如何使用 Laravel Query Builder 仅在一个 table 中创建不同的连接?
How to make a distinct in only one table of a join using Laravel Query Builder?
我有这个查询...我只想在 (tbl_holdings)
中进行区分
这是我的查询
$holdings = DB::table('tbl_perimetros')
->distinct()
->join('tbl_holdings', 'tbl_holdings.id', '=', 'tbl_perimetros.holdings_id')
->get();
使用groupBy
代替distinct
:
$holdings = DB::table('tbl_perimetros')
->groupBy('tbl_perimetros.holdings_id')
->get();
移除 ONLY_FULL_GROUP_BY:
将 modes
添加到 config/database。php
'mysql' => [
...
'modes' => [
'STRICT_ALL_TABLES',
'ERROR_FOR_DIVISION_BY_ZERO',
'NO_ZERO_DATE',
'NO_ZERO_IN_DATE',
'NO_AUTO_CREATE_USER',
],
],
我有这个查询...我只想在 (tbl_holdings)
中进行区分这是我的查询
$holdings = DB::table('tbl_perimetros')
->distinct()
->join('tbl_holdings', 'tbl_holdings.id', '=', 'tbl_perimetros.holdings_id')
->get();
使用groupBy
代替distinct
:
$holdings = DB::table('tbl_perimetros')
->groupBy('tbl_perimetros.holdings_id')
->get();
移除 ONLY_FULL_GROUP_BY:
将 modes
添加到 config/database。php
'mysql' => [
...
'modes' => [
'STRICT_ALL_TABLES',
'ERROR_FOR_DIVISION_BY_ZERO',
'NO_ZERO_DATE',
'NO_ZERO_IN_DATE',
'NO_AUTO_CREATE_USER',
],
],