【发布时间】:2016-02-04 04:36:44
【问题描述】:
我只是脚本编码的业余爱好者。我需要这里有人的帮助.. 我在这里遇到问题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,尝试在任何情况下显示它们,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";
}
【问题讨论】:
标签: php mysql sql arrays merge