我有一个表,列for_date
在数据库中按类型整数保存。为了用DateTime格式显示for_date
,我使用ActiveForm代码:
<?= $form->field($model, 'for_date')->textInput([
'class' => 'form-control datepicker',
'value' => $model->for_date ? date(Yii::$app->params['dateFormat'], $model->for_date) : date(Yii::$app->params['dateFormat'], time()),
'placeholder' => 'Time'])
->label('Attendance Date') ?>
但是当我创建并保存时,Yii通知This field must be integer
在模型文件中,我有2个函数在验证之前转换,但它仍然是错误的。
public function beforeValidate(){
if(!is_int($this->for_date)){
$this->for_date = strtotime($this->for_date);
$this->for_date = date('d/M/Y', $this->for_date);
}
return parent::beforeValidate();
}
public function afterFind(){
$this->for_date = 'Yii::t('app', Yii::$app->formatter->asDate($this->for_date));
$this->for_date = date('d/M/Y', $this->for_date);
return parent::afterFind();
}
我怎样才能使它正确保存到数据库与整数?
根据您的代码,for_date
在beforeValidate
运行后仍然处于日期格式,因为行:
$this->for_date = date('d/M/Y', $this->for_date);
删除这一行,它应该工作。
然而,您仍然会遇到问题,例如日期格式或有人输入无效日期,例如30/02/2015
。
我建议创建一个单独的属性,例如for_date_display
,并为此添加一个日期规则。然后在beforeSave
中将该日期转换为时间戳,并将for_date
设置为如下值:
public $for_date_display;
...
public function afterFind() {
$this->for_date_display = 'Yii::t('app', Yii::$app->formatter->asDate($this->for_date))
}
public function beforeSave($insert = true) {
$this->for_date = strtotime($this->for_date_display);
return parent::beforeSave($insert);
}
我找到解决办法了。在模型搜索中(我的例子是AttendanceSearch.php
),找到规则函数并将for_date
从integer
行移动到safe
public function rules()
{
return [
[['id', 'user_id', 'project_id', 'commit_time', 'workload_type'], 'integer'],
[['comment', 'for_date'], 'safe'],
];
}