Question

我尝试在使用ajax选中复选框时发布值，但是失败了，代码对我来说似乎是正确的..在url上看起来像这样，它不应该..

http://localhost/logsys/admin/roleaccess/%3C?%20$role[%27id%27];%20?%3E

我正在使用在xampp v 7.3.0上运行的codeigniter v 3.1.10

<script>
    $('.form-check-input').on('click', function() {
        const menuId = $(this).data('menu');
        const roleId = $(this).data('role');

        $.ajax({
            url: "<?= base_url('admin/changeaccess'); ?>",
            type: 'post',
            data: {
                menuId: menuId,
                roleId: roleId

            },
            success: function() {
                document.location.href = "<?= base_url('admin/roleaccess/'); ?>" + roleId;
            }
        });
    });
</script>

它应该传递允许用户访问某些菜单的值

Answer 1

<script>
    $('.form-check-input').on('click', function() {
        const menuId = $(this).data('menu');
        const roleId = $(this).data('role'); // roleId = <? $role[‘id’] ?> so php did not echo role id here

        $.ajax({
            url: "<?= base_url('admin/changeaccess'); ?>",
            type: 'post',
            data: {
                menuId: menuId,
                roleId: roleId

            },
            success: function() {
                document.location.href = "<?= base_url('admin/roleaccess/'); ?>" + roleId;
            }
        });
    });
</script>

要更改解决方案，请更改您的form-check-input元素

<... class=“form-check-input” data-role=“<?php echo $role[‘id’]; ?>” data-menu=“echo menu variable” ...>

如果这无法共享您的form-check-input按钮的html

使用ajax选中复选框时，发布值失败

1 个答案: