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從一個(gè)對(duì)象中刪除與其他對(duì)象項(xiàng)匹配的項(xiàng)

從一個(gè)對(duì)象中刪除與其他對(duì)象項(xiàng)匹配的項(xiàng)

MMMHUHU 2022-10-08 17:53:21
我有一個(gè)對(duì)象結(jié)構(gòu)如下"designProjects": [  {    "projectNumber": "number1",    "name": "test1"  },  {    "projectNumber": "number2",    "name": "test2"  },  {    "projectNumber": "number3",    "name": "test3"  },]我有另一個(gè)具有如下結(jié)構(gòu)的對(duì)象"allProjects": [  {    "project": {      "name": "test1",      "number": "number1"    },    "employee": {      "displayName": "name1"    },    "projectRoleName": "Editor"  },  {    "project": {      "name": "test2",      "number": "number2"    },    "employee": {      "displayName": "name2"    },    "projectRoleName": "Editor"  },]我有點(diǎn)像下面這樣的結(jié)果"designProjects": [  {    "projectNumber": "number3",    "name": "test3"  },]這里的結(jié)果 designProjects只有一個(gè),因?yàn)轫?xiàng)目編號(hào)和名稱與對(duì)象project數(shù)組匹配allprojects。有沒(méi)有辦法我們可以在react js. 任何建議或想法都會(huì)非常感謝我,非常感謝提前
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2 回答

?
回首憶惘然

TA貢獻(xiàn)1847條經(jīng)驗(yàn) 獲得超11個(gè)贊

您只需與 結(jié)合.filter使用.some,例如:


let d = {

  "designProjects": [

    {

      "projectNumber": "number1",

      "name": "test1"

    },

    {

      "projectNumber": "number2",

      "name": "test2"

    },

    {

      "projectNumber": "number3",

      "name": "test3"

    },

  ]

}


let a = {

  "allProjects": [

    {

      "project": {

        "name": "test1",

        "number": "number1"

      },

      "employee": {

        "displayName": "name1"

      },

      "projectRoleName": "Editor"

    },

    {

      "project": {

        "name": "test2",

        "number": "number2"

      },

      "employee": {

        "displayName": "name2"

      },

      "projectRoleName": "Editor"

    },

  ]

};


console.log(

    d.designProjects.filter((designProject) => {

        return !a.allProjects.some((project) => designProject.projectNumber === project.project.number && designProject.name === project.project.name);

     })

 );


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反對(duì) 回復(fù) 2022-10-08
?
慕村225694

TA貢獻(xiàn)1880條經(jīng)驗(yàn) 獲得超4個(gè)贊

您可以結(jié)合過(guò)濾器一些

filter用于根據(jù)條件返回一個(gè)新的過(guò)濾數(shù)組

some將成為條件并在找到匹配項(xiàng)后立即返回

const designProjects = [{

    "projectNumber": "number1",

    "name": "test1"

  },

  {

    "projectNumber": "number2",

    "name": "test2"

  },

  {

    "projectNumber": "number3",

    "name": "test3"

  },

];


const allProjects = [{

    "project": {

      "name": "test1",

      "number": "number1"

    },

    "employee": {

      "displayName": "name1"

    },

    "projectRoleName": "Editor"

  },

  {

    "project": {

      "name": "test2",

      "number": "number2"

    },

    "employee": {

      "displayName": "name2"

    },

    "projectRoleName": "Editor"

  },

]


const cleaned = designProjects.filter((x) => {

  return !allProjects.some(y => y.project.number === x.projectNumber);

});


console.info(cleaned);


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反對(duì) 回復(fù) 2022-10-08
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