来自平面JSON的PHP嵌套JSON

Question

我有一个数据库查询,它为我提供了一些员工数据的输出.我想使用此数据传递给生成组织结构图的插件.我正在提取的JSON对象中有一些字段是:

FirstName
LastName
EmployeeID
ManagerEmployeeID
Manager Name

数据作为扁平JSON对象返回,在层次结构中员工及其经理之间没有嵌套或关联.

由于我无法更改源数据(数据库查询)的输出,我试图找出一种嵌套数据的方法,以便JSON输出成为嵌套输出.

我的目标是获取此数组并基于ManagerID和EmployeeID将其嵌套,以便我可以创建树层次结构.

示例数据:

•   Tom Jones
   o    Alice Wong
   o    Tommy J.
•   Billy Bob
   o    Rik A.
     ?  Bob Small
     ?  Small Jones
   o    Eric C.

我的平面数据示例:

    {
        "FirstName": "Tom"
        "LastName": "Jones"
        "EmployeeID": "123"
        "ManagerEmployeeID": ""
        "Manager Name": ""
    },
    {
        "FirstName": "Alice"
        "LastName": "Wong"
        "EmployeeID": "456"
        "ManagerEmployeeID": "123"
        "Manager Name": "Tom Jones"
    },
    {
        "FirstName": "Tommy"
        "LastName": "J."
        "EmployeeID": "654"
        "ManagerEmployeeID": "123"
        "Manager Name": "Tom Jones"
    },
    {
        "FirstName": "Billy"
        "LastName": "Bob"
        "EmployeeID": "777"
        "ManagerEmployeeID": ""
        "Manager Name": ""
    },
    {
        "FirstName": "Rik"
        "LastName": "A."
        "EmployeeID": "622"
        "ManagerEmployeeID": "777"
        "Manager Name": "Billy Bob"
    },
    {
        "FirstName": "Bob"
        "LastName": "Small"
        "EmployeeID": "111"
        "ManagerEmployeeID": "622"
        "Manager Name": "Rik A."
    },
    {
        "FirstName": "Small"
        "LastName": "Jones"
        "EmployeeID": "098"
        "ManagerEmployeeID": "622"
        "Manager Name": "Rik A"
    },
    {
        "FirstName": "Eric"
        "LastName": "C."
        "EmployeeID": "222"
        "ManagerEmployeeID": "777"
        "Manager Name": "Billy Bob"
    }

示例所需输出:

[
  {
    "FirstName": "Tom",
    "LastName": "Jones",
    "EmployeeID": "123",
    "ManagerEmployeeID": "",
    "Manager Name": "",
    "employees": [
      {
        "FirstName": "Alice",
        "LastName": "Wong",
        "EmployeeID": "456",
        "ManagerEmployeeID": "123",
        "Manager Name": "Tom Jones"
      },
      {
        "FirstName": "Tommy",
        "LastName": "J.",
        "EmployeeID": "654",
        "ManagerEmployeeID": "123",
        "Manager Name": "Tom Jones"
      }
    ]
  },
  {
    "FirstName": "Billy",
    "LastName": "Bob",
    "EmployeeID": "777",
    "ManagerEmployeeID": "",
    "Manager Name": "",
    "employees": [
      {
        "FirstName": "Rik",
        "LastName": "A.",
        "EmployeeID": "622",
        "ManagerEmployeeID": "777",
        "Manager Name": "Billy Bob",
        "employees": [
          {
            "FirstName": "Bob",
            "LastName": "Small",
            "EmployeeID": "111",
            "ManagerEmployeeID": "622",
            "Manager Name": "Rik A."
          },
          {
            "FirstName": "Small",
            "LastName": "Jones",
            "EmployeeID": "098",
            "ManagerEmployeeID": "622",
            "Manager Name": "Rik A"
          }
        ]
      },
      {
        "FirstName": "Eric",
        "LastName": "C.",
        "EmployeeID": "222",
        "ManagerEmployeeID": "777",
        "Manager Name": "Billy Bob"
      }
    ]
  }
]

基本上我试图使用EmployeeID和ManagerEmployeeID作为两者之间的链接从平面对象创建嵌套的JSON输出.

用PHP解决这类问题的最佳方法是什么？

赏金更新:

以下是该问题的测试用例:https://eval.in/private/4b0635c6e7b059

您将看到名称中的最后一条记录Issue Here未显示在结果集中.它有一个managerID匹配根节点,应该在"Tom Jones" employees数组中.

Answer 1

我有以下实用程序类来完全按照您的需要进行操作.

class NestingUtil
{
    /**
     * Nesting an array of records using a parent and id property to match and create a valid Tree
     *
     * Convert this:
     * [
     *   'id' => 1,
     *   'parent'=> null
     * ],
     * [
     *   'id' => 2,
     *   'parent'=> 1
     * ]
     *
     * Into this:
     * [
     *   'id' => 1,
     *   'parent'=> null
     *   'children' => [
     *     'id' => 2
     *     'parent' => 1,
     *     'children' => []
     *    ]
     * ]
     *
     * @param array  $records      array of records to apply the nesting
     * @param string $recordPropId property to read the current record_id, e.g. 'id'
     * @param string $parentPropId property to read the related parent_id, e.g. 'parent_id'
     * @param string $childWrapper name of the property to place children, e.g. 'children'
     * @param string $parentId     optional filter to filter by parent
     *
     * @return array
     */
    public static function nest(&$records, $recordPropId = 'id', $parentPropId = 'parent_id', $childWrapper = 'children', $parentId = null)
    {
        $nestedRecords = [];
        foreach ($records as $index => $children) {
            if (isset($children[$parentPropId]) && $children[$parentPropId] == $parentId) {
                unset($records[$index]);
                $children[$childWrapper] = self::nest($records, $recordPropId, $parentPropId, $childWrapper, $children[$recordPropId]);
                $nestedRecords[] = $children;
            }
        }

        return $nestedRecords;
    }
}

使用您的代码:

$employees = json_decode($flat_employees_json, true);
$managers = NestingUtil::nest($employees, 'EmployeeID', 'ManagerEmployeeID', 'employees');
print_r(json_encode($managers));

输出:

[
  {
    "FirstName": "Tom",
    "LastName": "Jones",
    "EmployeeID": "123",
    "ManagerEmployeeID": "",
    "Manager Name": "",
    "employees": [
      {
        "FirstName": "Alice",
        "LastName": "Wong",
        "EmployeeID": "456",
        "ManagerEmployeeID": "123",
        "Manager Name": "Tom Jones",
        "employees": []
      },
      {
        "FirstName": "Tommy",
        "LastName": "J.",
        "EmployeeID": "654",
        "ManagerEmployeeID": "123",
        "Manager Name": "Tom Jones",
        "employees": []
      }
    ]
  },
  {
    "FirstName": "Billy",
    "LastName": "Bob",
    "EmployeeID": "777",
    "ManagerEmployeeID": "",
    "Manager Name": "",
    "employees": [
      {
        "FirstName": "Rik",
        "LastName": "A.",
        "EmployeeID": "622",
        "ManagerEmployeeID": "777",
        "Manager Name": "Billy Bob",
        "employees": [
          {
            "FirstName": "Bob",
            "LastName": "Small",
            "EmployeeID": "111",
            "ManagerEmployeeID": "622",
            "Manager Name": "Rik A.",
            "employees": []
          },
          {
            "FirstName": "Small",
            "LastName": "Jones",
            "EmployeeID": "098",
            "ManagerEmployeeID": "622",
            "Manager Name": "Rik A",
            "employees": []
          }
        ]
      },
      {
        "FirstName": "Eric",
        "LastName": "C.",
        "EmployeeID": "222",
        "ManagerEmployeeID": "777",
        "Manager Name": "Billy Bob",
        "employees": []
      }
    ]
  }
]

Edit1:修复以避免忽略某些员工

如果最后一项是具有有效经理的员工但经理不在列表中,则会被忽略,因为应该位于哪里？,它不是根,但没有有效的经理.

为避免这种情况,请在实用程序中的return语句之前添加以下行.

if (!$parentId) {
    //merge residual records with the nested array
    $nestedRecords = array_merge($nestedRecords, $records);
}

return $nestedRecords;

Edit2:将实用程序更新为PHP5.6

在PHP7中进行了一些测试后,该实用程序在php7.0中工作正常但在php5.6中没有,我不知道为什么,但是在数组引用和未设置中.我更新实用程序代码以使用php5.6和您的用例.

 public static function nest($records, $recordPropId = 'id', $parentPropId = 'parent_id', $childWrapper = 'children', $parentId = null)
    {
        $nestedRecords = [];
        foreach ($records as $index => $children) {
            if (isset($children[$parentPropId]) && $children[$parentPropId] == $parentId) {
                $children[$childWrapper] = self::nest($records, $recordPropId, $parentPropId, $childWrapper, $children[$recordPropId]);
                $nestedRecords[] = $children;
            }
        }

        if (!$parentId) {
            $employeesIds = array_column($records, $recordPropId);
            $managers = array_column($records, $parentPropId);
            $missingManagerIds = array_filter(array_diff($managers, $employeesIds));
            foreach ($records as $record) {
                if (in_array($record[$parentPropId], $missingManagerIds)) {
                    $nestedRecords[] = $record;
                }
            }
        }

        return $nestedRecords;
    }