dt250827 2014-09-04 12:50
浏览 33
已采纳

Yii将相关数据导入GridView

My view lets the user see a person's details, including the families they are in. This is done in the DB with a Person table, Family table, and a familyMembership link table between the two.

The relationship in the CActiveRecord model for Person is as such:

'families' => array(self::MANY_MANY, 'Family', 'familymembership(Person, Family)'),

In my person controller, I am wanting to pass a variable into the view that has the related data in a way that TbGridView (CGridView but Bootstrap) will accept (a data provider). The controller calls the person model's getFamilies() function:

public function getFamilies() {
    // returns an array of Family objects related to this Person model
    $familiesArray = $this->getRelated('families');
    // puts this array into a CArrayDataProvider
    $families = new CArrayDataProvider($familiesArray);
    return $families;
}

The return value goes back to the controller, which is then handed through in the renderPartial() call. The view has a TbGridView widget initialisation like this:

$this->widget('bootstrap.widgets.TbGridView', 
            array(
                //the CArrayDataProvider from the model function
                'dataProvider' => $families, 
                'columns' => array(
                    array(
                        'name' => 'Family Name',
                        // Example field from Family model
                        'value' => '$data->familyName' 
                    )
                )
        ));

However, in doing this I am getting the following error: 'Property "Family.id" is not defined. (D:\wamp\www\yii\framework\base\CComponent.php:130)'

The Family model does not have an id property, and I don't understand why the widget is looking for such.

What's going wrong here?

Thanks.

展开全部

  • 写回答

1条回答 默认 最新

  • dpl22899 2014-09-04 14:50
    关注

    Quoting the doc from Yii Class Reference, you have to provide a keyField:

    Elements in the raw data array may be either objects (e.g. model objects) or associative arrays (e.g. query results of DAO). Make sure to set the keyField property to the name of the field that uniquely identifies a data record or false if you do not have such a field.

    By default keyField will be set to "id", so you need to overwrite it with your Family model primary key :

    <?php
    $familyDataProvider = new CArrayDataProvider($rawData, array(
        'keyField' => 'yourPkNameHere',
    ));
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
编辑
预览

报告相同问题?

手机看
程序员都在用的中文IT技术交流社区

程序员都在用的中文IT技术交流社区

专业的中文 IT 技术社区,与千万技术人共成长

专业的中文 IT 技术社区,与千万技术人共成长

关注【CSDN】视频号,行业资讯、技术分享精彩不断,直播好礼送不停!

关注【CSDN】视频号,行业资讯、技术分享精彩不断,直播好礼送不停!

客服 返回
顶部