如何在JavaScript中迭代对象的特定键？

Question

我有一个对象，我想迭代该对象的一些特定键。如何实现？

考虑下面的片段：

如何迭代table1、table2和table3键，而不是所有键？

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"
};

- Karan Dhir

---

不要使用for (var key in table_object)，而是使用for (var key of ["table1", ...])？ - Bergi

{btsdaf} - Karan Dhir

{btsdaf} - Bergi

或许将“tables”和“tablenames”存储为两个单独的数组会更容易些。 - Bergi

{btsdaf} - Jared Smith

我认为这不是制作表格的正确方式。通过制作两个或更多的数组，您可以记录一个表格，并迭代行。 - Samū

Answer 1

您可以筛选键，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]); });

- Nina Scholz

---

/^table(\d+)$/将明确匹配以table开头，后跟一些数字的字符串。 - Samū

{btsdaf} - kamoroso94

/^([a-zA-Z0-9$]+)$/，或者 /^([a-zA-Z$]+)(\d+)$/。或者您可以更改正则表达式，或整个过滤函数。 - Samū

Answer 2

您需要指定要迭代的键：

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}`);
})

- Faly

---

{btsdaf} - Samū

Answer 3

你需要使用Object.keys来查找所有对象键，然后应用filter方法来过滤它们。

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);

- Mihai Alexandru-Ionut

---

{btsdaf} - Ilmari Karonen

{btsdaf} - Mihai Alexandru-Ionut

{btsdaf} - Bergi

Answer 4

你可以尝试像这样做：

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

- Shawn Janas

---

{btsdaf} - Samū

Answer 5

可能无法遍历有限键，但您可以拥有一个键数组，您想要迭代。然后循环该数组以从对象中获取相应的值

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])

})

- brk

---

{btsdaf} - Samū

Answer 6

我认为在这种情况下更好的做法是将您的对象制作成这样：

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 ] );
}

你可以使用这种方法制作表格。

但无论你的问题是什么：

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将是最佳选择。

- Samū

Answer 7

你不需要花哨的过滤器或正则表达式来完成如此简单的任务！忽略那些误导性的答案，开始使用JavaScript的全部威力！你应该在对象上使用Object.defineProperty()方法，并将你不想遍历的所有属性设置为enumerable: false。这样做还可以将你的命名约定与你的逻辑解耦。让我展示给你看：

// 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]);
}

不可枚举的属性仍然可以使用Object.getOwnPropertyNames()检索出来，与所有可枚举的属性一起。
但是，如果您计划实际使用这种第二种方法来遍历所有属性，当不再需要过滤时，我必须给您一些建议：

• enumerable可以设置回true，因此如果只是一次更改，我强烈建议还原它。

• 如果过滤经常来回切换，则应仔细检查对象的结构，因为可能有更适合您需求的选项（例如将数组放入对象中）。

- DoNotDownvote_JustUpvote

---

我考虑过那个方法，但它比筛选键的方法慢得多。 - Samū

另外，如果你真的关心性能，为什么不使用循环而非过滤函数？（https://jsperf.com/array-filter-performance）。当然，我并不是说这是最快的解决方案，但我认为这是最语义化正确的解决方案，最重要的是，它不会太慢以至于让你的浏览器卡死。;P - DoNotDownvote_JustUpvote