Question

以下sql语句在mysql查询中有效，但在尝试在Laravel中执行以下操作时会出现错误SQLSTATE[42000]: Syntax error or access violation: 1055 'dashboard.m.id' isn't in GROUP BY (SQL: SELECT m.id, ....：

$this->builder = DB::select
 ("SELECT m.id, m.member_num, m.first_name, m.last_name, COUNT(s.member_id) AS members 
FROM sessions AS s 
LEFT JOIN members AS m ON s.member_id = m.id 
WHERE s.created_at >= :dateFrom 
AND s.created_at < :dateTo 
GROUP BY s.member_id",            
['dateFrom' => "date('2017-11-01')", 
'dateTo' => "date('2017-12-01')"]);

任何人都可以看到问题，因为它在mysql中正常工作。

我只是想获得sessions的计数，并在member

中显示member_id个sessions的详细信息

Answer 1

您在Laravel中看到的错误与您选择未出现在SELECT * FROM Student LEFT JOIN fees ON student.Id = fees.StudentID where fees.StudentID is null; //intead of * you have to replace columns you want // for your question text i think ID in student is primary key and StudentId // in fees is foreign key so my answer is based on that only子句中的非聚合列有关。它在MySQL中工作的原因显然是因为MySQL在关闭GROUP BY的松散模式下运行，这可以容忍这种非标准语法。我怀疑你打算沿着这些方向使用查询：

ONLY_FULL_GROUP_BY

由于您正在聚合SELECT m.id, m.member_num, m.first_name, m.last_name, s.member_count FROM members m LEFT JOIN ( SELECT member_id, COUNT(*) AS member_count FROM sessions WHERE s.created_at >= :dateFrom AND s.created_at < :dateTo GROUP BY member_id ) s ON m.id = s.member_id表中的成员，并且仅计算该表中的字段，因此我建议在单独的子查询中进行聚合。然后，将其加入sessions以检索完整的成员记录。

Answer 2

要克服ONLY_FULL_GROUP_BY issue，您可以使用ANY_VALUE()来引用非聚合列。

DB::table('sessions')
    ->join('members', 'sessions.member_id', '=', 'members.id')
    ->select([
        'members.member_num',
        'members.first_name'
        'members.last_name',
        DB::raw('COUNT(sessions.member_id) AS members'),
        DB::raw('ANY_VALUE(members.id) AS id'),
    ])
    ->where([
        ['sessions.created_at', '>=', date('2017-11-01')],
        ['sessions.created_at', '<', date('2017-12-01')]
    ])
    ->groupBy('sessions.member_id')
    ->get();

Laravel原始选择错误;适用于mysql

2 个答案: