Javascript jQuery单击以呈现页面yii2
我有一列gridview。代码如下所示:Javascript jQuery单击以呈现页面yii2,javascript,jquery,yii2,Javascript,Jquery,Yii2,我有一列gridview。代码如下所示: [ 'format' => 'raw', 'value' => function ($model) { return "<p class='feedback'>" . $model->KOMENTAR . "</p><br><p class='feedback-date'>" . $model->TANGGAL . "</p>
[
'format' => 'raw',
'value' => function ($model) {
return "<p class='feedback'>" . $model->KOMENTAR . "</p><br><p class='feedback-date'>" . $model->TANGGAL . "</p><hr><div id='replay-" . $model->ID_KOMENTAR . "'><ul></ul><div id='replay-column'></div></div>";
},
]
我想在单击按钮时呈现页面,我尝试了一些代码,但它只是出现了错误
<?= $this->render('feedback_form', [
'model' => $model,
]) ?>
尝试以下简单的
$。ajax
var request = $.ajax({
url: "index.php?r=feedback/feedbacktest",
method: "GET""
});
request.done(function( data) {
$('#replay-column').append(data);
});
request.fail(function( jqXHR, textStatus ) {
alert( "Request failed: " + textStatus );
});
只需确保您的url
是正确的,然后查看是否有任何错误
var request = $.ajax({
url: "index.php?r=feedback/feedbacktest",
method: "GET""
});
request.done(function( data) {
$('#replay-column').append(data);
});
request.fail(function( jqXHR, textStatus ) {
alert( "Request failed: " + textStatus );
});