JSON 嵌套对象减少或展平

米哈尔·库比亚克

我需要将这个 JSON 扁平化，没有巢，只有贝类，然后是它的名称等。提前致谢！我尝试了一些东西，所以也许可以减少或映射，但我对解决问题的那些东西还不够了解，你们可以分享针对此类问题的任何在线资源吗？我用这些工作很多，掌握起来会很得心应手！

"data": [
        {
            "name": "Batch1",
            "description": "seed",
            "age": 2,
            "quantity": 1000,
            "source": "Hatchery",
            "hatchery": "robs hatchery",
            "location": "dingle",
            "shellfish": {
                "name": "oyster"
            },
            "grade_list": {
                "name": "Grade0"
            },
            "stock_type": {
                "name": "seeds"
            }
        },
        {
            "name": "Batch2",
            "description": "Batch2",
            "age": 20,
            "quantity": 15700,
            "source": "aka",
            "hatchery": "aka",
            "location": "dingle",
            "shellfish": {
                "name": "oyster"
            },
            "grade_list": {
                "name": "Grade1"
            },
            "stock_type": {
                "name": "mature"
            }
        },
        {
            "name": "5555",
            "description": "45",
            "age": 1,
            "quantity": 134,
            "source": "fhh",
            "hatchery": "hfhj",
            "location": "garden",
            "shellfish": {
                "name": "oyster"
            },
            "grade_list": {
                "name": "Grade0"
            },
            "stock_type": {
                "name": "seeds"
            }
        }
    ]

编辑：我需要这个没有巢，看起来像这样：

"data": [
        {
            "name": "Batch1",
            "description": "seed",
            "age": 2,
            "quantity": 1000,
            "source": "Hatchery",
            "hatchery": "robs hatchery",
            "location": "dingle",
            "shellfish": {
                "name": "oyster"
            },
            "grade_list": {
                "name": "Grade0"
            },
            "stock_type": {
                "name": "seeds"
            }
        },
        {
            "name": "Batch2",
            "description": "Batch2",
            "age": 20,
            "quantity": 15700,
            "source": "aka",
            "hatchery": "aka",
            "location": "dingle",
            "shellfish": {
                "name": "oyster"
            },
            "grade_list": {
                "name": "Grade1"
            },
            "stock_type": {
                "name": "mature"
            }
        },
        {
            "name": "5555",
            "description": "45",
            "age": 1,
            "quantity": 134,
            "source": "fhh",
            "hatchery": "hfhj",
            "location": "garden",
            "shellfish": "oyster",
            "grade_list": "Grade0",
            "stock_type": "seeds"
        }
    ]

马蒂亚斯·格诺·阿佐里尼

如果您想让数组中的所有元素的这个过程更加自动化，您可以使用 map 来更改所有具有 prop 的元素name。

const response = {
  "data": [
    {
      "name": "Batch1",
      "description": "seed",
      "age": 2,
      "quantity": 1000,
      "source": "Hatchery",
      "hatchery": "robs hatchery",
      "location": "dingle",
      "shellfish": {
        "name": "oyster"
      },
      "grade_list": {
        "name": "Grade0"
      },
      "stock_type": {
        "name": "seeds"
      }
    },
    {
      "name": "Batch2",
      "description": "Batch2",
      "age": 20,
      "quantity": 15700,
      "source": "aka",
      "hatchery": "aka",
      "location": "dingle",
      "shellfish": {
        "name": "oyster"
      },
      "grade_list": {
        "name": "Grade1"
      },
      "stock_type": {
        "name": "mature"
      }
    },
    {
      "name": "5555",
      "description": "45",
      "age": 1,
      "quantity": 134,
      "source": "fhh",
      "hatchery": "hfhj",
      "location": "garden",
      "shellfish": {
        "name": "oyster"
      },
      "grade_list": {
        "name": "Grade0"
      },
      "stock_type": {
        "name": "seeds"
      }
    }
  ]
}

const result = response.data.map((element) => {
  const keysWithNameProp = Object.keys(element).filter(key => element[key].name !== undefined);
  const copyOfElement = {...element};
  keysWithNameProp.forEach(prop => {
    copyOfElement[prop] = element[prop].name;
  })
  return copyOfElement;
});

console.log(result);

这段代码要做的第一件事是列出所有具有name道具的键。现在有了这个列表，我可以迭代它并更改对象道具以仅使用name.

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