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?
推送后删除重复项
如果你已经有一个包含重复项的数组,将对象数组转换为字符串数组,然后使用Set()函数消除重复项:
let arr_obj = [
{ name: "tom", text: "tasty" },
{ name: "tom", text: "tasty" }
]
let arr_str = arr_obj.map(JSON.stringify)
let arr_unique = [...new Set(arr_str)].map(JSON.parse)
推前检查
如果你到目前为止没有重复的元素,你想在推入一个新元素之前检查重复:
let arr_obj = [
{ name: "tom", text: "tasty" },
{ name: "tim", text: "tusty" }
]
let new_obj = { name: "tom", text: "tasty" }
let arr_str = arr_obj.map(JSON.stringify)
!arr_str.includes(JSON.stringify(new_obj)) && arr_obj.push(new_obj)
这个问题有点老了,但我的选择是
let finalTab = [{id: 1, name: 'dupont'}, {id: 2, name: 'tintin'}, {id: 3, name:'toto'}]; // Your array of object you want to populate with distinct data
const tabToCompare = [{id: 1, name: 'dupont'}, {id: 4, name: 'tata'}]; // A array with 1 new data and 1 is contain into finalTab
finalTab.push(
...tabToCompare.filter(
tabToC => !finalTab.find(
finalT => finalT.id === tabToC.id)
)
); // Just filter the first array, and check if data into tabToCompare is not into finalTab, finally push the result of the filters
console.log(finalTab); // Output : [{id: 1, name: 'dupont'}, {id: 2, name: 'tintin'}, {id: 3, name: 'toto'}, {id: 4, name: 'tata'}];
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 }])