0

日期问题更多。

基本上我试图用 Zend 的一些日期时间值填充一个表。我要填充的列的数据类型是“日期时间”。但是,无论我尝试使用哪个常量,该列始终填充为零。我试过了Zend_Date::ISO_8601,,无济于事。请帮忙。Zend_Date::DATETIMEZend_Date::now()

这是代码:

           if(!$exists){
                $date_time = Zend_Date::DATETIME;
                /*
                 * Create the new site record
                 */
                $site_row = $site_table->createRow();
                //if(isset($n_r['id']))    $row->id  = $n_r['id'];
                $site_row->name         = $n_r['name'];
                $site_row->active       = 1;

                $site_id = $site_table->total() + 1;
                $address_id = $address_table->total() + 1;

                $site_row->id       = $site_id;
                $site_row->address_id       = $address_id;

                /*
                 * Create the address record to go with it
                 */
                $address_row = $address_table->createRow();
                $address_row->postcode = $n_r['postcode'];
                $address_row->address_line1 = $n_r['name'];
                $address_row->start_date = $date_time;

                $address_row->id = $address_id;
                $address_row->save();
                $array[] = $site_row->toArray();
                $site_row->save();

                /*
                 * Create the new site2address record to go with it
                 */
                $site2address_row = $site2address_table->createRow();
                $site2address_row->site_id = $site_id;
                $site2address_row->address_id = $address_id;
                $site2address_row->start_date = $date_time;
            }else{
                $array[] = $exists;
            }
4

2 回答 2

1

mySQL 期望DATETIME列的输入是

YYYY-MM-DD HH:MM:SS

独立于您使用的语言环境。我看不到 Zend 常量,你可能需要自己格式化。

于 2010-09-30T11:26:33.170 回答
0

您可以使用 NOW() 之类的 mysql 函数以获得更方便的方式。如果你在下面的 mysql 代码可能对你有用。

  $address_row->start_date = new Zend_Db_Expr('NOW()');
于 2010-09-30T13:08:00.230 回答