【发布时间】:2019-07-29 08:09:37
【问题描述】:
我想在复选框中获取 id 的值。但是 id 中的变量总是给出第一个值。如何将 php 数组发送到 jquery 数组。谢谢
require "connection.php";
$sql = mysqli_query($con,"SELECT * FROM position ORDER BY position_align + 0 ASC");
$switch = array();
$pid = array();
while($row = mysqli_fetch_assoc($sql)){
$pcode = $row['position_code'];
$pdesc = $row['position_desc'];
$status = $row['status'];
?>
<tr>
<td><?php echo $pcode; ?></td>
<td><?php echo $pdesc; ?></td>
<td>
<div class="custom-control custom-switch">
<?php
if($status == 'active'){
?>
<div class="onoffswitch4">
<input type="checkbox" name="onoffswitch4" class="onoffswitch4-checkbox switch" onclick="change(this.value)" id="myonoffswitch" checked value="<?php echo $pcode; ?>">
<label class="onoffswitch4-label" for="pid[]">
<span class="onoffswitch4-inner"></span>
<span class="onoffswitch4-switch"></span>
</label>
</div>
<?php
}else{
?>
<div class="onoffswitch4">
<input type="checkbox" name="onoffswitch4" class="onoffswitch4-checkbox switch" onclick="change(this.value)" id="aaa[]" value="<?php echo $pcode; ?>">
<label class="onoffswitch4-label" for="myonoffswitch">
<span class="onoffswitch4-inner"></span>
<span class="onoffswitch4-switch"></span>
</label>
</div>
<?php
}
?>
$('.switch').on('change', function() {
if(this.checked) {
var id = this.value;
$.ajax({
url: "admin_change_status.php",
data: {
id:id
},
type: "POST",
success: function(result){
}
});
}else{
var id2 = this.value;
$.ajax({
url: "admin_change_status.php",
data: {
id2:id2
},
type: "POST",
success: function(result){
}
});
}
});
我希望将复选框中 id 的值传递给 jquery。
【问题讨论】:
-
您想在选中或所有值时获取复选框值?