我如何在laravel
问题描述:
这里结合两个或多个变量是一个例子我如何在laravel
变量1:
$encodedtoday = MemberProfile::whereDate('created_at', '=', $today)
->where('encoded_by', 'Mark')->count();
变量2:
$encodedtoday = MemberProfile::whereDate('created_at', '=', $today)
->where('encoded_by', 'Michael')->count();
变量3:
$encodedtoday = MemberProfile::whereDate('created_at', '=', $today)
->where('encoded_by', 'John')->count();
我知道这可以在数组或索姆完成ething。请告诉我
答
您可以通过名称编码创建数组,然后传递到其中的条件如下:
$encoded_by = ['Mark','Michael','John'];
$encodedtoday = MemberProfile::whereDate('created_at', '=', $today)
->whereIn('encoded_by', $encoded_by)->count();
return $encodedtoday;
答
尝试这样的:
$encodedtoday = MemberProfile::whereDate('created_at', '=', $today)
->whereIn('encoded_by', ['Mark','Michael','John'])->count();
希望这有助于
答
您可以定义 然后不断增加计数所有这样的:
$encodedtoday += MemberProfile::whereDate('created_at', '=', $today)
->where('encoded_by', 'Mark')->count();
$encodedtoday += MemberProfile::whereDate('created_at', '=', $today)
->where('encoded_by', 'Michael')->count();
$encodedtoday += MemberProfile::whereDate('created_at', '=', $today)
->where('encoded_by', 'John')->count();
所以在这里,$encodedtoday
将有最后的计数。
你举的例子恰好是东西,可以很容易地通过查询生成器来解决。它是唯一的用例还是更一般的问题? – apokryfos