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?
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'}];
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
正是出于这些原因,使用像underscore.js这样的js库。union:计算传入数组的并集:在一个或多个数组中出现的唯一项的列表。
_.union([1, 2, 3], [101, 2, 1, 10], [2, 1]);
=> [1, 2, 3, 101, 10]
您可以使用foreach检查数组,然后弹出项目,如果它存在,否则添加新的项目…
newItemValue &submitFields是键值对
> //submitFields existing array
> angular.forEach(submitFields, function(item) {
> index++; //newItemValue new key,value to check
> if (newItemValue == item.value) {
> submitFields.splice(index-1,1);
>
> } });
submitFields.push({"field":field,"value":value});