Ruby on Rails中按location_id合并数组对象的技术实现需求

通过查询获取两组数组filled与empty，合并得到data数组后，生成的结果中存在相同location_id的对象。需要合并这些同location_id的对象，保留有效的location_name，并整合onhandcylynder和emptycylynder字段值，得到预期的JSON输出。

原生成数组代码 #

filled  =  Product.on_hand_location(pid).to_a
empty   =  Product.on_hand_location_empty_cylinder(pid).to_a
data    = filled + empty 

result = data.map{ |k| 
  { 
    details: { 
          location_id: k['location_id'],
          "location_name"=>k['location_name'],
          "onhandcylynder"=>k['onhand'] == nil ?  0 :  k['onhand'],
          "emptycylynder"=> k['emptyonhand'] == nil ?  0 :  k['emptyonhand']
       } ,
  }
}

respond_with [ onhand: result ]

当前输出JSON（需合并同location_id对象） #

[{
    "onhand": [{
            "details": {
                "location_id": 1,
                "location_name": "Capitol Drive",
                "onhandcylynder": "4.0",
                "emptycylynder": 0
            }
        },
        {
            "details": {
                "location_id": 2,
                "location_name": "SM City Butuan",
                "onhandcylynder": "5.0",
                "emptycylynder": 0
            }
        },
        {
            "details": {
                "location_id": 1,
                "location_name": null,
                "onhandcylynder": 0,
                "emptycylynder": "2.0"
            }
        }
    ]
  }]

期望输出JSON #

[{
    "onhand": [{
        "details": {
            "location_id": 1,
            "location_name": "Capitol Drive",
            "onhandcylynder": "4.0",
            "emptycylynder": 0
        }
    },
    {
        "details": {
            "location_id": 2,
            "location_name": "SM City Butuan",
            "onhandcylynder": "5.0",
            "emptycylynder": "2.0"
        }
    }
   ]
}]

解决方案 #

修改代码，通过group_by按location_id分组，再合并每组内的字段值：

filled = Product.on_hand_location(pid).to_a
empty = Product.on_hand_location_empty_cylinder(pid).to_a
data = filled + empty

# 按location_id分组，合并每组数据
grouped_data = data.group_by { |item| item['location_id'] }.values.map do |group|
  # 合并字段：保留有效location_name，累加数量字段
  merged = group.reduce do |acc, item|
    {
      'location_id' => acc['location_id'],
      'location_name' => acc['location_name'] || item['location_name'],
      'onhand' => (acc['onhand'] || 0).to_f + (item['onhand'] || 0).to_f,
      'emptyonhand' => (acc['emptyonhand'] || 0).to_f + (item['emptyonhand'] || 0).to_f
    }
  end

  # 转换为目标输出格式
  {
    details: {
      location_id: merged['location_id'],
      location_name: merged['location_name'],
      onhandcylynder: merged['onhand'].zero? ? 0 : merged['onhand'].to_s,
      emptycylynder: merged['emptyonhand'].zero? ? 0 : merged['emptyonhand'].to_s
    }
  }
end

respond_with [onhand: grouped_data]

关键修改说明 #

• 分组：使用group_by { |item| item['location_id'] }将相同location_id的数据归为一组
• 合并字段：
  • location_name：优先保留已有非空值，若当前值为空则取后续项的非空值
  • onhand和emptyonhand：转换为浮点数后累加，避免字符串拼接错误
• 格式转换：最后将合并后的数据转换为目标结构，处理数值为0时的显示逻辑

内容的提问来源于stack exchange，提问作者coolshox