如何使用"prep_url"从我在Laravel视图中的codeigniter? [英] How can I use "prep_url" from codeigniter inside my View in Laravel?
问题描述
我将来自codeigniter的功能"prep_url"添加到我的laravel项目中,如下所示:
I add the function "prep_url" from codeigniter to my laravel project like here:
如果我在控制器中这样操作:
If I do it in my Controller like this:
$website = example.com;
$url = prep_url($website);
然后在我看来,我可以这样做:
and then in my view I can do it like this:
{{$url}}
它向我显示了以下内容: http://example.com
it shows me this: http://example.com
一切正常:)
但是我有问题
我有桌子用户,那里的所有用户都有一个网站
I have the table users and there all users have a website
在我的控制器中,我这样做:
in my controller I do it like this:
$user = DB::table('users')->get();
return view('user/einzelansicht' ['user' => $user] );
在我看来,我可以向用户展示
and in my view I can show my user with
{{$user[1]->name}}
{{$user[1]->website}}
如何在此处使用prep_url?
how can I use prep_url here?
因为在我看来,我已经这样做了:
because I already did this in my view:
{{prep_url($user[1]->website);}}
但是它告诉我不允许使用分号,并且如果没有分号,它将不起作用
but it tells me semicolons are not allowed and without the semicolon it does not work
然后我在控制器中尝试了以下方法:
then I tried in my controller this:
$website = $user->website;
$url = prep_url($website);
但这也不起作用
那我该怎么办呢?
推荐答案
也许您的网站"属性不是字符串.
Maybe your "website" attribute is not a string.
尝试
prep_url(json_encode($u[1]->website));
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