很难理解这一点），也许您想将ORDER BY参数用于GROUP_CONCAT()函数？例如：

SELECT GROUP_CONCAT(fix_ident SEPARATOR '-' ORDER BY ...) AS fix_seq

但是，我不清楚您需要什么顺序（ORDER BY原始查询中的唯一子句是完全多余的）。

于 2013-05-15T10:04:29.610 回答