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如何循環(huán)遍歷一個對象形成形成條件

如何循環(huán)遍歷一個對象形成形成條件

回首憶惘然 2023-03-18 17:52:50
我有一個對象，如下所示：{    Condition0: "5"    Condition1: "6"    LogicalOperator0: "&&"    Operator0: "<"    Operator1: "!="    Question0: "How do you rate our services?"    Question1: "How likely are you to recommend our services to others?"}我想安排它形成一個條件Question0 Operator0 Condition0 LogicalOperator0 Question1 Operator1 Condition1 因此結(jié)果形成如下所示的比較運算符：How do you rate our services? < 5 && How likely are you to recommend our services to others? != 6任何人都請協(xié)助在 JS 中實現(xiàn)這一目標(biāo)。
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3 回答

?
GCT1015

TA貢獻(xiàn)1827條經(jīng)驗 獲得超4個贊

您可以為屬性和數(shù)字部分采用嵌套循環(huán)，并將所有部分收集在一個數(shù)組中。


let data = { Condition0: "5", Condition1: "6", LogicalOperator0: "&&", Operator0: "<", Operator1: "!=", Question0: "How do you rate our services?", Question1: "How likely are you to recommend our services to others?" },

    keys = ['Question', 'Operator', 'Condition', 'LogicalOperator'],

    result = [],

    i = 0;


outer: while (true) {

    for (const part of keys) {

        const key = `${part}${i}`;

        if (!(key in data)) break outer;

        result.push(data[key]);

    }

    i++;

}


console.log(result.join(' '));


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反對 回復(fù) 2023-03-18
?
當(dāng)年話下

TA貢獻(xiàn)1890條經(jīng)驗 獲得超9個贊

將對象分配給變量并從那里訪問它：


var someName = {

    Condition0: "5"

    Condition1: "6"

    LogicalOperator0: "&&"

    Operator0: "<"

    Operator1: "!="

    Question0: "How do you rate our services?"

    Question1: "How likely are you to recommend our services to others?"

}

//Accessing the values would look like:

//someName.question0 + somename.operator0 + somename.condition0...

如果您遍歷該對象，則只能按照創(chuàng)建它的順序訪問它。您似乎需要以不同的順序訪問它。


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反對 回復(fù) 2023-03-18
?
蠱毒傳說

TA貢獻(xiàn)1895條經(jīng)驗 獲得超3個贊

您可以將顯示屬性的順序存儲到一個數(shù)組中并操作該數(shù)組


const obj = {

  Condition0: '5',

  Condition1: '6',

  LogicalOperator0: '&&',

  Operator0: '<',

  Operator1: '!=',

  Question0: 'How do you rate our services?',

  Question1: 'How likely are you to recommend our services to others?'

}


const order = [

  'Question0',

  'Operator0',

  'Condition0',

  'LogicalOperator0',

  'Question1',

  'Operator1',

  'Condition1'

]


const res = order.map(prop => obj[prop]).join(' ')


console.log(res)


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反對 回復(fù) 2023-03-18
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