如何将两个变量传递给JavaScript函数以根据该值更改下拉列表?
$(document).ready(function() {
$("#date").change(function() {
$(this).after('<div id="loader"><img src="img/loading.gif" alt="loading category" />
</div>');
$.get('loadb.php?date=' + $(this).val(), function(data) {
$("#time").html(data);
$('#loader').slideUp(200, function() {
$(this).remove();
});
});
});
});
</script>
<div class="sub-page-left">
<br/>
<br/>
<div id='cssmenu'>
<form id="myform" action="spotDate.php" method="post">
<?php
echo '<input type="hidden" name="NID" size="20" value="'.$NID.'"/>';
?>
Choose Date: <select name="date" id="date" class="dropdown"><OPTION VALUE=0>Choose Date<? $option ?>
<?php
$q2 = "select DISTINCT(AvailableDate) from FreeTimeSlots where PASSLeaderID = $SID and CourseID = $CourseID";
$r2 = mysqli_query($dbc, $q2);
while ($row2 = mysqli_fetch_array($r2)) {
echo "<option value='"" . $row2['AvailableDate'] . "'">" . $row2['AvailableDate'] . "</option>";
}
?>
</select>
Time: <select name="time" id="time" class="dropdown"><OPTION VALUE=0>Choose Time<? $option ?>
</select><br /><br />
此代码适用于我仅传递日期值的地方,但我想根据日期和隐藏的 ID 进行更新!!
试试这段代码:
$.get('loadb.php?date=' + $(this).val()+'&nid='+$('#NID').val(), function(data) {
$("#time").html(data);
$('#loader').slideUp(200, function() {
$(this).remove();
});
});
您也可以在 url 中传递隐藏字段:'loadb.php?date=' + $(this).val() + '&nid='+$("input[name=NID]").val()