如果 projectId 匹配子項和 projectId 分組，則數據轉換

Question

我需要將所有 projectId 組合在一起，如果 projectId 相同，則孩子應該在當前項目中。 請參閱下圖了解預期的最終結果。

感謝是否有人可以幫助我。 我正在嘗試在構建應用程序功能的同時學習 go。 謝謝你。

當前 Json 格式：

[
  {
    "projectIds": "1aa",
    "children": [
      {
        "email": "alyssayo@xxx.com",
        "permissionIds": null,
        "roleIds": [
          "raa",
          "baa",
          "caa"
        ],
        "projectIds": "1aa"
      }
    ]
  },
  {
    "projectIds": "3aa",
    "children": [
      {
        "email": "alyssayo@xxx.com",
        "permissionIds": null,
        "roleIds": [
          "raa",
          "baa",
          "caa"
        ],
        "projectIds": "3aa"
      }
    ]
  },
  {
    "projectIds": "1aa",
    "children": [
      {
        "email": "chiuewww@xxx.com",
        "permissionIds": null,
        "roleIds": [
          "baa",
          "caa"
        ],
        "projectIds": "1aa"
      }
    ]
  },
  {
    "projectIds": "2aa",
    "children": [
      {
        "email": "chiuewww@xxx.com",
        "permissionIds": null,
        "roleIds": [
          "baa",
          "caa"
        ],
        "projectIds": "2aa"
      }
    ]
  },
  {
    "projectIds": "3aa",
    "children": [
      {
        "email": "chiuewww@xxx.com",
        "permissionIds": null,
        "roleIds": [
          "baa",
          "caa"
        ],
        "projectIds": "3aa"
      }
    ]
  },
  {
    "projectIds": "1aa",
    "children": [
      {
        "email": "lalaqq@xxx.com",
        "permissionIds": null,
        "roleIds": [
          "caa"
        ],
        "projectIds": "1aa"
      }
    ]
  }
]

預期 Json 格式：

const data = [
  {
    projectIds: "1aa",
    children: [
      {
        email: "alyssayo@xxx.com",
        permissionIds: null,
        roleIds: ["raa","baa","caa"],
        projectIds: "1aa",
      },
      {
        email: "chiuewww@xxx.com",
        permissionIds: null,
        roleIds: ["baa","caa"],
        projectIds: "1aa",
      },
      {
        email: "lalaqq@xxx.com",
        permissionIds: null,
        roleIds: ["caa"],
        projectIds: "1aa",
      }
    ]
  },
  {
    projectIds: "2aa",
    children: [
      {
        email: "chiuewww@xxx.com",
        permissionIds: null,
        roleIds: ["baa","caa"],
        projectIds: "2aa",
      }
    ]
  },
  {
    projectIds: "3aa",
    children: [
      {
        email: "alyssayo@xxx.com",
        permissionIds: null,
        roleIds: ["raa","baa","caa"],
        projectIds: "3aa",
      },    
      {
        email: "chiuewww@xxx.com",
        permissionIds: null,
        roleIds: ["baa","caa"],
        projectIds: "3aa",
      }
    ]
  }
];

Answer 1

你可以使用數組減少和數組查找，

 const data = [ { "projectIds": "1aa", "children": [ { "email": "alyssayo@xxx.com", "permissionIds": null, "roleIds": [ "raa", "baa", "caa" ], "projectIds": "1aa" } ] }, { "projectIds": "3aa", "children": [ { "email": "alyssayo@xxx.com", "permissionIds": null, "roleIds": [ "raa", "baa", "caa" ], "projectIds": "3aa" } ] }, { "projectIds": "1aa", "children": [ { "email": "chiuewww@xxx.com", "permissionIds": null, "roleIds": [ "baa", "caa" ], "projectIds": "1aa" } ] }, { "projectIds": "2aa", "children": [ { "email": "chiuewww@xxx.com", "permissionIds": null, "roleIds": [ "baa", "caa" ], "projectIds": "2aa" } ] }, { "projectIds": "3aa", "children": [ { "email": "chiuewww@xxx.com", "permissionIds": null, "roleIds": [ "baa", "caa" ], "projectIds": "3aa" } ] }, { "projectIds": "1aa", "children": [ { "email": "lalaqq@xxx.com", "permissionIds": null, "roleIds": [ "caa" ], "projectIds": "1aa" } ] } ] const findMatch = (arr, projectIds) => arr.find(item => item.projectIds === projectIds) const grouppingArray = (originalArr) => { return Array.isArray(originalArr)? originalArr.reduce((previousObj, obj) => { if (findMatch(previousObj, obj.projectIds)) { findMatch(previousObj, obj.projectIds).children.push(...obj.children) } else { previousObj.push(obj) } return previousObj }, []): 'Need an array' } console.log(grouppingArray(data))

Answer 2

您可以嘗試 array.reduce() 其中 initialData 是您當前的數據

const data = initialData.reduce(function (r, a) {
        r[a.projectIds] = r[a.projectIds] || [];
        r[a.projectIds].push(a);
        return r;
    }, Object.create(null));