使用php和mysql的菜单问题

Question

我有一个表，其中有两列，分别名为rugby_union和matchleague。我已经创建了2个查询来显示截然不同的rugby_union和matchleague。当我在菜单上有下拉菜单，它必须显示比赛联盟从rugby_union只。下面是我使用的代码：

$sql="select distinct(union_league) from teams where league_complete ='No' ";
$query = mysql_query($sql);
while($row = mysql_fetch_array($query)){ 
$matchleaguecomp[] = $row; 
}

$sql1="select distinct(rugby_union) AS rugby_union from teams where league_complete ='No' ";
$query1 = mysql_query($sql1);
while($row1 = mysql_fetch_array($query1)){ 
$union_list[] = $row1; 
}

导航栏中的PHP代码：

<?php                       
    for($i=0;$i<count ($union_list);$i++){  
?>
<li>
    <a href="#"><?php echo $union_list[$i][0]; ?></a>
    <ul class="dropdown-menu" 
         style="position: absolute; left: 160px; top: 0px; display: none;
                height: <?php if(count($matchleaguecomp)>0) {
                                  echo '260px;'; 
                              } else { 
                                  echo (count($matchleaguecomp)*26).'px'; 
                              } ?>  
                overflow: auto;">
        <?php
            for($i=0;$i<count($matchleaguecomp);$i++){                      
                //$fixture_name=$url->encode("name=".$matchleaguecomp[$i][0]);
            $fixture_name="name=".$matchleaguecomp[$i][0];
        ?>
        <li>
            <a href="<?php echo $include_path.'new_logs.php?'.$fixture_name; ?>">
                <?php echo $matchleaguecomp[$i][0]; ?>
            </a>
        </li>
        <?php  } ?>
    </ul>
</li>
<?php  } ?> 

提前感谢，我不明白为什么它不显示联盟的工会列表

Answer 1

首先，注意@tadman关于mysql_query已经过时的说法……这是你固定的代码：

$sql="select distinct(union_league) as `union_league` from teams where league_complete ='No' ";
$query = mysql_query($sql);
while($row = mysql_fetch_array($query)){ 
$matchleaguecomp[] = $row['union_league']; // you need to get the returned value from the array
}

$sql1="select distinct(rugby_union) AS rugby_union from teams where league_complete ='No' ";
$query1 = mysql_query($sql1);
while($row1 = mysql_fetch_array($query1)){ 
$union_list[] = $row1['rugby_union']; // you need to get the returned value from the array
}