來自平面JSON的PHP嵌套JSON

Question

我有一個數據庫查詢，它為我提供了一些員工數據的輸出。 我想使用此數據傳遞給生成組織結構圖的插件。 我正在提取的JSON對象中有一些字段是：

FirstName
LastName
EmployeeID
ManagerEmployeeID
Manager Name

數據作為扁平JSON對象返回，在層次結構中員工及其經理之間沒有嵌套或關聯。

由於我無法更改源數據（數據庫查詢）的輸出，我試圖找出一種嵌套數據的方法，以便JSON輸出成為嵌套輸出。

我的目標是獲取此數組並基於ManagerID和EmployeeID將其嵌套，以便我可以創建樹層次結構。

示例數據：

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

我的平面數據示例：

    {
        "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"
    }

示例所需輸出：

[
  {
    "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"
      }
    ]
  }
]

實際上，我試圖使用EmployeeID和ManagerEmployeeID作為兩者之間的鏈接，從扁平對象創建嵌套的JSON輸出。

用PHP解決這類問題的最佳方法是什么？

賞金更新：

以下是該問題的測試用例： https ： //eval.in/private/4b0635c6e7b059

您將看到名為Issue Here最后一條記錄未顯示在結果集中。 這有一個與根節點匹配的managerID ，應該在“Tom Jones”的employees數組中。

Answer 1

我有以下實用程序類來完全按照您的需要進行操作。

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;
    }
}

使用您的代碼：

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

輸出：

[
  {
    "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：修復以避免忽略某些員工

如果最后一項是具有有效經理的員工但經理不在列表中，則會被忽略，因為應該位於哪里？，它不是根，但沒有有效的經理。

為避免這種情況，請在實用程序中的return語句之前添加以下行。

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

return $nestedRecords;

Edit2：將實用程序更新為PHP5.6

在PHP7中進行一些測試后，該實用程序在php7.0中工作正常，但在php5.6中沒有，我不知道為什么，但是在數組引用和未設置中。 我更新實用程序代碼以使用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;
    }

Answer 2

這是從你的小提琴直接翻譯到PHP：

function makeTree($data, $parentId){
    return array_reduce($data,function($r,$e)use($data,$parentId){
        if(((empty($e->ManagerEmployeeID)||($e->ManagerEmployeeID==(object)[])) && empty($parentId)) or ($e->ManagerEmployeeID == $parentId)){
            $employees = makeTree($data, $e->EmployeeID);
            if($employees) $e->employees = $employees;
            $r[] = $e;
        }
        return $r;
    },[]);
}

它與您的測試輸入一起正常工作。 請參閱https://eval.in/private/ee9390e5e8ca95 。

用法示例：

$nested = makeTree(json_decode($json), '');
echo json_encode($nested, JSON_PRETTY_PRINT);

@rafrsr解決方案非常靈活，但問題是foreach的unset() 。 它在迭代時修改數組，這是一個壞主意。 如果刪除unset() ，它可以正常工作。

Answer 3

你可以在這里使用遞歸的神奇力量。 請參考以下示例。 正如您在此處所看到的那樣，正在調用getTreeData 。

function getTreeData($data=[], $parent_key='', $self_key='', $key='')
{
    if(!empty($data))
    {

        $new_array = array_filter($data, function($item) use($parent_key, $key) {

            return $item[$parent_key] == $key;
        });

        foreach($new_array as &$array)
        {
            $array["employees"] = getTreeData($data, $parent_key, $self_key, $array[$self_key]);

            if(empty($array["employees"]))
            {
                unset($array["employees"]);
            }
        }

        return $new_array;
    }
    else
    {
        return $data;
    }
}

$employees = json_decode($employees_json_string, true);

$employees_tree = getTreeData($employees, "ManagerEmployeeID", "EmployeeID");

Answer 4

這是一個巧妙的技巧，利用了對象通過引用傳遞的事實。

array_column首先用於通過EmployeeID重新生成輸入數組。 然后使用array_map迭代條目並將每個項目重新分配給相應管理器的employees數組。 由於條目是對象（stdClass） ，因此正在修改相同的輸入項。

通過引用添加到employees數組的項目在根級別為空。 然后使用array_filter刪除這些空條目。 最后，通過array_values刪除臨時EmployeeID鍵控。

$input = json_decode($flat_employees_json, true);
$input = array_column($input, null, "EmployeeID");
$input = array_map(function ($entry) {
    return (object) $entry;
}, $input);
$output = array_values(array_filter(array_map(function ($entry) use ($input) {
    if (!empty($entry->ManagerEmployeeID)) {
        $input[$entry->ManagerEmployeeID]->employees[] = $entry;
        return null;
    }
    return $entry;
}, $input)));
echo json_encode($output, JSON_PRETTY_PRINT);

在線嘗試。

PHP 7支持array_column的對象數組，因此輸入初始化可以簡化為：

$input = array_column(json_decode($flat_employees_json), null, "EmployeeID");