Laravel 5.1 ajax url参数是url

I am working with Laravel 5.1 for the first time and I cannot understand why I am getting 404s on an ajax call that passes a URL to the server PHP script as a parameter.

我第一次使用Laravel 5.1，我无法理解为什么我要在ajax调用中获得404s，该调用将URL作为参数传递给服务器PHP脚本。

I am executing an Ajax call that is being handled by a route as follows:

我正在执行的Ajax调用由以下路径处理:

Route::get('ajax/{act}', ['uses' => 'AjaxController@helpers', 'as' => 'ajax.helpers']);

I want the variable {act} to hold the sring of key / value pairs I pass. I decode these in the PHP at the server end. The Ajax PHP script contains a variety of helpers and I do not want to create a Laravel rout for each.

我希望变量{act}包含我传递的键/值对。我在服务器端的PHP中解码这些。Ajax PHP脚本包含各种助手，我不想为每个助手创建一个Laravel路由器。

In my app, the user will input a url in a form field, which I capture in a variable called website

在我的应用程序中，用户将在表单字段中输入url，我将在一个名为website的变量中捕获该字段

My ajax call needs to accept:

我的ajax电话需要接受:

var url = '/ajax/act=url&u=' + website;

I am doing this to build the url I then pass to a jQuery $.getJSON call:

我这样做是为了构建url，然后将url传递给jQuery $。getJSON电话:

var url = '/ajax/act=url&&u=' + encodeURIComponent(website);

I would expect the encodeURIcompponent() function to make this work, but it returns 404 when any of the parameters contain / characters prior to the encodeURIComponent(). My base url works perfectly without the additional url as a parameter.

我希望使用encodeURIcompponent()函数来完成这项工作，但是在encodeURIComponent()之前的任何参数包含/字符时，它都会返回404。我的基本url在没有附加url作为参数的情况下工作得很好。

But passing a url as a variable value, it throws 404.

但将url作为变量值传递时，会抛出404。

This is what the url in ajax call looks like that returns 404:

这就是ajax调用中的url返回404的样子:

http://my.app/ajax/act=url&u=http%3A%2F%2Fgoogle.com

This url works perfectly (I have removed the // from http://google.com:

这个url非常好用(我已经从http://google.com删除了//)。

http://my.app/ajax/act=url&u=http%3Agoogle.com

It also fails when there is additional path items in the variable url as it contains additional / characters, like as follows:

当变量url中有额外的路径项时，它也会失败，因为它包含了额外的/字符，如下所示:

http://google.com/subfolder

How do I pass the full url as a parameter in the ajax call? Thanks!

如何将完整url作为ajax调用中的参数传递?谢谢!

2 个解决方案

#1

I think you're confusing route parameters and query parameters. Your route is defined as ajax/{action}. In this case, {action} is a route parameter, but you're trying to stuff query parameters into it.

我认为您混淆了路由参数和查询参数。您的路由被定义为ajax/{action}。在本例中，{action}是一个route参数，但您试图将查询参数填充到其中。

For example, if you access the url http://my.app/ajax/act=url&u=google.com, this will work because you've hit the route ajax/{action}, where {action} is act=url&u=google.com. That is the value that will get passed to your AjaxController@helpers function. However, since this data is passed in as a route parameter, it is not in the request input. $request->all() will be empty.

例如，如果您访问url http://my.app/ajax/actt = url&ui =google.com，这将有效，因为您访问了ajax/{动作}路径，其中{动作}为act=url&u=google.com。这是传递给ajaxcontroller@helper函数的值。但是，由于该数据作为路由参数传入，所以它不在请求输入中。请求- >所有美元()将是空的。

However, if you access the url http://my.app/ajax/act=url&u=http://google.com, this will not work, as you do not have this route defined. This does not map to the ajax/{action} route; this route would be mapped to ajax/{action}//google.com, which you do not have defined (hence the 404).

但是，如果您访问url http://my.app/ajax/act= url&ui =http://google。这不会映射到ajax/{action}路径;该路径将映射到ajax/{action}/ google.com，您还没有定义它(因此是404)。

I think what you're really looking for is this: http://my.app/ajax/url?u=http%3A%2F%2Fgoogle.com. This will hit your ajax/{action} route with url as the {action} route parameter and the url value will be in the query parameters. Inside your AjaxController@helpers function, you can access the url via $request->input('u');.

我想你真正想要的是:http://my.app/ajax/url?这将以url作为{action}路由参数命中ajax/{action}路由，url值将在查询参数中。在ajaxcontroller@helper函数中，可以通过$request->输入('u')访问url。

Edit

If you really need this data to come in as a route parameter, another option you have to make sure your route parameter consumes everything, including slashes:

如果您确实需要这些数据作为路由参数输入，那么您必须确保您的路由参数使用所有内容，包括斜线:

Route::get('ajax/{action}', ['uses' => 'AjaxController@helpers', 'as' => 'ajax.helpers'])
    ->where('action', '.*');

If you do this, however, this route will catch everything that falls under http://my.app/ajax/....

但是，如果您这样做，该路径将捕获属于http://my.app/ajax/…的所有内容。

#2

You are using relative paths with your url variable (as you've a leading / in your URL), that could be causing the 404 not found error.

您正在使用url变量的相对路径(就像您的url中有一个引导/)，这可能导致404没有找到错误。

Try storing your base url using Laravel's helper method url() in a hidden field in your view (you could do it in your master view if you wanted).

尝试使用Laravel的helper方法url()将基本url存储在视图中的一个隐藏字段中(如果需要，可以在主视图中进行)。

<input id='baseUrl' type='hidden' value='{{ url() }}' />

You can grab the value as simple as creating a JS helper function as such:

您可以简单地获取值，创建一个JS助手函数，如下所示:

var baseUrl = function(){
    return $('#baseUrl').val();
};

Then append that url into the beginning of your variable url like so:

然后将该url附加到变量url的开头，如下所示:

var url = baseUrl() + '/ajax/act=url&&u=' + encodeURIComponent(website);