Question

我有一个结果集，需要显示在屏幕上。但问题是在使用echo命令显示结果集的每一行时，订单会发生变化。任何人都可以说为什么会发生这种情况并提供我是一种克服它的方法。这是我的实际打印输出。

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));
 }

Answer 1

您的查询看起来过于复杂。它出现，以std_star_leg（sequence_num和start_ident分组）查找分组最大transition_ident记录，不包括那些已经存在的记录一个匹配的非自引用deploy_star，然后将结果再次分组，并将所有匹配的fix_ident值连接成一个字符串？

如果是这样，以下大大简化的查询应该达到相同的结果：

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，虽然您之前在最外面的选择列表中选择了x.fix_ident，但我省略了这样的列，因为服务器会从fix_seq中的那些值中不确定地选择它。

现在，关于您的问题（显示与fix_ident字符串GROUP_CONCAT()中出现fix_seq值的顺序相关 - 尽管它是非常难以理解你的问题），或许你想将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样的排序（原始查询中唯一的ORDER BY子句完全是多余的。）

结果集顺序在php中更改

1 个答案: