Question

我从数据库中获取数据的查询是，

select mailboxtoolno,DATE_FORMAT(maileventdate,'%d %b %Y') as 
date,DATE_FORMAT(maileventdate,'%H:%i:%s') as time,mailtype from 
domiciliation_mailbox where reg_id =".$regid." order by id DESC

如何将其更改为laravel eloquent model

我一直试图改变它，

 $timeline= mailbox::select('mailboxtoolno','DATE_FORMAT(maileventdate,"%d %b %Y") as date','DATE_FORMAT(maileventdate,"%H:%i:%s") as time','mailtype')
->where('reg_id', '=',$reg_id )
->paginate(10);

但得到了一个错误，

Unknown column 'DATE_FORMAT(maileventdate,"%d %b %Y")' in 'field list'

如何在laravel中以日期格式获取正确的值

Answer 1

Laravel不支持复杂的选择表达式，因此您必须使用Raw Expressions。试试这种方式：

    $timeline= mailbox::select('mailboxtoolno',DB::raw('DATE_FORMAT(maileventdate,"%d %b %Y") as date'),DB::raw('DATE_FORMAT(maileventdate,"%H:%i:%s") as time'),'mailtype')
->where('reg_id',$reg_id )
->orderBy('id','DESC')
->paginate(10);

为了在此查询中使用->orderBy()，您必须手动设置严格模式以通过comprobations忽略顺序。在database.php配置数据库连接数组参数中执行此操作：

'strict'    => true,
        'modes'     => [
            'STRICT_TRANS_TABLES',
            'NO_ZERO_IN_DATE',
            'NO_ZERO_DATE',
            'ERROR_FOR_DIVISION_BY_ZERO',
            'NO_AUTO_CREATE_USER',
            'NO_ENGINE_SUBSTITUTION'
        ],

或设置strict => false（我不会这样做）

Answer 2

请改用raw语句。例如：

$user = User::select(DB::raw('count(*) as user_count, status'))->where('status', '=', 'active');

实际上，Laravel在DateTime类型的字段中有mutator。因此，您可以正常选择它并在以后格式化。实施例;

$user = User::find(2);
$date = $user->created_at->format('d M Y'); // you can display it with any format you want with this way.

更多信息请参阅official documentation和this

日期格式在Laravel中查询

2 个答案: