php - 在 PHP 中下拉

Question

我正在尝试从具有特定值的表中拉取一个链接，并从同一个表中拉取相同的链接但值不同。但是两个下拉菜单都出现在链接中。

<div class="col-sm-9">
        <select    style="width:350px;" name="featuredname" >
        <?php
        $selectunfeatured=mysql_query("select * from `properties` where `featured`='0' and `block`='0' and `active`='1' and `type`='1'");

        while($selectunfeaturedarray=mysql_fetch_array($selectunfeatured))
        {
        echo '<option value='.$selectunfeaturedarray['property_id'].'>'.$selectunfeaturedarray['property_title'].'</option>';
        }
        ?>

型号:

                <select    style="width:350px;" name="featuredname" >
        <?php
        $selectunfeatured=mysql_query("select * from `properties` where `featured`='0' and `block`='0' and `active`='1' and `type`='2'");

        while($selectunfeaturedarray=mysql_fetch_array($selectunfeatured))
        {
        echo '<option value='.$selectunfeaturedarray['property_id'].'>'.$selectunfeaturedarray['property_title'].'</option>';
        }
        ?>

        </select>   

Answer 1

当您在 <select> 中执行相同的操作时elements 我建议你创建一个函数来提取你需要的所有数据，方法是传递 type值作为函数参数:

<?php
function getData($type){
    $db=mysql_connect('database_host','database_user','database_pass');
    mysql_select_db('database_name',$db);    
    $selectunfeatured = mysql_query("select * from `properties` where `featured`='0' and `block`='0' and `active`='1' and `type`='$type'");    
    while ($selectunfeaturedarray = mysql_fetch_assoc($selectunfeatured)) {
        $data[] = $selectunfeaturedarray;
    }    
    return $data;
}
?>

并生成<option>元素的做如下:

<div class="col-sm-9">
    <select style="width:350px;" name="featuredname[male]">
        <?php foreach(getData(1) as $male):?>
            <option value="<?=$male['property_id'];?>"><?=$male['property_title'];?></option>        
        <?php endforeach;?>
    </select> 

    <select style="width:350px;" name="featuredname[female]">
        <?php foreach(getData(2) as $female):?>
            <option value="<?=$female['property_id'];?>"><?=$female['property_title'];?></option>        
        <?php endforeach;?>
    </select>   
</div>

注意:在您的示例中，select元素具有相同的 name属性。在这种情况下，仅选择第二个值 select将被保存。如果你想得到它们，那么你必须把方括号 [] 在每个name的结尾属性值 name="featuredname[]"或 name="featuredname[male]" && name="featuredname[female]"