[英]Add/Show dynamic form fields depending on select value
我正在使用Jquery Mobile创建一个弹出窗体,显示用户可以选择的select语句。 我使用ajax使select语句动态化。 我已经获得了要显示的数据并创建了一个新的select语句。 它似乎没有正确格式化。
<?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:
首先,从一开始就将所有选择框放在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>
优良作法是为没有javascript(优雅降级)的用户提供替代方案。
然后,在您的javascript文件中,隐藏应该在开始时隐藏的输入字段。 将事件处理程序绑定到第一个选择字段的change事件,并使用Ajax调用来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();
$.ajax({
type: "POST",
url: "getdatabaseresults.php",
data: {"class": selectedClass },
success: function(result){
//remove old options
selectElement.empty();
//add new options
selectElement.append(result);
}
});
};
});
在PHP文件中,处理Ajax调用并返回想要的结果:
<?php
if(isset($_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()
),因此它可以用于不同的ajax调用。
注意 :有许多不同的方法可以实现这一点,这只是一个例子。
我认为问题在于你没有在循环后关闭你的select
标签。 此外,建议最后只进行一次写操作。 因此:
<?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
?>
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