laravel eloquent JOIN ON 多个条件

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【中文标题】laravel eloquent JOIN ON 多个条件【英文标题】：laravel eloquent JOIN ON multiple condition 【发布时间】：2015-05-17 18:02:13 【问题描述】：

我有一个表调用bank，它存储国家银行名称、代码等

我有一个表调用 site_banks，它存储银行帐户、银行持有人姓名和外国 bank_id，它们指的是 bank

我有一个表调用deposit_records，它存储了存款时间、存款收据、状态等以及引用site_banks的外国bank_id /p>

对于deposit_records，status字段，值为1表示成功

现在我想计算每个 site_banks

的总收入

但我不希望每一行 site_banks 调用一个查询来计数，这将有很多 sql 查询调用

所以我想我可以使用 join，下面是 site_banks 模型文件，scopeGroupAllSiteBanksByBranch 是我要收集所有 site_banks 的地方，加入SUM，返回数组

但即使我也有大约 200 个 site_banks 行，很多 deposit_records 行是 status = '1'，但是 scopeGroupAllSiteBanksByBranch 总是只返回一行，我不明白为什么，我复制 sql 查询在 mysql 中运行，结果相同，只返回 1 行。

<?php

use Illuminate\Database\Eloquent\SoftDeletingTrait;

class SiteBanks extends \LaravelBook\Ardent\Ardent 

    use SoftDeletingTrait;

    /**
     * The database table used by the model.
     *
     * @var string
     */
    protected $table = 'site_banks';

    protected $dates = ['deleted_at'];

    protected $fillable = array('bank_id', 'bank_name', 'bank_account', 'bank_holder', 'bank_username', 'bank_password', 'min_deposit', 'max_deposit', 'daily_max_deposit');

    public static $rules = array(
        'bank_id' => 'required|exists:banks,id',
        'bank_name' => 'required|min:1',
        'bank_account' => 'required|min:1|numberonly',
        'bank_holder' => 'required|min:1',
        'bank_username' => 'required|min:1',
        'bank_password' => 'required|min:1',
        'min_deposit' => 'required|fund',
        'max_deposit' => 'required|fund',
        'daily_max_deposit' => 'required|fund',
    );

    public function bank() 
        return $this->belongsTo('Banks', 'bank_id');
    

    public function transactions() 
        return $this->hasMany('Deposit', 'bank_id', 'id');
    

    public function transactionsDone() 
        return $this->hasMany('Deposit', 'bank_id', 'id')->where('status', '=', 1);
    

    public function transactionsBanned() 
        return $this->hasMany('Deposit', 'bank_id', 'id')->where('status', '=', 2);
    

    public function transactionsPending() 
        return $this->hasMany('Deposit', 'bank_id', 'id')->where('status', '=', 0);
    

    public function setBankPasswordAttribute($value) 
        $this->attributes['bank_password'] = Crypt::encrypt($value);
    

    public function getBankPasswordAttribute() 
        return Crypt::decrypt($this->attributes['bank_password']);
    

    public function scopeGetAvailableBanks($query) 
        $query->where(DB::raw("(current_deposit < daily_max_deposit) OR (current_deposit IS NULL) OR (DATE(last_deposit) != DATE(NOW()) OR last_deposit IS NULL) OR (DATE(last_deposit) = DATE(NOW()) && current_deposit < daily_max_deposit)"));
        $query->groupBy('bank_id')->orderBy('id', 'ASC');
    

    public function scopeGroupAllSiteBanksByBranch($query) 
        $bank = \SiteBanks::withTrashed()
            ->leftJoin('deposit_records', function($q) 
                $q->on('deposit_records.bank_id', '=', 'site_banks.id');
                $q->where('deposit_records.status', '=', 1, 'and');
            )
            ->select(array('site_banks.*', DB::raw('SUM(`deposit_records`.`deposit_amount`) AS `total_income`')))
            ->get();
//        $bank = \SiteBanks::withTrashed()->get();
        $bank->load('bank');

        $banks = array();
        foreach($bank as $key => $var) 
            $arr = array();

            $arr = $var->toArray();
            $arr['income'] = $var->total_income;
            $banks[$var->bank->bank_name][] = $arr;
        

        return $banks;

SQL 查询

选择site_banks.*, SUM(deposit_records.deposit_amount) AS total_income from site_banks left join deposit_records on deposit_records.bank_id = site_banks.id 和 deposit_records.status = '1'

var_dumpscopeGroupAllBanksByBranch

的结果 
array (size=1)
  'BANKNAME' => 
    array (size=1)
      0 => 
        array (size=20)
          'id' => int 1
          'bank_id' => int 4
          'bank_name' => string 'BANKNAME' (length=6)
          'bank_account' => string '123456789' (length=9)
          'bank_holder' => string 'ACCOUNT HOLDER NAME' (length=11)
          'bank_username' => string 'username' (length=8)
          'bank_password' => string 'hehethisispassword' (length=18)
          'min_deposit' => string '400.00' (length=6)
          'max_deposit' => string '9999.99' (length=7)
          'daily_max_deposit' => string '25500.00' (length=8)
          'last_deposit' => string '2015-02-05 03:04:00' (length=19)
          'current_deposit' => string '0.00' (length=4)
          'created_by' => int 1
          'updated_by' => null
          'created_at' => string '2015-02-04 08:21:16' (length=19)
          'updated_at' => string '2015-02-04 08:21:16' (length=19)
          'deleted_at' => null
          'total_income' => string '1722.00' (length=7)
          'bank' => 
            array (size=7)
              ...
          'income' => string '1722.00' (length=7)

【问题讨论】：

【参考方案1】：

这里是解决方案。您可以加入多个条件。

$query = self::select('websites.id','websites.website_url','cron_tracking_details.cron_type');
    $query = $query->leftJoin('cron_tracking_details',function ($join)
        $join->on(function ($queryone)
            $queryone->on('websites.id','=','cron_tracking_details.website_id');
            $queryone->where('cron_tracking_details.cron_type','=',"sitemap_create");
            $queryone->Where('cron_tracking_details.cron_date','<','2020-11-19');
        );
    );
    $query = $query->orderBy('cron_tracking_details.id','desc');
    $query = $query->limit(1);
    return $query = $query->get();

这段代码将像这样准备查询。

select `websites`.`id`, `websites`.`website_url`, `cron_tracking_details`.`cron_type` from `websites` left join `cron_tracking_details` on (`websites`.`id` = `cron_tracking_details`.`website_id` and `cron_tracking_details`.`cron_type` = ? and `cron_tracking_details`.`cron_date` < ?) order by `cron_tracking_details`.`id` desc limit 1

【讨论】：

【参考方案2】：

你可以按照代码解决你的问题

SQL 查询

LEFT JOIN bookings  
           ON rooms.id = bookings.room_type_id AND (bookings.arrival = ? OR bookings.departure = ? )

Laravel 多条件连接 

 ->join('bookings', function($join) use ($key1, $key2)

    $join->on('rooms.id', '=', 'bookings.room_type_id');
    $join->on(function($query) use ($key1, $key2)
            
             $query->on('bookings.arrival', '=', $key1);
             $query->orOn('departure', '=',$key2);
            );

)

【讨论】：

【参考方案3】： 
public function scopeGroupAllSiteBanksByBranch($query) 
        $bank = \SiteBanks::withTrashed()
            ->leftJoin('deposit_records', function($q) 
                $q->on('deposit_records.bank_id', '=', 'site_banks.id');
                $q->where('deposit_records.status', '=', 1, 'and');
            )
            ->select(array('site_banks.*', DB::raw('SUM(`deposit_records`.`deposit_amount`) AS `total_income`')))
            ->groupBy('site_banks.id')
            ->get();

我找到了解决方案，groupBy('site_banks.id') 然后一切就绪。

【讨论】：

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