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如何根據(jù)數(shù)組對象javascript中的條件檢查返回布爾值

JavaScript

開心每一天1111 2023-05-25 17:22:37

我有對象數(shù)組 obj1，obj2，如果對象名稱不能有All鍵名則返回 true如果對象名只能有一個(gè)All鍵名，其他對象返回false如果對象的名稱只能有一個(gè)All鍵名且沒有其他對象返回 true如果對象的名稱有太多鍵名All返回 false基于以上條件如何在 javascript 中執(zhí)行。function checkObj(ob){ var result = ob.filter(e=>e.name=="All"); if(result.length !== 1){ return false; } else{ return true; }}var obj1=[ {id:1, name: "All", value:"all"}, {id:2, name: "Sun", value:"sun"}, {id:3, name: "Mon", value:"mon"},]var obj2=[ {id:4, name: "Thur", value:"thur"}, {id:8, name: "Mon", value:"mon"},]var obj3=[ {id:1, name: "Thur", value:"thur"}, {id:5, name: "All", value:"all"}, {id:2, name: "Mon", value:"mon"}, {id:6, name: "All", value:"all"}]var obj4=[{id:1, name: "All", value:"all"}]var r1= this.checkObj(obj1);var r2=this.checkObj(obj2);var r3=this.checkObj(obj3);var r4=this.checkObj(obj4);Expected Output:falsetruefalsetrue

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2 回答

瀟瀟雨雨

TA貢獻(xiàn)1833條經(jīng)驗(yàn) 獲得超4個(gè)贊

您可以檢查數(shù)組是否包含任何對象name = "All"。

那么，return true如果

數(shù)組有name = "All"對象，數(shù)組的長度是1

或者數(shù)組沒有name = "All"

function checkObj(arr) {

const hasAll = arr.some(o => o.name === "All")

return (arr.length === 1 && hasAll) || !hasAll

}

這是一個(gè)片段：

function checkObj(arr) {

const hasAll = arr.some(o => o.name === "All")

return (arr.length === 1 && hasAll) || !hasAll

}

const obj1=[{id:1,name:"All",value:"all"},{id:2,name:"Sun",value:"sun"},{id:3,name:"Mon",value:"mon"},],

obj2=[{id:4,name:"Thur",value:"thur"},{id:8,name:"Mon",value:"mon"},],

obj3=[{id:1,name:"Thur",value:"thur"},{id:5,name:"All",value:"all"},{id:2,name:"Mon",value:"mon"},{id:6,name:"All",value:"all"}],

obj4=[{id:1,name:"All",value:"all"}];

console.log( checkObj(obj1) )

console.log( checkObj(obj2) )

console.log( checkObj(obj3) )

console.log( checkObj(obj4) )

反對回復(fù) 2023-05-25

九州編程

TA貢獻(xiàn)1785條經(jīng)驗(yàn) 獲得超4個(gè)贊

你可以Array#every在計(jì)數(shù)超過一時(shí)短路。

function checkArray(array) {

? ? return array.every((count => ({ name }) => (count += name === "All") < 2)(0));

}

var array1 = [{ id: 1, name: "All", value: "all" }, {? id: 2, name: "Sun", value: "sun" }, { id: 3, name: "Mon", value: "mon" }],

? ? array2 = [{ id: 4, name: "Thur", value: "thur" }, { id: 8,? name: "Mon", value: "mon" }],

? ? array3 = [{ id: 1, name: "Thur", value: "thur" }, { id: 5, name: "All",? value: "all" }, { id: 2, name: "Mon",? value: "mon" }, { id: 6, name: "All", value: "all" }];

console.log(checkArray(array1), checkArray(array2), checkArray(array3));

反對回復(fù) 2023-05-25