如果两个值都不存在,我如何推入数组?这是我的数组:

[
    { name: "tom", text: "tasty" },
    { name: "tom", text: "tasty" },
    { name: "tom", text: "tasty" },
    { name: "tom", text: "tasty" },
    { name: "tom", text: "tasty" }
]

如果我试图再次推入数组的名字:“tom”或文本:“tasty”,我不希望发生任何事情…但如果这两个都不存在那么我就输入。push()

我该怎么做呢?


当前回答

这里你有一种方法可以在一行中为两个数组做这件事:

const startArray = [1,2,3,4]
const newArray = [4,5,6]

const result = [...startArray, ...newArray.filter(a => !startArray.includes(a))]

console.log(result);
//Result: [1,2,3,4,5,6]

其他回答

你可以用一个自定义方法扩展Array原型:

// check if an element exists in array using a comparer function
// comparer : function(currentElement)
Array.prototype.inArray = function(comparer) { 
    for(var i=0; i < this.length; i++) { 
        if(comparer(this[i])) return true; 
    }
    return false; 
}; 

// adds an element to the array if it does not already exist using a comparer 
// function
Array.prototype.pushIfNotExist = function(element, comparer) { 
    if (!this.inArray(comparer)) {
        this.push(element);
    }
}; 

var array = [{ name: "tom", text: "tasty" }];
var element = { name: "tom", text: "tasty" };
array.pushIfNotExist(element, function(e) { 
    return e.name === element.name && e.text === element.text; 
});

如果不在列表中,则添加

对于一个简单值的列表,它是一行程序…

[...new Set([...someArray, someElement])]

JavaScript的用法:

var myArray = ['bill','bob']
var alreadyIn = [...new Set([...myArray, 'bob'])] // ['bill','bob']
var notAlreadyIn = [...new Set([...myArray, 'peter'])] // ['bill','bob','peter']

TypeScript文本(注意include vs includes):

interface Array<T> {
  include(element: T): Array<T>
}
Array.prototype.include = function (element: any): any[] {
  return [...new Set([...this, obj])]
}

...但对于对象来说,情况就复杂多了

[...new Set([...someArray.map((o) => JSON.stringify(o)),
    JSON.stringify(someElement)]).map((o) => JSON.parse(o))

TypeScript文本处理任何事情:

Array.prototype.include = function (element: any): any[] {
  if (element && typeof element === 'object')
    return [
      ...new Set([
        ...this.map((o) => JSON.stringify(o)),
        JSON.stringify(element),
      ]),
    ].map((o) => JSON.parse(o))
  else return [...new Set([...this, element])]
}

像这样的吗?

var item = "Hello World";
var array = [];
if (array.indexOf(item) === -1) array.push(item);

与对象

var item = {name: "tom", text: "tasty"}
var array = [{}]
if (!array.find(o => o.name === 'tom' && o.text === 'tasty'))
    array.push(item)

我知道这是一个非常老的问题,但如果你使用ES6,你可以使用一个非常小的版本:

[1,2,3].filter(f => f !== 3).concat([3])

非常简单,首先添加一个过滤器,删除项目-如果它已经存在,然后通过concat添加它。

下面是一个更现实的例子:

const myArray = ['hello', 'world']
const newArrayItem

myArray.filter(f => f !== newArrayItem).concat([newArrayItem])

如果你的数组包含对象,你可以像这样调整过滤器函数:

someArray.filter(f => f.some(s => s.id === myId)).concat([{ id: myId }])

我更喜欢使用原生js Array.prototype.some(),即使在jqenv中 文档:w3s some or mdn some

let arr = [
    { name: "tom", text: "tasty" },
    { name: "tom", text: "tasty" }
];
let oneMore = { name: "tom", text: "tasty" };
!arr.some(i => i.name == oneMore.name && i.text == oneMore.text)
  && arr.push(oneMore);