单击提交按钮时重新加载页面后如何滚动到特定 div?

lan*_*lot 5 javascript php ajax jquery

我正在尝试做一些奇特的事情。我已经成功编写了代码,以便在填写小表单并单击提交按钮时显示数据库中的结果。结果显示在表单的正下方。但我觉得如果页面能自动向下滚动到包含已填写表单结果的 div,那就太好了。我想我必须使用 jquery 或 ajax 来实现这一点。由于我对它们一无所知,因此我在互联网上搜索复制粘贴代码。但它们都不起作用。

单击提交按钮时,页面将重新加载以从数据库中获取结果。我已经获得了当页面从网络重新加载时向下滚动到 div 的代码,但问题是......即使您提交按钮,滚动也会发生没有点击。因此,有人可以给出仅在单击提交按钮时以及重新加载页面后向下滚动到 div 的代码吗?

代码是这样的

<form name="searchdonor" action="" id="form" method="POST" align="center" enctype="application/x-www-form-urlencoded">  
--- ------ ------  
--- ------ ------  
--- ------ ------

</form>  
<input type="submit" name="submit" id="submit" value="Submit"  onclick="return(searchval()); "> 
//the div i want to scroll down is below one 
<div class="col-sm-8" id="what">
</div>  
Run Code Online (Sandbox Code Playgroud)

这是 html 和 php-mysqli 中的完整代码

<div class="col-sm-4">
         <h1 class="register-title">Search a Donor</h1>
              <div id="wrapper">
              <div id="chatbox">
      <form name="searchdonor" action="" id="form" method="POST" align="center" enctype="application/x-www-form-urlencoded">
        <table id="results">
                <tr><td><h4>Country:</h4></td><td>&nbsp&nbsp&nbsp&nbsp&nbsp
                 <select id="slct1" name="country" onchange="populate1(this.id,'slct2')">
                  <option value="">--Select--</option>
                  <option value="India">India</option>
                 </select></td></tr>
                <tr><td><h4>State:</h4></td><td>&nbsp&nbsp&nbsp&nbsp&nbsp
                 <select id="slct2" name="state" onchange="populate2(this.id,'slct3')">
                  <option value="">--Select--</option>
                 </select></td></tr>
                <tr><td><h4>District:</h4></td><td>&nbsp&nbsp&nbsp&nbsp&nbsp
                 <select id="slct3" name="district" onchange="populate3(this.id,'slct4')">
                  <option value="">--Select--</option>
                 </select></td></tr>
                <tr><td><h4>City:</h4></td><td>&nbsp&nbsp&nbsp&nbsp&nbsp
                 <select id="slct4" name="city">
                  <option value="">--Select--</option>
                 </select></td></tr>
                <tr><td><h4>Blood group:</h4></td><td>&nbsp&nbsp&nbsp&nbsp&nbsp
                 <select name="bloodgroup">
        <option value="">--Select--</option>
        <option value="A+">A+</option>
        <option value="A-">A-</option>
        <option value="B+">B+</option>
        <option value="B-">B-</option>
        <option value="O+">O+</option>
        <option value="O-">O-</option>
        <option value="AB+">AB+</option>
        <option value="AB-">AB-</option>
                </select></td></tr>

      </form>
      </table><br />

    <input type="submit" name="submit" id="submit" value="Submit"  onclick="return(searchval());">
    </div>


              </div>
         </div>

         <div class="col-sm-2">
         </div>
     </div>
     <div class="dropdownwrap">
     <?php 
     if(isset($_POST['submit'])){

              $country=$_POST['country'];
              $state=$_POST['state'];
              $district=$_POST['district'];
              $city=$_POST['city'];
              $bloodgroup=$_POST['bloodgroup'];
              ?>
     <div class="row"><br />
        <div class="col-sm-2">
        </div>
        <div class="col-sm-8" id="what">
        <a class="what"></a>
        <?php echo "<h4 align='center'> Donors in <b>".$city."</b> for <b>".$bloodgroup."</b> are</h4>";?>
        </div>
        <div class="col-sm-2">
        </div>
     </div>
       <div class="row">
        <div class="col-sm-2">
        </div>
        <div class="col-sm-8">
          <div class="table-responsive">

           <table id="tablepaging" class="table" align="center">
            <thead><hr />
             <tr>
                <th><b>Full Name</b></th>
                <th><b>Contact Number</b></th>
             </tr>
            </thead>
            <tbody>
              <?php 


              $connect = mysqli_connect('localhost', 'root', '', 'blood'); if (mysqli_connect_errno()) echo "Failed to connect to MySQL: " . mysqli_connect_error();
              $query="SELECT * FROM users WHERE country='$country' && state='$state' && district='$district' && city='$city' && bloodgroup='$bloodgroup' && activity='available'";
              $result=mysqli_query($connect,$query) or die('Error, query failed');
              mysqli_close($connect);
              if (mysqli_num_rows($result) == 0) {
              echo"<h3 align=\"center\">Sorry, No Donors Found</h3>";
              }
              elseif($result)
              {
              while ($row = mysqli_fetch_array($result)){

              echo"<tr>";
              echo"<td>".$row["firstname"]."&nbsp".$row["lastname"]."</td>";
              echo"<td>".$row["phonenumber"]."<br />".$row["secondnumber"]."</td>";
              }
              echo"</tr>";

              }}
              ?>
            </tbody>
           </table>
            <div id="pageNavPosition" align="center"></div>
          </div>
         </div>



         <div class="col-sm-2">
         </div>


</div>       

</div>

</div>
Run Code Online (Sandbox Code Playgroud)

PiT*_*NjA 5

您可以在 URL 中使用锚点。

或者,如果你想要流畅的动画,只需在if(isset($_POST['submit']))条件中插入 JS 代码即可。这样,只有在单击提交按钮并且页面重新加载时才会发生滚动。

我还建议您使用$(function() { /**/ });jQuery 语法,以便仅在加载 DOM 时才会发生滚动。

<?php

if(isset($_POST['submit']))
{

    //[...]
    //sql query and display
    //[...]

    ?>
    <script>
        $(function() {
            $('html, body').animate({
                scrollTop: $("#myDiv").offset().top
            }, 2000);
         });
    </script>
    <?php

}

?>
Run Code Online (Sandbox Code Playgroud)

滚动到特定 div 的代码可在此处找到: 平滑滚动到 div id jQuery