将MySql下拉列表中的值传递给其他下拉列表,而不在Db php上发布

我想你错过了什么。下面是代码片段

<?php 

$connection = mysqli_connect(DB_HOST, DB_USER, DB_PASS,DATABASE); if (mysqli_connect_errno()) echo "Failed to connect to MySQL: " . mysqli_connect_error(); 

echo '<p>Company <span>*</span></p>
  <span class="icon-case"><i class="fa fa-comments-o"></i></span>
  <select name="company" id="company">';

    echo '<option value="">Please select</option>';

      $query="SELECT name,company_id FROM `companies` ORDER BY `name`ASC";
                $result=mysqli_query($connection,$query) or die ("Query to get data from sector table failed: ".mysql_error());



        while ($row=mysqli_fetch_array($result)) {
        $company_id=$row["company_id"];


        $company_name=$row["name"];
        echo "<option value=".$company_id.">".$company_name."   </option>";
        }



    echo '</select>

  </div>  

  <p>Contact in Company <span>*</span></p>
  <span class="icon-case"><i class="fa fa-comments-o"></i></span>
  <select name="contact" id="contact">';
    <option value=""> Please select</option>
    </select>';

?>

<script   src="https://code.jquery.com/jquery-3.3.1.min.js"   integrity="sha256-FgpCb/KJQlLNfOu91ta32o/NMZxltwRo8QtmkMRdAu8="    crossorigin="anonymous"></script>
<script>
$(document).ready(function(){
    $('#company').on('change',function(){
        var companyid = $(this).val();
        alert('hello');
        if(companyid){
            $.ajax({
                type:'POST',
                url:'ajaxData.php',
                data:'id='+companyid,
                success:function(html){
                    $('#contact').append(html);

                }
            }); 
        }
    });
});

</script>

Ajaxdata。php

<?php /* Connect to MySQL and select the database. */ 
$connection = mysqli_connect('localhost', 'root', '','demo'); if (mysqli_connect_errno()) echo "Failed to connect to MySQL: " . mysqli_connect_error();
?>
<?php 
$company_id=$_POST['id'];
echo $query="SELECT * FROM `contact` WHERE company_id='".$company_id."'";

 $result=mysqli_query($connection,$query) or die ("Query to get data from sector table failed: ".mysqli_connect_error());

 while ($row=mysqli_fetch_array($result)) {
        $company_id=$row["company_id"];
        $contact_id=$row["contact_id"];
        echo "<option value=\"$company_id\"> $contact_id</option>";
        }

我建议使用ajax来获得所需的联系人。下面是可能有用的代码片段

$(document).ready(function(){
    $('#company').on('change',function(){
        var companyid = $(this).val();
        if(companyid){
            $.ajax({
                type:'POST',
                url:'ajaxData.php',
                data:'id='+companyid,
                success:function(html){
                    $('#conatct').html(html);

                }
            }); 
        }
    });
});

公司变动时,我们可以打ajax电话。下面是Ajax的参数

1. 类型:应为POST或GET
2. Url:将根据seelct公司从数据中提取php文件
3. 数据:所选公司
4. 成功:在何处显示提取的数据。

因为公司是以post格式在php文件中提供的 $公司\u id=$\u职位['id']; 基于此,我们可以从数据库中提取数据。