php 如何在jquery中为laravel route传递多个参数

jjhzyzn0 于 2022-10-30 发布在 PHP

关注(0)|答案(2)|浏览(75)

var url = "{{ route('order-detail',':slug') }}";
console.log(url);

我想在slug之后再添加一个参数id：

var url = "{{ route('order-detail',':slug',':id') }}";
console.log(url);

这是错误：
缺少[Route：订单详细信息] [URI：订单/详细信息/{slug}/{id}] [缺少参数：id]。（查看：）（请参阅：

2条答案

首先，创建两个变量slug和id

var slug = ...;
var id = ...;
// you should know how to get this, you didn't show your full code
// so I can't know

接下来，创建url变量，如下所示：

var url = "{{ route('order-detail', ['slug' => ':slug', 'id' => ':id']) }}";
url = url.replace(':id', id);
url = url.replace(':slug', slug);

现在，这应该会为您提供一个合适的路径url

尝试

var url = "{{ route('order-detail', [':slug', ':id']) }}";