php - 将 2 个查询中的 2 个条件数组合并到 1 个查询中

Question

我只是脚本编码的业余爱好者。我需要这里有人的帮助.. 我在这里遇到了麻烦，我试图从我的代码中显示 2 个条件合并，请看这里...

<ul class="nav navbar-nav">
   <?php               
   $main=mysql_query("SELECT * FROM mainmenu WHERE aktif='Y'");
   while($r=mysql_fetch_array($main)){
   $t=$r[''];
   $tm="<a href='$r[link]' class='dropdown-toggle' data-toggle='dropdown' role='button' aria-expanded='false'>$r[nama_menu]<span class='caret'></span></a>";
   $th="<a href='$r[link]'>$r[nama_menu]</a>";
   if ($t!= ""){
   $tombol=$th;
   }else{
   $tombol=$tm;
   } 
   echo "<li class='dropdown'>$tombol
         <ul class='dropdown-menu' role='menu'>";
    	 $sub=mysql_query("SELECT * FROM submenu, mainmenu  
         WHERE submenu.id_main=mainmenu.id_main
         AND submenu.id_main=$r[id_main]");
    	 while($w=mysql_fetch_array($sub)){
   echo " <li><a href='$w[link_sub]'>$w[nama_sub]</a></li>";
    	 }
   echo "</ul></li>";}
   ?>
</ul>

我已将其拆分为，尝试在任何条件下显示它们，这更相关，但我无法合并它，我不知道我要做什么，请看这个

这是第一个条件-->

<ul class="nav navbar-nav">
<?php               
    $main=mysql_query("SELECT  DISTINCT    a.* FROM mainmenu a
    INNER JOIN submenu b ON a.id_main = b.id_main AND a.aktif = 'Y'");
    while($r=mysql_fetch_array($main)){
    echo "<li class='dropdown'><a href='$r[link]' class='dropdown-toggle' data-toggle='dropdown' role='button' aria-expanded='false'>$r[nama_menu]<span class='caret'></span></a><ul class='dropdown-menu' role='menu'>";
    $sub=mysql_query("SELECT * FROM submenu, mainmenu  
    WHERE submenu.id_main=mainmenu.id_main 
    AND submenu.id_main=$r[id_main]");
    while($w=mysql_fetch_array($sub)){
    echo " <li><a href='$w[link_sub]'>$w[nama_sub]</a></li>";
    }
   echo "</ul></li>";} 
?>
</ul>

这是第二个条件 -->

<ul class="nav navbar-nav">
<?php      
  $menu=mysql_query("SELECT  DISTINCT    a.* FROM mainmenu a
  LEFT OUTER JOIN submenu b ON a.id_main = b.id_main WHERE b.id_main is null AND a.aktif = 'Y'");
  while($s=mysql_fetch_array($menu)){
  echo "<li class='dropdown'><a href='$s[link]'>$s[nama_menu]</a></li>";} ?>
</ul>

将 INNER JOIN 与 LEFT OUTER JOIN 合并时我想要的是:

if (bla, bla, bla){
echo "Show INNER JOIN";
}else{
echo "Show LEFT OUTER JOIN";
} 

Answer 1

通常，当两个查询使用相同的表时，使用 OR 连接两组条件很简单。但在本例中，一个查询具有 JOIN，另一个查询具有 LEFT OUTER JOIN。

我认为你可以像这样使用UNION:

SELECT  DISTINCT    a.* 
FROM mainmenu a
INNER JOIN submenu b ON a.id_main = b.id_main AND a.aktif = 'Y'
UNION 
SELECT  DISTINCT    a.* 
FROM mainmenu a
LEFT OUTER JOIN submenu b ON a.id_main = b.id_main 
WHERE b.id_main is null AND a.aktif = 'Y'

您也许可以将其重写为:

SELECT  DISTINCT    a.* 
FROM mainmenu a
LEFT OUTER JOIN submenu b ON a.id_main = b.id_main OR b.id_main is null
WHERE a.aktif = 'Y'

没有样本数据和结果，我无法为您测试和确认。