javascript - 如何从数组中删除不包含特定字符串的字段的所有元素？

Question

这是从 ElasticSearch 查询获得的一些结果..

"buckets": [{
    "key": "another_service_name:0.0.1",
    "doc_count": 105,
    "containers": {
        "doc_count_error_upper_bound": 0,
        "sum_other_doc_count": 0,
        "buckets": [{
            "key": "7eaf4933b0366f7212f30f9f3c315672ea18e026922d0651b29b844763be6b0a",
            "doc_count": 105,
            "someField": {
                "value": 1.2708647376015072
            }
        }]
    }
}, {
    "key": "my_service_name:0.0.1",
    "doc_count": 200,
    "containers": {
        "doc_count_error_upper_bound": 0,
        "sum_other_doc_count": 0,
        "buckets": [{
            "key": "4c27931147386d49e90f581a1d5cb22f53cda00b16107864cad9eff32f61d776",
            "doc_count": 100,
            "someField": {
                "value": 77.9621390914917
            }
        }, {
            "key": "392e7e6981cd4c24c526ffda03977b7984ebcae58194680c82940af4acec4748",
            "doc_count": 100,
            "someField": {
                "value": 76.63604362487793
            }
        }]
    }
}]

是否有一个简单的 1-liner 可以从外部 buckets 数组中删除没有包含 my_service_name 的 key 的任何元素？

我知道我可以循环数组，使用 indexOf 和 splice ..但是有更短的方法吗？

我希望结果看起来像这样..

"buckets": [{
    "key": "my_service_name:0.0.1",
    "doc_count": 200,
    "containers": {
        "doc_count_error_upper_bound": 0,
        "sum_other_doc_count": 0,
        "buckets": [{
            "key": "4c27931147386d49e90f581a1d5cb22f53cda00b16107864cad9eff32f61d776",
            "doc_count": 100,
            "someField": {
                "value": 77.9621390914917
            }
        }, {
            "key": "392e7e6981cd4c24c526ffda03977b7984ebcae58194680c82940af4acec4748",
            "doc_count": 100,
            "someField": {
                "value": 76.63604362487793
            }
        }]
    }
}]

Answer 1

您可以使用Array.prototype.filter方法:

var queryResults = /* your result as above */;

var myServices = queryResults.filter(function(element) {
    return element.key.indexOf('my_service_name') >= 0;
})

在 ES6 中，可以将其缩短为:

let queryResults = /* your result as above */;

let myServices = queryResults.filter(element => element.key.includes('my_service_name'))