我正在React中构建一个纸牌游戏应用程序,并尝试过滤当前数组中是否有多个具有“纸牌”的数组的值为6。例如:
let arr = [{type: "span", key: "51", ref: null, props: {children: "6"}}, { type: "span", key: "52", ref: null, props: {children: "7"}}, { type: "span", key: "53", ref: null, props: {children: "6"}}]
let arr2 = [{ type: "span", key: "50", ref: null, props: {children: "6"}}]
let result = arr.filter((card) => {
return arr2 === card
})
console.log(result) //returns []
//I want it to return the items that have props:{children: "6"}
然后我需要从arr中删除该项目并将其放在arr2中。那我会做这样的事情吗?
this.setState(({arr2, arr}) => {
return {
arr2: [...arr2, ...arr.slice(0, 1)],
arr: [...arr.slice(1, arr.length)]
};
});
答案 0 :(得分:2)
您可以在some
内使用filter
来获取基于props值的公共元素。
let arr = [{
type: "span",
key: "51",
ref: null,
props: {
children: "6"
}
}, {
type: "span",
key: "52",
ref: null,
props: {
children: "7"
}
}, {
type: "span",
key: "53",
ref: null,
props: {
children: "6"
}
}]
let arr2 = [{
type: "span",
key: "50",
ref: null,
props: {
children: "6"
}
}]
let result = arr.filter((card, index) => {
return arr2.some(function(card2){
return card2.props.children === card.props.children
});
})
arr = arr.filter((card, index) => {
return arr2.some(function(card2){
return card2.props.children !== card.props.children
});
})
arr2 = arr2.concat(result)
console.log("arr list are:")
console.log(arr)
console.log("arr2 list are:")
console.log(arr2)
希望它会有所帮助:)
答案 1 :(得分:0)
let arr = [
{type: "span", key: "51", ref: null, props: {children: "6"}},
{ type: "span", key: "52", ref: null, props: {children: "7"}},
{ type: "span", key: "53", ref: null, props: {children: "6"}}
];
let arr2 = [
{ type: "span", key: "50", ref: null, props: {children: "6"}}
];
let result = arr.filter(
card => card.props.children === '6'
? arr2.push(card)
: null
); //If children: 6 then add it to arr2
arr = arr.filter(card => card.props.children !== '6'); //Remove from arr the ones that has children: 6
console.log(result);
console.log(arr2);
console.log(arr);