Question

我想加入两个包含两列的表，它在phpmyadmin工作。但它没有与laravel合作。那么如何在laravel结构中转换它。

SELECT 
    `j`.*,
    `ts`.*,
    `ps`.*,
    `pn`.*,
    `c`.*,
    `planner`.*,
    `ps`.`name` AS `section_name`,
    `ps`.`id` AS `section_id`,
    `planner`.`date` AS `planner_date`,
    `pn`.`id` AS `planner_note_id`
FROM
    `planner`
        INNER JOIN
    `timeslots` AS `ts` ON `ts`.`id` = `planner`.`timeslot_id`
        INNER JOIN
    `planner_sections` AS `ps` ON `ps`.`id` = `planner`.`planner_section_id`
        LEFT JOIN
    `planner_notes` AS `pn` ON `pn`.`date` = `planner`.`date` and `pn`.`contract_id` = `planner`.`contract_id`
        INNER JOIN
    `contracts` AS `c` ON `c`.`id` = `planner`.`contract_id`
        INNER JOIN
    `jobs` AS `j` ON `j`.`contract_id` = `c`.`id`
WHERE
    `c`.`id` = 57        
        AND MONTH(`planner`.`date`) = 6
        AND `planner`.`date` >= '2017-06-13'
        AND `planner`.`deleted_at` IS NULL
ORDER BY `planner`.`date` ASC

但是当我把它放在laravel查询结构中时，它无法正常工作。

$planner_list = Planner::join("timeslots as ts","ts.id","planner.timeslot_id")
                ->join("planner_sections as ps","ps.id","planner.planner_section_id")
                ->leftJoin("planner_notes as pn","pn.date","planner.date")
                ->leftJoin("planner_notes as pn","pn.contract_id","planner.contract_id")
                ->join("contracts as c","c.id","planner.contract_id")
                ->join("jobs as j","j.contract_id","c.id")
                ->where('c.id','=',$contract_id)
                ->where('planner.date','>=',$current_date)
                ->whereMonth('planner.date',$current_month)
                ->select(['j.*','ts.*','ps.*','pn.*','c.*','planner.*','ps.name as section_name','ps.id as section_id','planner.date as planner_date','pn.id as planner_note_id'])
                ->orderBy('planner.date','ASC')
                ->orderBy('planner.timeslot_id','ASC')
                ->orderBy('section_id','ASC')
                ->get();

以上的谈话对我不起作用。

它也没有合作，

->leftJoin("planner_notes as pn","pn.date","planner.date","pn.contract_id","planner.contract_id")

修改： - 不，我没有收到任何错误。它只是考虑到最后的条件。

ON `pn`.`contract_id` = `planner`.`contract_id`

我还有一个困惑，如果我写两个leftjoin然后On条件可以覆盖或不覆盖？

EDIT-2： - 它只返回最后一个条件。查询日志查询是： -

SELECT 
    `j`.*,
    `ts`.*,
    `ps`.*,
    `pn`.*,
    `c`.*,
    `planner`.*,
    `ps`.`name` AS `section_name`,
    `ps`.`id` AS `section_id`,
    `planner`.`date` AS `planner_date`,
    `pn`.`id` AS `planner_note_id`
FROM
    `planner`
        INNER JOIN
    `timeslots` AS `ts` ON `ts`.`id` = `planner`.`timeslot_id`
        INNER JOIN
    `planner_sections` AS `ps` ON `ps`.`id` = `planner`.`planner_section_id`
        LEFT JOIN
    `planner_notes` AS `pn` ON `pn`.`contract_id` = `planner`.`contract_id`
        INNER JOIN
    `contracts` AS `c` ON `c`.`id` = `planner`.`contract_id`
        INNER JOIN
    `jobs` AS `j` ON `j`.`contract_id` = `c`.`id`
WHERE
    `c`.`id` = 57        
        AND MONTH(`planner`.`date`) = 6
        AND `planner`.`date` >= '2017-06-13'
        AND `planner`.`deleted_at` IS NULL
ORDER BY `planner`.`date` ASC

Answer 1

Laravel加入多个条件

LEFT JOIN
    `planner_notes` AS `pn` ON `pn`.`date` = `planner`.`date` and `pn`.`contract_id` = `planner`.`contract_id`

它相当于：

->leftJoin('planner_notes as pn', function($join){
            $join->on('pn.date', '=', 'planner.date');
            $join->on('pn.contract_id','=','planner.contract_id'); 
        })

laravel.com参考：method_on

在联接中添加“on”子句。

On子句可以链接，例如

$ join-＆gt; on（'contacts.userid'，'='，'users.id'） - ＆gt; on（'contacts.infoid'，'='，'info.id'）

将生成以下SQL：

on contacts.user_id = users.id和contacts.info_id = info.id

Answer 2

你可以做到

->leftJoin('planner_notes as pn', function($join){
            $join->on('pn.date', '=', 'planner.date');
            $join->on('pn.contract_id','=','planner.contract_id');    
        })

或

将表的别名更改为

->leftJoin("planner_notes as pn","pn.date",'=',"planner.date")
->leftJoin("planner_notes as pnotes","pnotes.contract_id",'=',"planner.contract_id")

左连接2列

2 个答案: