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?
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])]
}
可以使用带有回调函数及其"this"参数的findIndex方法。
注意:旧的浏览器不知道findIndex,但是一个polyfill是可用的。
示例代码(注意,在原始问题中,只有当一个新对象的数据都不在之前的推送对象中时,它才会被推送):
var a=[{name:"tom", text:"tasty"}], b;
var magic=function(e) {
return ((e.name == this.name) || (e.text == this.text));
};
b={name:"tom", text:"tasty"};
if (a.findIndex(magic,b) == -1)
a.push(b); // nothing done
b={name:"tom", text:"ugly"};
if (a.findIndex(magic,b) == -1)
a.push(b); // nothing done
b={name:"bob", text:"tasty"};
if (a.findIndex(magic,b) == -1)
a.push(b); // nothing done
b={name:"bob", text:"ugly"};
if (a.findIndex(magic,b) == -1)
a.push(b); // b is pushed into a