Question

我在mongodb中有三个收藏品，如

    employeeTitleTbl contains 

    {
    "_id": ObjectId("5a7c2027d2ccda04d0006070"),
    "EmployeeTitleName": "Librarian",
    "EmployeeTitleCode": "ji8766",
   ..
    }

 DepartmentsTbl contains

  {
   "_id": ObjectId("5a817402d2ccda13240015a5"),
   "DepartmentCode": "149",
   "DepartmentName": "Biology",
  ..
 }

 EmployeesTbl contains 

  {
   "_id": ObjectId("5a81957bd2ccda13240015a8"),
    "EmployeeNumber": "25789998",
   "JobTitle": ObjectId("5a7c2008d2ccda04d000606f"),
   "Department": ObjectId("5a8173e6d2ccda13240015a4"),
       ...
  }

我对mongodb很新，而且驱动程序也更新了3.6，因此很难在php中找到最新的聚合方法。所以我调整了以下代码，用于添加部门和员工职位的实际值，如＆＃34; EmployeeTitleName＆＃34;和＆＃34; DepartmentName＆＃34;在EmployeesTbl。

  public function fetchAll()
{         
        $pipeline = array(
                            array(
                                '$lookup' => array(
                                'from' => 'DepartmentsTbl',
                                'localField' => 'Department',
                                'foreignField' => '_id',
                                'as' => 'DepartmentDetails'
                                      )
                                  ),

                                array(
                                '$lookup' => array(
                                'from' => 'employeeTitleTbl',
                                'localField' => 'JobTitle',
                                'foreignField' => '_id',
                                'as' => 'JobTitleDetails'
                                      )
                                  ),
                         );

                 try
                 {                    

                    $cursor = $this->db->EmployeesTbl->aggregate($pipeline);

                 }

                 catch(Exception $e){

                 }

    return $cursor->toArray();
}

请用最新方法帮助我!!!

输出就像

array（2）{[0] =＆gt; object（MongoDB \ Model \ BSONDocument）＃29（1）{[＆＃34; storage＆＃34;：＆＃34; ArrayObject＆＃34;：private] =＆gt; array（25）{[＆＃34; _id＆＃34;] =＆gt; object（MongoDB \ BSON \ ObjectId）＃10（1）{[＆＃34; oid＆＃34;] =＆gt; string（24）＆＃34; 5a8176c0d2ccda13240015a6＆＃34; } [＆＃34; EmployeeNumber＆＃34;] =＆gt; string（8）＆＃34; 25799989＆＃34; .... ........... [＆＃34; DepartmentDetails＆＃34;] =＆GT; object（MongoDB \ Model \ BSONArray）＃27（1）{[＆＃34; storage＆＃34;：＆＃34; ArrayObject＆＃34;：private] =＆gt; array（0）{}} [＆＃34; JobTitleDetails＆＃34;] =＆gt; object（MongoDB \ Model \ BSONArray）＃28（1）{[＆＃34; storage＆＃34;：＆＃34; ArrayObject＆＃34;：private] =＆gt; array（0）{}}}} [1] = .........

Answer 1

您需要使用$ lookup来连接两个或更多表格，例如..

 db.employeeTitle.aggregate([{
  $lookup: {
    from: 'employees',
    localField: 'EmployeeTitleName',
    foreignField: '_id',
    as: 'employeeData'
  }
 },
 { $unwind: '$employeeData' },
 {
   $project: { profile: '$employeeData' }     
 }    
])

它连接两个表employeeTitle和员工，并为您提供有关员工的完整信息。

如何在php中进行mongo db聚合连接操作？

1 个答案: