Question

我正在做一个搜索引擎。在该模块中，我必须在菜单栏中选择城市。我从城市名称中取出城市名称到菜单栏。如果我选择一个特定的城市，我必须在该菜单中显示相同的内容。怎么做？帮助我...

<?php 
    $sql="SELECT DISTINCT CITY_ID,CITY_TITLE FROM CITY";
    $res=mysql_query($sql);
?>

<header>
    <nav class="headerTop">
        <div class="container">
            <div class="row">
                <div class="headerTopHolder">
                    <div class="dropdown locationDropdown">
                        <a id="bangaloreclick" class="btn dropdown-toggle" data-toggle="dropdown" aria-haspopup="true" aria-expanded="true">
                            Bangalore
                            <span class="glyphicon glyphicon-menu-down"></span>
                        </a>
                        <ul class="dropdown-menu" aria-labelledby="dropdownMenu1" id="insideareas">
                        <?php while($cityrow=mysql_fetch_array($res)) { ?>
                            <li>
                                <a href="#"><?php echo $cityrow['CITY_TITLE'];?></a>
                            </li>
                        <?php } ?>

                        </ul>
                    </div>

Answer 1

你是说这个？

$('#insideareas a').on("click",function(){
  alert($(this).html())

});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<ul class="dropdown-menu" aria-labelledby="dropdownMenu1" id="insideareas">
   <li><a href="#">City Name 1</a></li>
   <li><a href="#">City Name 2</a></li>
   <li><a href="#">City Name 3</a></li>
   <li><a href="#">City Name 4</a></li>
</ul>

如何使用PHP在菜单中显示所选城市

1 个答案: