Laravel print error if bad query

Question

I'm laravel newbie, I have written a simple code:

Route::get('users/{uname}/{uid}', function($uname, $uid)
{
    $u = DB::table('users')->where('login', $uname)->first();

    return View::make('users', 
        array(
            'username'  => $u->login,
            'userid'    => $u->uid
        )
    );
});

I have 2 questions regarding this code,

First: Is this code valid?

Second: How can I show 404 error when query is not found? For example if I write site.com/users/badusername/badid then show 404 error, not blank page.

Thanks in advance !

P.S how can i make $u query to two wheres? For example WHERE login = ? AND uid = ?

jedrzej.kurylo · Accepted Answer · 2015-08-03 08:54:36Z

3

First: code looks ok. You could also make use of a little un-documented feature of dynamic where methods, and call it like that:

DB::table('users')->whereLogin($uname)->first();

Second: In order to throw 404 error, you can simply call

App::abort(404);

Third: You can chain the where() methods, e.g.:

$u = DB::table('users')->where('login', $uname)->where('uid', $uid)->first();

The query above doesn't make much sense though, as uid and probably username as well are unique and fetching just by one of them should be enough. If you have a Eloquent model for your users, you could also use model's find() or findOrFail() methods to search by primary key. Especially findOrFail() is very useful, as it throws 404 exception if model cannot be found, so no need to wrap it in the if statement.

answered Aug 3, 2015 at 8:54

jedrzej.kurylo

41.1k10 gold badges108 silver badges112 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

violator667 · Accepted Answer · 2015-08-03 08:54:26Z

0

Assumming you have User model. Try:

Route::get('users/{uname}/{uid}', function($uname, $uid)
{
    $u = User::where('login', $uname)->where('uid',$uid)->first();

    if(!$u->isEmpty()) {
          return View::make('users',array('username'  => $u->login,
                                          'userid'    => $u->uid));
    }else{
         App::abort(404);
    }
});

answered Aug 3, 2015 at 8:54

violator667

4994 silver badges19 bronze badges

3 Comments

Ash Over a year ago

If you're using Eloquent the best way is to use findOrFail. In this case firstOrFail would be more appropriate.

violator667 Over a year ago

Yes, I know but I use it intentionaly not to mess with caching an exception - to make code cleaner and simplier as OP said he is newbie

Ash Over a year ago

Well your way is not simple neither is it cleaner. I was just improving your answer with better ways. Also the OP didn't mention using Eloquent you're making an assumption so you're still likely to be wrong.

Romain Lanz · Accepted Answer · 2015-08-03 08:57:19Z

0

Yes, you're code is valid, but you can simplified it.

get('users/{uname}/{uid}', function($uname, $uid)
{
    $user = User::where('login', $uname)->where('uid', $uid)->first();

    if (! $user) abort(404);

    return view('users', compact('user'));
});

To use this code you need to use Laravel 5+ and to have a model for your User table.

Also, as you can see, I have write another condition to handle the uid.

Note that I send $user to the view.

edited Aug 3, 2015 at 8:57

answered Aug 3, 2015 at 8:53

Romain Lanz

8855 silver badges14 bronze badges

2 Comments

jedrzej.kurylo Over a year ago

It won't throw an exception. It will return null if a matching model is not found. Only eloquent findOrFail() throws exceptions

Romain Lanz Over a year ago

Correct! Laravel 4 automatism... Thanks, I've edited my answer.

Collectives™ on Stack Overflow

Laravel print error if bad query

3 Answers 3

Comments

3 Comments

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

3 Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related