Laravel：在 JOIN with Fluent 中使用复杂条件
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Laravel: Use complex condition in JOIN with Fluent
提问by jmadsen
For reasons of a complex schema & a library that requires either Fluent or Eloquent to be used (not just raw DB::query() ), I need to create:
由于复杂的模式和需要使用 Fluent 或 Eloquent 的库（不仅仅是原始 DB::query() ），我需要创建：
LEFT JOIN `camp_to_cabin`
ON `camper_to_cabin`.`cabin_id` = `camp_to_cabin`.`cabin_id`
AND `camp_to_cabin`.`camp_id` =1 OR `camp_to_cabin`.`camp_id` IS NULL
as the join clause; I've tried the callbacks & everything else I can think of, but cannot get the proper syntax to generate.
作为连接子句；我已经尝试了回调和我能想到的所有其他内容，但无法获得正确的语法来生成。
I have tried:
我试过了：
->left_join('camp_to_cabin', function ($join){
$join->on( 'camper_to_cabin.cabin_id', '=', 'camp_to_cabin.cabin_id')
$join->on( 'camp_to_cabin.camp_id', '=', 1)
$join->on( 'camp_to_cabin.camp_id', '=', null)
})
but it puts backticks around my 1 & null (I know the null bit isn't right - experimenting) that I can't get rid of; otherwise it looks pretty close
但它在我的 1 & null 周围加上了反引号（我知道 null 位不正确 - 正在试验）我无法摆脱；否则它看起来很接近
Any help?
有什么帮助吗？
TIA
TIA
Thanks, Phil - final answer is:
谢谢，菲尔 - 最终答案是：
->left_join('camp_to_cabin', function ($join) use ($id){
$join->on( 'camper_to_cabin.cabin_id', '=', 'camp_to_cabin.cabin_id');
$join->on( 'camper_to_cabin.cabin_id', '=', DB::raw($id));
$join->or_on( 'camper_to_cabin.cabin_id', 'IS', DB::raw('NULL'));
})
回答by Phill Sparks
Looks like you need to use DB::raw()otherwise ->on()expects two columns. Something like...
看起来你需要使用DB::raw()否则->on()需要两列。就像是...
->left_join('camp_to_cabin', function ($join){
$join->on( 'camper_to_cabin.cabin_id', '=', 'camp_to_cabin.cabin_id')
$join->on( 'camp_to_cabin.camp_id', '=', DB::raw(1))
$join->or_on( 'camp_to_cabin.camp_id', '=', DB::raw(null))
})
