laravel How to get database values using ajax?

Question

I want to post this form to Controller using ajax.

<form class="delete_confirm" action="{{ route('comment.delete_confirm', $mypage->id) }}">
  <input type="hidden" name="_token" value="{{ csrf_token() }}">
  <input type="password" name="pass_con" placeholder="비밀번호를 입력하세요.">
  <input type="submit" class="pass_submit" value="확인">
</form>

JS

$del_conForm.on('submit', function(){

        $.ajax({
            url: 'comment/{id}',
            type: 'post',
            data: $(this).serialize(),
            success: function(res){
                if (!res) {
                    alert('not correct');
                    return false;
                }
                else{
                    if (confirm("Click OK to continue?")) {
                        $delForm.submit();
                    }
                }
            }
        });

    });

I want to know $mypage->id in HTML form is also submit controller.

Controller

public function delete_confirm($id) {

    $mypage = mypage::findOrFail($id);

    $password = $mypage->password;

    $data = Input::get('pass_con');

    if ($data == $password) {
        return 'suc';
    }else{
        return false;
    }
}

Route

Route::get('comment/{id}', ['as'=>'comment.delete_confirm', 'uses'=>'MyPageController@delete_confirm']);
Route::post('comment/{id}', ['as'=>'comment.delete_confirm', 'uses'=>'MyPageController@delete_confirm']);

Request::ajax() part is work successfully, but $password returns nothing. Can controller delete_confirm method receive $id ?

Can you provide us with your route definition please? The parameter id needs to be declared there, too. — Rene M.
– Rene M., Commented Nov 17, 2016 at 9:43
Looks ok, I would use a step debuger to check if id is filled and if the query returns a result and is the result the result you expect. — Rene M.
– Rene M., Commented Nov 17, 2016 at 10:10

R.K.Saini · Accepted Answer · 2016-11-17 11:05:23Z

1

Try this :

$del_conForm.on('submit', function(){
    url = $(this).attr("action");
    $.ajax({
        url: url,
        type: 'post',
        dataType : 'json'
        data: $(this).serialize(),
        success: function(res){
            if (!res.status) {
                alert('not correct');
                return false;
            }
            else{
                if (confirm("Click OK to continue?")) {
                    $delForm.submit();
                }
            }
        }
    });

});

// your controller code should look like this

public function delete_confirm($id) {

    $mypage = mypage::findOrFail($id);

    $password = $mypage->password;

    $data = Input::get('pass_con');
    $response = array();
    if ($data == $password) {
       $response['status'] = true;
    }else{
       $response['status'] = false;
    }
    return response()->json($response);
}

edited Nov 17, 2016 at 11:05

answered Nov 17, 2016 at 9:46

R.K.Saini

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11 Comments

R.K.Saini Over a year ago

try echoing $id in controller and you need to remove quotes around $id. $mypage = mypage::where('id', $id)->get();

R.K.Saini Over a year ago

Do one more thing check what url your ajax request hit using developer tool or simply put alert(url); inside submit event

jungmyung Over a year ago

removing quoutes around $id, $id is echoing successfully

R.K.Saini Over a year ago

Try to var_dump($my_page) and check what it return. May be password is empty or null

jungmyung Over a year ago

$mypage = mypage::findOrFail($id); var_dump($mypage); brings array normally.

|

LF-DevJourney · Accepted Answer · 2016-11-17 09:40:57Z

0

$mypage = mypage::find($id);

$password = $mypage->password;

answered Nov 17, 2016 at 9:40

LF-DevJourney

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Comments

JérémyCasper · Accepted Answer · 2016-11-17 09:44:36Z

0

There is one error in your code, you put $id between quote so it's like a string and not the variable. Also, you could change the way you use the request

(This is assuming you're using Laravel 5)

public function delete_confirm(Request $request, $id) {

$mypage = mypage::where('id', $id)->get();

$password = $mypage->password;

if($request->ajax()) {

    $data = $request->get('pass_con');

    return $password;
}

}

Also, I don't know what you really want to do, but you can use Auth::attempt in order to valid a password (see https://laravel.com/docs/5.3/authentication )

You can also put some var_dump($id) in your code to check if the variable $id is really setted.

answered Nov 17, 2016 at 9:44

JérémyCasper

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Comments

Veshraj Joshi · Accepted Answer · 2016-11-17 09:57:39Z

0

// use dedication methods to make it workable most of the time
$(document).on('submit','.delete_confirm', function(e){
    e.preventDefault();
    $this = $(this);
    $.ajax({
        url: $($this).attr('action'),
        // use method for get,post and others and keep in mind all method values should be in block letters
        method: 'POST',
        // error is right here you just use this keyword but under ajax this refers to the ajax object of jquery so you have to create $this object like above
        data: $($this).serialize(),
        success: function(res){
            console.log(res);

        }
    });

});

answered Nov 17, 2016 at 9:57

Veshraj Joshi

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1 Comment

R.K.Saini Over a year ago

why you write $($this).attr('action')? $this is already a jQuery object

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