Question

我有一个对象，我想迭代该对象的某些特定键。怎么做到这一点？

考虑下面的代码段：

如何迭代table1，table2和table3键而不是所有键？

＆＃13;

var table_object = {
  table1: "hello world",
  table1_name: "greetings_english.html",
  table2: "hola",
  table2_name: "greetings_spanish.html",
  table3: "Bonjour",
  table3_name: "greetings_french.html"
};

＆＃13;

Answer 1

您可以过滤密钥，然后迭代其余部分。

var table_object = { table1: "hello world", table1_name: "greetings_english.html", table2: "hola", table2_name: "greetings_spanish.html", table3: "Bonjour", table3_name: "greetings_french.html" };

Object
    .keys(table_object)
    .filter(function (k) { return !/_/.test(k); })
    .forEach(function (k) { console.log(table_object[k]); });

Answer 2

您必须指定必须迭代的键：

var table_object = {
  table1: "hello world",
  table1_name: "greetings_english.html",
  table2: "hola",
  table2_name: "greetings_spanish.html",
  table3: "Bonjour",
  table3_name: "greetings_french.html"
};

var keysToIterateThrough = ["table1", "table2", "table3"];
keysToIterateThrough.forEach(key => {
    var value = table_object[key];
    console.log(`key: ${key} => value: ${value}`);
})

Answer 3

您必须使用Object.keys才能找到所有对象键，然后应用filter方法以过滤它们。

var table_object = {
  table1: "hello world",
  table1_name: "greetings_english.html",
  table2: "hola",
  table2_name: "greetings_spanish.html",
  table3: "Bonjour",
  table3_name: "greetings_french.html"
};
var keysToIterate = ["table1", "table2", "table3_name"];
let values=Object.keys(table_object)
                .filter(a=>keysToIterate.includes(a))
                .map(a=> table_object[a]);
console.log(values);

Answer 4

可以尝试类似

的内容

var keys = ['table1', 'table2', 'table3']
Object.keys(table_object).forEach(function(key) {
    if (keys.indexOf(key) != -1) {
        var table_value = table_object[key]
    }
})

Answer 5

可能无法迭代有限的键，但您可以拥有要迭代的键数组。然后遍历该数组以从对象

获取相应的值

＆＃13;

var objKeys = ['table1', 'table2', 'table3']

var table_object = {
  table1: "hello world",
  table1_name: "greetings_english.html",
  table2: "hola",
  table2_name: "greetings_spanish.html",
  table3: "Bonjour",
  table3_name: "greetings_french.html"
};

objKeys.forEach(function(item) {
  console.log(table_object[item])

})

＆＃13;

Answer 6

我认为在这种情况下更好的方法就是让你的对象像这样：

var table = {
    "hello world": "greetings_english.html",
    "hola": "greetings_spanish.html",
    "bonjour": "greetings_french.html"
};

for( var i in table ) {
    console.log( i );
    console.log( table[ i ] );
}

或者你可以创建两个数组：

var greetings = [
    "hello world",
    "hola",
    "bonjour"
];
var names = [
    "greetings_english.html",
    "greetings_spanish.html",
    "greetings_french.html"
];

for( var i = 0; i < greetings.length; i ++ ) {
    console.log( greetings[ i ] );
    console.log( names[ i ] );
}

您可以使用此方法制作表格

但无论如何你的问题：

var table = {
    table1: "hello world",
    table1_name: "greetings_english.html",
    table2: "hola",
    table2_name: "greetings_spanish.html",
    table3: "bonjour",
    table3_name: "greetings_french.html"
};

// Now there are three methods

console.log( "--- Method 1 ---" );
Object.keys( table )
      .filter( function( key ) {
          return /^table(\d+)$/.test( key );
      } )
      .forEach( function( key ) {
          console.log( key );
          console.log( table[ key ] );
          console.log( table[ key + "_name" ] );
      } );
      
console.log( "--- Method 2 ---" );
for ( var i in table ) {
    if ( /^table(\d+)$/.test( i ) ) {
        console.log( i );
        console.log( table[ i ] );
        console.log( table[ i + "_name" ] );
    }
}
      
console.log( "--- Method 3 ---" );
var keys = [
    "table1",
    "table2",
    "table3"
];
for ( var i = 0; i < keys.length; i ++ ) {
    console.log( keys[ i ] );
    console.log( table[ keys[ i ] ] );
    console.log( table[ keys[ i ] + "_name" ] );
}

方法2将是最好的。

Answer 7

您不需要花哨的过滤器或正则表达式来完成这么简单的任务！忽略那些误导性的答案，并开始使用JavaScript的全部功能！
您应该在对象上使用Object.defineProperty()方法，并设置为enumerable: false您不想迭代的所有属性。通过这种方式，您还可以将命名约定与逻辑分离。让我告诉你：

// Defining iterable properties. This ones will be returned
// by Object.keys()

var table_object = {
  table1: "hello world",
  table2: "hola",
  table3: "Bonjour",
  // It works even if you declare them in advance
  // table1_name: "greetings_english.html",
  // table2_name: "greetings_spanish.html",
  // table3_name: "greetings_french.html",
};

// Declaring the not iterable properties. They can still be
// accessed, but they will not be iterated through

Object.defineProperty(table_object, "table1_name", {
  // delete the next line if you have already declared it
  value: "greetings_english.html", 
  enumerable: false
});

Object.defineProperty(table_object, "table2_name", {
  // delete the next line if you have already declared it
  value: "greetings_spanish.html",
  enumerable: false
});

Object.defineProperty(table_object, "table3_name", {
  // delete the next line if you have already declared it
  value: "greetings_french.html",
  enumerable: false
});

// Iterating through the object using for ... in, which iterates
// over the keys

for (var key in table_object) {
  console.log(table_object[key]);
}

仍然可以使用Object.getOwnPropertyNames()检索不可枚举的属性以及所有可枚举的属性。
但是，如果您打算在不再需要过滤时实际使用第二种方法迭代所有属性，我会给你一些建议：

enumerable可以设置回true，因此，如果是一次性更改，我强烈建议您将其还原。
如果过滤来回频繁，那么你应该检查对象结构的两倍，因为可能有更合适的选项满足你的需求（例如数组到对象中）。

如何在JavaScript中迭代Object的特定键？

7 个答案: