Php 将查询转换为Laravel

我从中得到这个查询，有人能把它转换成laravel查询吗

SELECT * FROM Position ORDER BY 
CASE  WHEN position_value = 1 THEN created_at END ASC,
CASE WHEN position_value = 2 THEN created_at END DESC

我已经这样做了，但仍然收到错误消息

$position = Position::orderBy(DB::raw('(CASE WHEN position_value = 1 THEN created_at END asc, CASE WHEN position_value = 2 THEN created_at END desc)'))->paginate(29);

SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'asc, CASE WHEN position_value = 2 THEN created_at END desc) asc limit 29 offset 0' at line 1 (SQL: select * from `position` order by (CASE WHEN position_value = 1 THEN created_at END asc, CASE WHEN position_value = 2 THEN created_at END desc) asc limit 29 offset 0)

这是错误消息

$position = Position::orderBy(DB::raw('(CASE WHEN position_value = 1 THEN created_at END asc, CASE WHEN position_value = 2 THEN created_at END desc)'))->paginate(29);

SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'asc, CASE WHEN position_value = 2 THEN created_at END desc) asc limit 29 offset 0' at line 1 (SQL: select * from `position` order by (CASE WHEN position_value = 1 THEN created_at END asc, CASE WHEN position_value = 2 THEN created_at END desc) asc limit 29 offset 0)

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谢谢

删除多余的括号并使用

orderByRaw

方法

$position = Position::orderByRaw('CASE WHEN position_value = 1 THEN created_at END asc, CASE WHEN position_value = 2 THEN created_at END desc')
           ->paginate(29);

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为什么否决我的noob问题（尝试删除括号？问题似乎是Elount希望排序顺序（asc，desc）作为orderBy方法的第二个参数传递，就像这样

orderBy（'created_at'，'desc'）；

。它起作用了！我正在删除额外的括号，现在我在合并orderBy（DB:：raw（'FIELD（position，1,2）'），'asc'时遇到了问题）这感谢我使用$position=position:：orderByRaw（'CASE WHEN position\u value=1然后在asc末尾创建，CASE WHEN position\u value=2然后在末尾创建'）->paginate（29）；我删除了最后一个'desc'，因为它有自己的默认值。如何以正确的方式进行注释，语法从不突出显示