php - Laravel 5.6 Eloquent whereDate 和 WhereRaw 错误

Question

我正在尝试查找今天创建的所有记录，并以列 functional_id 的 1 开头。

我得到以下语法错误:

SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'WHERE functional_id LIKE "1%"' at line 1 (SQL: select count(*) as aggregate from image_requests where date(created_at) = 2019-01-24 00:00:00 and WHERE functional_id LIKE "1%")

我的 Eloquent 代码:

$counter = ImageRequest::whereDate('created_at', $date)->whereRaw('WHERE functional_id LIKE "1%"')->count();

Answer 1

这应该有效:

$counter = ImageRequest::whereDate('created_at', $date)
    ->where('functional_id', 'like', '1%')
    ->count();

我认为问题在于额外的 "。但是，在这种情况下不需要使用 whereRaw()。where() 绰绰有余，而且不太容易出现这样的错误 :)

编辑:根据您的评论，如果您想要检查第十二个字符是否为 1，以下应该可行:

$counter = ImageRequest::whereDate('created_at', $date)
    ->whereRaw('SUBSTRING(funtional_id, 12, 1) = 1')
    ->count();