Question

例如，以下对象：

var fruits = {
    "red" : "apple",
    "blue" : "blueberry",
    "yellow" : "banana"
}

我知道我可以使用delete fruits["red"]按键名删除它，但是如何通过水果名称删除对象项呢？

Answer 1

你尝试过这样的事吗？

function deleteByValue(val) {
    for(var f in fruits) {
        if(fruits[f] == val) {
            delete fruits[f];
        }
    }
}

根据Rocket的评论，您可能需要检查hasOwnProperty以确保不删除对象原型的成员：

function deleteByValue(val) {
    for(var f in fruits) {
        if(fruits.hasOwnProperty(f) && fruits[f] == val) {
            delete fruits[f];
        }
    }
}

Answer 2

var key = null;
for (var k in fruits){
  if (fruits[k] === 'apple'){
    key = k;
    break;
  }
}
if (key != null)
  delete fruits[key];

迭代找到相应键的对象，然后将其删除（如果找到）。

Answer 3

这个怎么样？

function delteByValue(a){
  fruits.foreach( function( k, v ) {
    if (fruits[v] == a){
      delete fruits[k];
    }
  });
}

Answer 4

我认为创建一个函数并覆盖Object.prototype是一个好主意：

/**
 *  @autor Javier Cobos
 *  @param value The value to look for
 *  @return true if founded deleted, false if not
 */        
Object.prototype.removeByValue = function(value){
         var i;
         for(i in this){
            if(this.hasOwnProperty(i))
                if(value === this[i]){
                   delete(this[i]);
                   return true;
                }
         }   
         return false;
        }

// Example
     var fruits = {
        "red" : "apple",
        "blue" : "blueberry",
        "yellow" : "banana"
    }

        fruits .removeByValue("apple");

这样我们脚本中的每个Object都有一个新方法：）

Answer 5

我知道你现在有几个可以接受的答案......我将为此包含一个通用函数......

// NOTE, replace "Object.removePropertyByValue" 
// with say "jQuery.removePropertyByValue" or "_.removePropertyByValue"
// to create a jQuery or Underscore extension.
Object.removePropertyByValue = function(obj, val) {
  //store list of properties to remove
  var propsToRemove = [];

  for (var prop in obj) {
    if (obj.hasOwnProperty(prop) && obj[prop] === val) {
      //save the property name here, deleting while enumerating is a bad idea.
      propsToRemove.push(prop);
    }
  }

  //remove all the items stored.
  while (propsToRemove.length) delete obj[propsToRemove.pop()];
}

从此处您应该可以致电：Object.removePropertyByValue(fruits, "red");

Answer 6

不知道这在处理方面是否有效，但使用过滤器可以在三行中完成：

var fruits = {
    "red" : "apple",
    "blue" : "blueberry",
    "yellow" : "banana"
}

var appleless_keys = Object.keys(fruits).filter(this_fruit => fruits[this_fruit] !== "apple");
appleless_obj = {};
appleless_keys.forEach(key => appleless_obj[key] = fruits[key]);
console.dir(appleless_obj);

或作为一项功能：

var fruits = {
    "red" : "apple",
    "blue" : "blueberry",
    "yellow" : "banana"
}

function remove_fruit(fruit_to_remove,fruits){
  var new_keys = Object.keys(fruits).filter(this_fruit => fruits[this_fruit] !== fruit_to_remove);
  new_obj = {};
  new_keys.forEach(key => new_obj[key] = fruits[key]);  
  return new_obj;
}

console.dir(remove_fruit("apple",fruits));

Answer 7

使用密钥删除进行简化，

function stripProperty(o, v) {
    return  (delete o[Object.keys(o).splice(Object.values(o).indexOf(v), 1)])?o:0;
}

    var fruits = {
        "red" : "apple",
        "blue" : "blueberry",
        "yellow" : "banana"
    }

    function stripProperty(o, v) {
        return  (delete o[Object.keys(o).splice(Object.values(o).indexOf(v), 1)])?o:0;
    }   


    console.log(stripProperty(fruits, 'banana'));

用法，

var fruits = {
    "red" : "apple",
    "blue" : "blueberry",
    "yellow" : "banana"
}

console.log(stripProperty(fruits, 'apple'))

Answer 8

您可以使用拼接方法

fruits.splice($.inArray("red", fruits), 1);

但是这当然使用jQuery。

您也可以使用此扩展程序：

Array.prototype.remove = function () {
    var what, a = arguments, L = a.length, ax;
    while (L && this.length) {
        what = a[--L];
        while ((ax = this.indexOf(what)) != -1) {
            this.splice(ax, 1);
        }
    }
    return this;
}

Answer 9

使用此功能

function delobj(x,n,v){
    n = "elem."+n;
    var u = x.filter(function(elem, index, self){
        if(eval(n) != v){
            return index == self.indexOf(elem);
        }
    })
    return u;
}

像这样使用

delobj(object "object name",string "name of element",string "value of element")

Answer 10

var fruits = {
    "red" : "apple",
    "blue" : "blueberry",
    "yellow" : "banana"
}
delete fruits.red; // or use => delete fruits['red'];
console.log(fruits);

这会删除fruits.red

如何按值删除javascript对象项？

10 个答案: