--从一个案例分享status的过滤的原理
tab页面status,是如何点击不同Tab传递不同的status的?
js文件里,关键:
{field: 'workauthorstatus', title: 'workauthorstatus', searchList: {"submited": 'submited', "passed": 'passed',"rejected": 'rejected'}, formatter: Table.api.formatter.status},
html里:
<ul class="nav nav-tabs" data-field="workauthorstatus">
<li class="active"><a href="#t-submited" data-value="submited" data-toggle="tab">待审批的作者</a></li>
<li class="{:$Think.get.workauthorstatus === 'passed' ? 'active' : ''}"><a href="#t-passed" data-value="passed"
data-toggle="tab">已审批的作者</a></li>
<li class="{:$Think.get.workauthorstatus === 'rejected' ? 'active' : ''}"><a href="#t-rejected"
data-value="rejected" data-toggle="tab">被拒绝的作者</a></li>
</ul>
controller里,从菜单点击进来,默认显示某个状态的数据,$filter =='{}' ,$where =‘’空的都是不行的。
$filter = $this->request->get("filter", '');
if($filter =='{}'){ //从菜单点击进来,默认显示submited的数据
$list = $this->model
->with('group')
->where('workauthorstatus','submited')
->order($sort, $order)
->paginate($limit);
}else{
$list = $this->model
->with('group')
->where($where)
->order($sort, $order)
->paginate($limit);
}