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?
不确定速度,但stringification + indexOf是一个简单的方法。首先将数组转换为字符串:
let strMyArray = JSON.stringify(myArray);
然后,对于一系列属性-值对,您可以使用:
if (strMyArray.indexOf('"name":"tom"') === -1 && strMyArray.indexOf('"text":"tasty"') === -1) {
myArray.push({ name: "tom", text: "tasty" });
}
查找整个对象更简单:
if (strMyArray.indexOf(JSON.stringify(objAddMe) === -1) {
myArray.push(objAddMe);
}
使用数组是很容易做到的。函数findIndex,它以函数作为参数:
var arrayObj = [{name:"bull", text: "sour"},
{ name: "tom", text: "tasty" },
{ name: "tom", text: "tasty" }
]
var index = arrayObj.findIndex(x => x.name=="bob");
// here you can check specific property for an object whether it exist in your array or not
index === -1 ? arrayObj.push({your_object}) : console.log("object already exists")
我有这个问题,我做了一个简单的原型,使用它,如果你喜欢它
Array.prototype.findOrPush = function(predicate, fallbackVal) {
let item = this.find(predicate)
if(!item){
item = fallbackVal
this.push(item)
}
return item
}
let arr = [{id: 1}]
let item = arr.findOrPush(e => e.id == 2, {id: 2})
console.log(item) // {id: 2}
// will not push and just return existing value
arr.findOrPush(e => e.id == 2, {id: 2})
conslog.log(arr) // [{id: 1}, {id: 2}]