开发者问题收集

从对象内部的对象数组中删除一个对象

2021-07-22
128
javascript object

我不确定如何提出这个问题，但我会尽力。 我想从父对象内的数组 produits 中删除一个对象，这取决于 families.name ，该对象必须等于我想要删除的初始对象的字符串（我已经实现了这一部分）。

现在我被卡住了，不知道如何更新初始 Parent 对象，我能够从数组 produits 中删除该对象。

父对象： 

let Parent = {
    "Id": "60f03c42a512e04c4b0161e7",
    "familles": [{
            "name": "LOV D+",
            "parent": "QB",
            "produits": [{
                    "uid": "ITEM-20210715-20758",
                    "type": "fabricant",
                    "parent_uid": null,
                    "version_application": "1.1.0",
                    "creation_date": null,
                    "titu": "REST1",
                    "ref_cstb": "162367462",
                    "gamme": "SUPERGRES",
                    "serie": "PREMIERE",
                    "description": "300x300 - 20 - Blue",
                    "usines": null
                },
                {
                    "uid": "QB32N-20210720-75584",
                    "type": "fabricant",
                    "parent_uid": null,
                    "version_application": "1.1.0",
                    "creation_date": "2021-07-20T14:34:04.49Z",
                    "titu": "REST1",
                    "ref_cstb": "ref-18-taieb",
                    "gamme": "SUPERGRES T_20",
                    "serie": "WIND",
                    "description": null,
                    "usines": null
                }
            ],
            "version_schema": null
        },
        {
            "name": "LOV F+",
            "parent": "QB",
            "produits": [{
                "uid": "ITEM-20210720-33547",
                "type": "fabricant",
                "parent_uid": null,
                "version_application": "1.1.0",
                "creation_date": "2021-07-20T14:46:37.649Z",
                "titu": "REST1",
                "ref_cstb": "ref19-taieb",
                "gamme": "SUPERGRES T_20",
                "serie": "STOCKHOLM",
                "description": null,
                "usines": null
            }],
            "version_schema": null
        }
    ]
}

我想要删除的对象 

let obj1 = {
    "uid": "ITEM-20210715-20758",
    "type": "fabricant",
    "parent_uid": null,
    "version_application": "1.1.0",
    "creation_date": null,
    "titu": "REST1",
    "ref_cstb": "162367462",
    "gamme": "SUPERGRES",
    "serie": "PREMIERE",
    "description": "300x300 - 20 - Blue",
    "usines": null,
    "group": "QB - LOV D+",
    "parent": "QB"
}

这样我的父对象将如下所示 

let Parent = {
    "Id": "60f03c42a512e04c4b0161e7",
    "familles": [{
            "name": "LOV D+",
            "parent": "QB",
            "produits": [{
                "uid": "ITEM-20210720-75584",
                "type": "fabricant",
                "parent_uid": null,
                "version_application": "1.1.0",
                "creation_date": "2021-07-20T14:34:04.49Z",
                "titu": "REST1",
                "ref_cstb": "ref-18-taieb",
                "gamme": "SUPERGRES T_20",
                "serie": "WIND",
                "description": null,
                "usines": null
            }],
            "version_schema": null
        },
        {
            "name": "LOV F+",
            "parent": "QB",
            "produits": [{
                "uid": "ITEM-20210720-33547",
                "type": "fabricant",
                "parent_uid": null,
                "version_application": "1.1.0",
                "creation_date": "2021-07-20T14:46:37.649Z",
                "titu": "REST1",
                "ref_cstb": "ref19-taieb",
                "gamme": "SUPERGRES T_20",
                "serie": "STOCKHOLM",
                "description": null,
                "usines": null
            }],
            "version_schema": null
        }
    ]
}

我到目前为止所做的： 

    Remove(obj, parentObj) {

        let a = e.group.split("-").pop().slice(1)
        console.log(a) // Will Display LOV D+ 
        let newArray = this.certificat.familles.find((x: {
            name: any;
        }) => x.name === a); // Will display the proper object that have the name = LOV D+
        newArray = newArray.produits.filter(({
            uid
        }) => {
            return uid !== e.uid;
        }); // This will delete the object from the array, but it will not update the parent object, I'm missing something here 
    }

Remove(obj1, parent)
3个回答
let Parent = {
    "Id": "60f03c42a512e04c4b0161e7",
    "familles": [{
            "name": "LOV D+",
            "parent": "QB",
            "produits": [{
                    "uid": "ITEM-20210715-20758",
                    "type": "fabricant",
                    "parent_uid": null,
                    "version_application": "1.1.0",
                    "creation_date": null,
                    "titu": "REST1",
                    "ref_cstb": "162367462",
                    "gamme": "SUPERGRES",
                    "serie": "PREMIERE",
                    "description": "300x300 - 20 - Blue",
                    "usines": null
                },
                {
                    "uid": "QB32N-20210720-75584",
                    "type": "fabricant",
                    "parent_uid": null,
                    "version_application": "1.1.0",
                    "creation_date": "2021-07-20T14:34:04.49Z",
                    "titu": "REST1",
                    "ref_cstb": "ref-18-taieb",
                    "gamme": "SUPERGRES T_20",
                    "serie": "WIND",
                    "description": null,
                    "usines": null
                }
            ],
            "version_schema": null
        },
        {
            "name": "LOV F+",
            "parent": "QB",
            "produits": [{
                "uid": "ITEM-20210720-33547",
                "type": "fabricant",
                "parent_uid": null,
                "version_application": "1.1.0",
                "creation_date": "2021-07-20T14:46:37.649Z",
                "titu": "REST1",
                "ref_cstb": "ref19-taieb",
                "gamme": "SUPERGRES T_20",
                "serie": "STOCKHOLM",
                "description": null,
                "usines": null
            }],
            "version_schema": null
        }
    ]
}


let obj1 = {
    "uid": "ITEM-20210715-20758",
    "type": "fabricant",
    "parent_uid": null,
    "version_application": "1.1.0",
    "creation_date": null,
    "titu": "REST1",
    "ref_cstb": "162367462",
    "gamme": "SUPERGRES",
    "serie": "PREMIERE",
    "description": "300x300 - 20 - Blue",
    "usines": null,
    "group": "QB - LOV D+",
    "parent": "QB"
}

function filterParent(obj, parentObj) {
  
    let a = obj.group.split("-").pop().slice(1) 
    let newArray = parentObj.familles.map((x) => x.name !== a ? x : {...x, produits:x.produits.filter(matchUID)}); 
  
  function matchUID(produit){
    return produit.uid !== obj.uid
  }
  return {...parentObj, familles:newArray}
}

Parent = filterParent(obj1, Parent)

console.log(Parent)
Mohamed Farid
2021-07-22

就这样。这将根据你的需要，从 Parent 中拼接具有匹配 name 和 uid 的对象。 

let Parent = {
  "Id": "60f03c42a512e04c4b0161e7",
  "familles": [{
      "name": "LOV D+",
      "parent": "QB",
      "produits": [{
          "uid": "ITEM-20210715-20758",
          "type": "fabricant",
          "parent_uid": null,
          "version_application": "1.1.0",
          "creation_date": null,
          "titu": "REST1",
          "ref_cstb": "162367462",
          "gamme": "SUPERGRES",
          "serie": "PREMIERE",
          "description": "300x300 - 20 - Blue",
          "usines": null
        },
        {
          "uid": "QB32N-20210720-75584",
          "type": "fabricant",
          "parent_uid": null,
          "version_application": "1.1.0",
          "creation_date": "2021-07-20T14:34:04.49Z",
          "titu": "REST1",
          "ref_cstb": "ref-18-taieb",
          "gamme": "SUPERGRES T_20",
          "serie": "WIND",
          "description": null,
          "usines": null
        }
      ],
      "version_schema": null
    },
    {
      "name": "LOV F+",
      "parent": "QB",
      "produits": [{
        "uid": "ITEM-20210720-33547",
        "type": "fabricant",
        "parent_uid": null,
        "version_application": "1.1.0",
        "creation_date": "2021-07-20T14:46:37.649Z",
        "titu": "REST1",
        "ref_cstb": "ref19-taieb",
        "gamme": "SUPERGRES T_20",
        "serie": "STOCKHOLM",
        "description": null,
        "usines": null
      }],
      "version_schema": null
    }
  ]
}

let obj1 = {
  "uid": "ITEM-20210715-20758",
  "type": "fabricant",
  "parent_uid": null,
  "version_application": "1.1.0",
  "creation_date": null,
  "titu": "REST1",
  "ref_cstb": "162367462",
  "gamme": "SUPERGRES",
  "serie": "PREMIERE",
  "description": "300x300 - 20 - Blue",
  "usines": null,
  "group": "QB - LOV D+",
  "parent": "QB"
}

function Remove(obj, parentObj) {
  let famille = parentObj.familles.findIndex((x) => x.name === obj.group.split(" - ").pop());
  let produit = parentObj.familles[famille].produits.findIndex((x) => x.uid === obj.uid);
  parentObj.familles[famille].produits.splice(produit, 1);
}

Remove(obj1, Parent);

console.log(Parent);
Timothy Alexis Vass
2021-07-22

您可以搜索特定元素/对象的索引，然后使用

delete Parent.familles[0].produits[0] 

const fruits = ["Banana", "Orange", "Apple", "Mango"];
      console.log(fruits.indexOf("Apple"))
将其删除
Rajanand
2021-07-22