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laravel 5.6-db select对数组上的成员函数join()的调用

laravel-5 laravel

0

Sasha · 技术社区 · 8 年前

我有以下疑问:

$clinics =  \DB::select('SELECT * FROM
(SELECT *,
 (100 * acos(cos(radians(' . $lat . ')) * cos(radians(lat)) *
 cos(radians(lng) - radians(' . $lng . ')) +
 sin(radians(' . $lat . ')) * sin(radians(lat))))
 AS distance
 FROM clinics) AS distances
 WHERE distance < 100
 ORDER BY distance')
 ->join('countries', 'clinics.country_id', '=', 'countries.id');

运行时出现以下错误:

Symfony \ Component \ Debug \ Exception \ FatalThrowableError (E_ERROR)
Call to a member function join() on array

有可能表演吗加入有这个问题吗? 我试着加上加入在select查询中,但我得到一个错误:

Column not found: 1054 Unknown column 'clinics.country_id' in 'on clause' (SQL: SELECT * FROM (SELECT *, (100 * acos(cos(radians(43.1557012)) * cos(radians(lat)) * cos(radians(lng) - radians(22.5856811)) + sin(radians(43.1557012)) * sin(radians(lat)))) AS distance FROM clinics) AS distances JOIN countries ON countries.id = clinics.country_id WHERE distance < 100 ORDER BY distance)

有疑问的问题是:

$clinics =  \DB::select('SELECT * FROM
    (SELECT *,
     (100 * acos(cos(radians(' . $lat . ')) * cos(radians(lat)) *
     cos(radians(lng) - radians(' . $lng . ')) +
     sin(radians(' . $lat . ')) * sin(radians(lat))))
     AS distance
     FROM clinics) AS distances
     JOIN countries ON countries.id = clinics.country_id
     WHERE distance < 100
     ORDER BY distance');

1 回复 | 直到 8 年前

1

2

Namoshek 8 年前

根据我所看到的,子查询不是必需的,您可以将查询转换成Laravel符号,如:

$clinics = Clinic::join('countries', 'countries.id', '=', 'clinics.country_id')
    ->selectRaw('clinics.*, countries.*, (
        100 * acos(
            cos(radians(?)) *
            cos(radians(lat)) *
            cos(radians(lng) - radians(?)) +
            sin(radians(?)) * sin(radians(lat))
        )
    ) as distance', [$lat, $lng, $lat])
    ->having('distance', '<', 100)
    ->orderBy('distance')
    ->get();

您可能需要明确列出要选择的列 countries 桌子,因为 countries.* 您将覆盖例如 id 你的专栏 clinics 表。我补充说 国家* 因为我想这就是你加入后想做的。