Yii2 submit link with Ajax

Question

I'm trying to create votes for my content in Yii2, so I'm trying to insert data with Ajax using "a" tag and widget but with ajax it's not working , without ajax it's working perfect. but when I use Ajax it's not working and I have no errors in console. my code for view is :

<script type="text/javascript">
    $('document').ready(function(){
        $('#btn-vote-up' + <?= $pst_id ?>).on("click", function(e){
            $.ajax({
                type: "POST",
                data: {"value" : "Like"},
                success: function(msg) {
                    console.log (<?= $pst_id ?>);
                    $('#note-up' + <?= $pst_id ?>).load(' #note-up' + <?= $pst_id ?>);
                }               
            });
            e.preventDefault();
        });
    });
</script>


    <?= Html::a('<i class="ly-ic-favorite-plus"></i>', '#', [
                'class' => 'btn-logout',
                'id' => 'btn-vote-up'.$pst_id,
                'data'  => [
                    'params' => [
                        'value' => 'Like',
                        'pstIDL' => $pst_id,
                        ],
                    ],
        ]) ;
    ?>
    <?php yii\widgets\Pjax::begin(['id' => 'note-up'.$pst_id]) ?>
        <span><?= $total_up ?></span>
    <?php yii\widgets\Pjax::end() ?>

And my widget code is :

if (Yii::$app->request->post('value') == 'Like')
            {
                $pstIDL = Yii::$app->request->post('pstIDL');

                $modelLIKEPOST = $this->findLikePost($pstIDL);

                AxVotePost::VoteUP($modelLIKEPOST, $this->usr_rid, $this->chn_id, $pstIDL);
                header('Location: ' .Url::current() );
                exit;
            }

saying it not working is not enough, what happens, when you send an ajax call, do you receive a 404 bad request or anything? and above all you are not sending the $pstIDL value when making an ajax call and only value is being sent. — Muhammad Omer Aslam
– Muhammad Omer Aslam, Commented Aug 1, 2018 at 15:04
the answer help me; thank you so much for your time this is the 3rd question you try to help me; thank you so much — Bynd
– Bynd, Commented Aug 1, 2018 at 15:36
:) we are all here to help each other @Bynd , you are always welcome. — Muhammad Omer Aslam
– Muhammad Omer Aslam, Commented Aug 1, 2018 at 15:42

Muhammad Omer Aslam · Accepted Answer · 2018-08-01 15:41:02Z

1

You are not sending the $pstIDL value when making an ajax call and only value is being sent, you can use var data=$(this).data('params') to get both pstIDL, and the value json and send it along the data in the call see below

<script type="text/javascript">
    $('document').ready(function(){
        $('#btn-vote-up' + <?= $pst_id ?>).on("click", function(e){
           var data= $(this).data('params');
            $.ajax({
                type: "POST",
                data: data,
                success: function(msg) {
                    console.log (<?= $pst_id ?>);
                    $('#note-up' + <?= $pst_id ?>).load(' #note-up' + <?= $pst_id ?>);
                }               
            });
            e.preventDefault();
        });
    });
</script>

edited Aug 1, 2018 at 15:41

answered Aug 1, 2018 at 15:10

Muhammad Omer Aslam

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3 Comments

Bynd Over a year ago

I tried your code but still not working; console.log pstIDL it's give "undefined".

Bynd Over a year ago

I changed you code from var pstIDL=$(this).data('pstIDL'); to : var pstIDL = $(this).data('params'); and it's working; please can you edit your answer and change only var pstIDL=$(this).data('pstIDL'); to : var pstIDL = $(this).data('params');

Muhammad Omer Aslam Over a year ago

yeah you can do it that way to get the whole json string and send as is @Bynd

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