Php 关于URL中参数的疑问

本人将以下表格寄至以下地址：

http://localhost/myapp/web/index.php?r=cashbook%2Faccomplishment

谁处理并生成以下URL：

http://localhost/myapp/web/index.php?r=cashbook%2Faccomplishment&Cashbook%5Bcategory_id%5D=18

但我只需要在dropDownList中选择的category_id值。也就是说，URL应该是：

http://localhost/myapp/web/index.php?r=cashbook%2Faccomplishment&category_id=19

如何保持这种状态或仅获取视图中的类别id添加注册表：

不要使用ActiveForm:：begin。。。还有你所有的代码。不要使用这个：

echo Html::activeDropDownList($model, 'id_category', ArrayHelper::map(
            Category::find()->where(['user_id' => Yii::$app->user->identity->id])
                            ->orderBy("desc_category ASC")
                            ->all(), 'id_category', 'desc_category'), ['onchange'=>'submit(this.value);','prompt'=>'-- Select --']);

当我从/？category\u id=更改为&category\u id=时工作。非常感谢你

$this->registerJs('var submit = function (val){
if (val > 0) {
    window.location.href = "' . Url::to(['/cashbook/accomplishment']) . '/?category_id=" + val;
}
}', View::POS_HEAD);

echo Html::activeDropDownList($model, 'id_category', ArrayHelper::map(
            Category::find()->where(['user_id' => Yii::$app->user->identity->id])
                            ->orderBy("desc_category ASC")
                            ->all(), 'id_category', 'desc_category'), ['onchange'=>'submit(this.value);','prompt'=>'-- Select --']);