Question

基本上我是一个刚开始学习PHP和MySQL的新资源管理器，我要做的是使依赖项下降。好的第一个下拉列表不错，但是第二个下拉列表与第一个下拉列表没有关联，有人可以帮忙吗？

谢谢。

index.php

<?php
  $connect = new mysqli ("localhost", "root", "root", "carsdb");
?>

<!doctype html>
<html lang="en">
  <head>
  <body>
    <!-- Required meta tags -->
    <meta charset="utf-8">
    <meta name="viewport" content="width=device-width, initial-scale=1, shrink-to-fit=no">

    <!-- Bootstrap CSS -->
    <link rel="stylesheet" href="https://stackpath.bootstrapcdn.com/bootstrap/4.2.1/css/bootstrap.min.css" integrity="sha384-GJzZqFGwb1QTTN6wy59ffF1BuGJpLSa9DkKMp0DgiMDm4iYMj70gZWKYbI706tWS" crossorigin="anonymous">

        <title>Car Selection</title>  
<div class = "container" style = "width:700px;">

    <form action = "" method = "POST">

    <div class = "jumbotron">
  <h1>Car Selection</h1><br>

  <label>Brand:</label> <select id = "Brand">
      <option value = "">Car Brand Selection</option>
        <?php
          $query = "SELECT Brand FROM Brand";
          $result = $connect->query($query);
            while ($row = $result->fetch_assoc())
            {
              echo "<option value = '{$row["Brand"]}'>{$row["Brand"]}</option>";
            }
        ?>
    </select><br>

    <label>Model:</label> <select id = "Model">
        <option value = "">Car Model Selection</option>
    </select><br><br>

      <input type = "submit" name="insert" class = "btn btn-success btn-sm" value="Next">

  </form>

</div>
</div>  

<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<script>
      $("#Brand").change(function(){
        bid=$(this).val();
        $.post("fetch.php",{id:bid}, function(data){
          $("#Model").html(data);
        });
      }); 
</script>

    </body>    
</html>

fetch.php

<?php
$connect = new mysqli ("localhost", "root", "root", "carsdb");
$query = "SELECT Model FROM Model where CODE = {$_POST["id"]}";
$result = $connect->query($query);
echo "<option value = ''>Select</option>";
    while ($row = $result->fetch_assoc())
    {
        echo "<option value = '{$row["Code"]}'>{$row["Model"]}</option>";
    }
?>

Answer 1

而不是-

echo "<option value = '{$row["Code"]}'>{$row["Model"]}</option>";

尝试----

echo '<option value="'.$row['Code'].'">'.$row['Model'].'</option>';

依赖下拉列表无法正常工作

1 个答案: