Question

我有一个对象数组，我正在尝试在对象中找到值并删除已在存在对象中找到的值。例如，

当前JSON对象：

exist=[{"x":6811,"y":15551,"a":["aa","ab","ac"]},{"x":6811,"y":15552,"a":["bb"]},{"x":6812,"y":15551,"a":["aa","cc"]}]

我想找到值为aa的“a”键

最后的结果是

exist= [{"x":6811,"y":15552,"a":["bb"]}]
found= [{"x":6811,"y":15551,"a":["aa","ab","ac"]},{"x":6812,"y":15551,"a":["aa","cc"]}]

Answer 1

你可以不用下划线做到这一点：

let exist= [
    { x: 6811, y: 15551, a: ['aa', 'ab', 'ac'] },
    { x: 6811, y: 15552, a: ['aa', 'bb'] },
    { x: 6812, y: 15551, a: ['cc'] },
];

const found = exist.filter(({ a }) => a.includes('aa'));
exist = exist.filter(({ a }) => !a.includes('aa'));

console.log('found:', found);
console.log('exist:', exist);

Answer 2

您可以将.reject 与 .contains和_.result组合使用

exist=[
{"x":6811,"y":15551,"a":["aa","ab","ac"]},{"x":6811,"y":15552,"a":["bb"]},{"x":6812,"y":15551,"a":["aa","cc"]}, {"x":6812,"y":15551}
]

exist = _.reject(exist, function(item){ 
    return _.contains(_.result(item, "a"), "aa"); 
})

console.log(exist);

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下划线删除值而不替换原始变量

2 个答案: