Thinkphp5 表单提交额外参数和页面跳转参数传递url

1. 表单提交

<input type="hidden" name="project_name" value="$project_name"/>

在控制器中获取

$project_name=input("post.project_name");

 

2. php中跳转

$this->success('新增项目成功',url("Version/index",array('project_name'=>$project_name)));die;