[英]How to make controller remember ajax post data Yii2
我有多個模式窗口,從其中一個窗口我想發送幾個$ models到另一個窗口,就像父親有幾個兒子一樣……而我們無法保存兒子,因為我們不知道父親的身份證和姓名。
JS
$("#modal-deposit").submit(function($form) {
$form.preventDefault();
$.ajax({
type: "POST",
url: "index.php?r=family/create", // сreatedeposit создает запись.
data: $("#form-deposit").serialize(), // Сериализует the form's elements.
return false;
});
行動
public function actionCreate()
{
$family_model = new family();
$model_deposit = new FinanceIncomeExpenses();
if ($family_model->load(Yii::$app->request->post())
&& $family_model->save())
{
if ($model_deposit->sum != null)// one of the required fields
{
$model_deposit->family_id = $family_model->id;
$model_deposit->save();
}
} else {
if(Yii::$app->getRequest()->isAjax){
return $this->renderAjax('create', [
'family_model' => $family_model,
'model_deposit' => $model_deposit,
]);
}else{
return $this->render('create', [
'family_model' => $family_model,
'model_deposit' => $model_deposit,
]);
}
}
}
對於一個你需要改變
url: "index.php?r=family/create",
至
url: "<?php echo \Yii::$app->getUrlManager()->createUrl('family/create') ?>",
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