我想在symfony2中使用复选框。我想使用复选框值动态更新表(0/1)中的字段值。
这是我的错误代码:
index.html.twig:
<div class="slider demo" id="slider-1">
{% if plate.share == true %}
<input type="checkbox" value="1" checked>
{% else %}
<input type="checkbox" value="1">
{% endif %}
</div>
<script type="text/javascript">
$("input[type='checkbox']").on('click', function(){
var checked = $(this).attr('checked');
if (checked) {
var value = $(this).val();
$.post("{{ path('plate_share', { 'id': plate.id }) }}", { value:value }, function(data){
if (data == 1) {
alert('the sharing state was changed!');
};
});
};
});
</script>
的routing.yml
plate_share:
pattern: /{id}/share
defaults: { _controller: "WTLPlateBundle:Plate:share" }
PlateController.php:
public function shareAction($id)
{
if($_POST && isset($_POST['value'])) {
$link = mysql_connect('127.0.0.1', 'admin', 'wtlunchdbpass');
if (!$link) {
print(0);
}
mysql_select_db('wtlunch');
$value = mysql_real_escape_string($POST['value']);
$sql = "INSERT INTO table (value) VALUES ('$value')";
if (mysql_query($sql, $link)) {
print(1);
}
else {
print(0);
}
}
}
但是这个解决方案是错误的,不起作用。 是否可以创建表单并仅使用复选框提交?
有想法吗?感谢。
例如控制器中的编辑表单操作:
public function editAction($id)
{
$user = $this->container->get('security.context')->getToken()->getUser();
if (!is_object($user) || !$user instanceof UserInterface) {
throw new AccessDeniedException('This user does not have access to this section.');
}
$em = $this->getDoctrine()->getManager();
$entity = $em->getRepository('WTLPlateBundle:Plate')->find($id);
if (!$entity) {
throw $this->createNotFoundException('Unable to find Plate entity.');
}
$editForm = $this->createEditForm($entity);
$deleteForm = $this->createDeleteForm($id);
return $this->render('WTLPlateBundle:Plate:edit.html.twig', array(
'entity' => $entity,
'edit_form' => $editForm->createView(),
'delete_form' => $deleteForm->createView(),
));
}