根据选择值添加/显示动态表单字段 - Add/Show dynamic form fields depending on select value

我使用Jquery Mobile创建一个弹出式表单，显示选择语句供用户选择。我使用ajax使选择语句动态。我已经得到了要显示的数据并创建了一个新的选择语句。它的格式似乎不正确。

带有before和After的Form图片

弹出式表单代码

<?php
$q = intval($_GET['q']);
include_once('session.php');
include_once('dbConnect.php');
$sql="SELECT Enrollment.c_id, Enrollment.s_id, users.f_name, users.l_name
FROM Enrollment
INNER JOIN users ON Enrollment.s_id = users.s_id
WHERE c_id=$q";
$result = mysqli_query($dbc, $sql);
echo "<label for='selectuser' class='select'>Select user:</label>";
echo "<select name='selectuser' id='selectuser' data-native-menu='false'>";
echo "<option>Choose Users:</option>";
echo "<option value='instructor'>All Instructors</option>";
echo "<option value='students'>All Students</option>";
while($row = mysqli_fetch_array($result))
  {
        $s_id = $row['s_id'];
        $f_name = $row['f_name'];
        $l_name = $row['l_name'];
        echo "<option value='$s_id'>$f_name $l_name</option>";
  }
echo "</select>";
?>

更简单的方法是:

HTML

首先，把所有的选择框从头放到html中:

<select name="selectclass" id="selectclass" data-native-menu="false">
   <option value='default'>Select Class:</option>
   <?php echo $allClassOptions; ?>
</select>
<select name="selectuser" id="selectuser" data-native-menu="false">
   <option value='default'>Select User:</option>
   <?php echo $allUsers; ?>
</select>

为没有javascript的用户提供另一种选择(优雅的降级)是很好的做法。

Javascript

然后，在javascript文件中隐藏开始时应该隐藏的输入字段。将事件处理程序绑定到第一个选择字段的更改事件，并使用Ajax调用填充第二个选择字段的选项字段。

var selectElement = $("#selectuser");
selectElement.hide();
$("#selectclass").on("change", function(){
    var selectedClass = this.value;
    if(selectedClass != "default"){
       selectElement.show();
       $.ajax({
           type: "POST",
           url: "getdatabaseresults.php",
           data: {"class": selectedClass },
           success: function(result){
                //remove old options
                selectElement.empty();
                //add new options
                selectElement.append(result);
           }
       });
    };
});

PHP

在PHP文件中，处理Ajax调用并返回所需的结果:

<?php
if(isset($_SERVER["HTTP_X_REQUESTED_WITH"]) && strtolower($_SERVER["HTTP_X_REQUESTED_WITH"]) == "xmlhttprequest"){
    //this is an Ajax call!
    $selectedClass = $_POST["class"];
    $options = "<option value='default'>Select User:</option>";
    //do whatever you want with the data
    //database calls and whatnot
    $stmt = mysqli_prepare($dbc, "SELECT * FROM users WHERE c_id = ?");
    mysqli_stmt_bind_param($stmt, "s", $selectedClass);
    mysqli_stmt_execute($stmt);
    mysqli_stmt_bind_result($stmt, $row);
    while(mysqli_stmt_fetch($stmt)) {
        $user = $row['username'];
        $options.= "<option value='$user'>$user</option>";
    }
    mysqli_stmt_close($stmt);
    echo $options;
}
?>

这个php文件可以扩展(例如使用switch())，以便它可以用于不同的ajax调用。

注意:有许多不同的方法来实现这一点，这只是一个例子。

我认为问题是你没有关闭你的select标签后循环。另外，建议在最后只写一次。比如:

<?php
$q = intval($_GET['q']);
include_once('session.php');
include_once('dbConnect.php');
$sql="SELECT Enrollment.c_id, Enrollment.s_id, users.f_name, users.l_name
FROM Enrollment
INNER JOIN users ON Enrollment.s_id = users.s_id
WHERE c_id=$q";
$result = mysqli_query($dbc, $sql);
$text = "<label for='selectuser' class='select'>Select user:</label>";
$text .= "<select name='selectuser' id='selectuser' data-native-menu='false'>";
$text .= "<option>Choose Users:</option>";
$text .= "<option value='instructor'>All Instructors</option>";
$text .= "<option value='students'>All Students</option>";
while($row = mysqli_fetch_array($result))
{
    $s_id = $row['s_id'];
    $f_name = $row['f_name'];
    $l_name = $row['l_name'];
    $text .= "<option value='$s_id'>$f_name $l_name</option>";
}
$text .= "</select>"
echo $text
?>