我有一个巨大的收藏品,我正在寻找集合中某个地方的钥匙。获取包含该键/索引的所有对象的引用列表或完整路径的可靠方法是什么?我使用jQuery和lodash,如果它有帮助,你可以忘记无限指针递归,这是一个纯粹的JSON响应。
fn({ 'a': 1, 'b': 2, 'c': {'d':{'e':7}}}, "d");
// [o.c]
fn({ 'a': 1, 'b': 2, 'c': {'d':{'e':7}}}, "e");
// [o.c.d]
fn({ 'aa': 1, 'bb': 2, 'cc': {'d':{'x':9}}, dd:{'d':{'y':9}}}, 'd');
// [o.cc,o.cc.dd]
fwiw lodash有一个_.find函数,可以找到两个嵌套深度的嵌套对象,但之后似乎失败了。 (例如http://codepen.io/anon/pen/bnqyh)
答案 0 :(得分:46)
这应该这样做:
function fn(obj, key) {
if (_.has(obj, key)) // or just (key in obj)
return [obj];
// elegant:
return _.flatten(_.map(obj, function(v) {
return typeof v == "object" ? fn(v, key) : [];
}), true);
// or efficient:
var res = [];
_.forEach(obj, function(v) {
if (typeof v == "object" && (v = fn(v, key)).length)
res.push.apply(res, v);
});
return res;
}
答案 1 :(得分:22)
纯JavaScript解决方案如下所示:
function findNested(obj, key, memo) {
var i,
proto = Object.prototype,
ts = proto.toString,
hasOwn = proto.hasOwnProperty.bind(obj);
if ('[object Array]' !== ts.call(memo)) memo = [];
for (i in obj) {
if (hasOwn(i)) {
if (i === key) {
memo.push(obj[i]);
} else if ('[object Array]' === ts.call(obj[i]) || '[object Object]' === ts.call(obj[i])) {
findNested(obj[i], key, memo);
}
}
}
return memo;
}
以下是您使用此功能的方法:
findNested({'aa': 1, 'bb': 2, 'cc': {'d':{'x':9}}, dd:{'d':{'y':9}}}, 'd');
,结果将是:
[{x: 9}, {y: 9}]
答案 2 :(得分:5)
这将深入搜索一个对象(hay)的数组(针)值,然后返回一个包含结果的数组...
search = function(hay, needle, accumulator) {
var accumulator = accumulator || [];
if (typeof hay == 'object') {
for (var i in hay) {
search(hay[i], needle, accumulator) == true ? accumulator.push(hay) : 1;
}
}
return new RegExp(needle).test(hay) || accumulator;
}
答案 3 :(得分:2)
这样的东西可以工作,将其转换为对象并递归。
function find(jsonStr, searchkey) {
var jsObj = JSON.parse(jsonStr);
var set = [];
function fn(obj, key, path) {
for (var prop in obj) {
if (prop === key) {
set.push(path + "." + prop);
}
if (obj[prop]) {
fn(obj[prop], key, path + "." + prop);
}
}
return set;
}
fn(jsObj, searchkey, "o");
}
小提琴:jsfiddle
答案 4 :(得分:2)
如果您可以用纯JS(或结合lodash)编写一个递归函数(按性能)将是最好的,但是如果您希望从一侧跳过递归并想获得一个简单的可读代码(这可能不是最佳性能),那么您可以将lodash#cloneDeepWith用于必须递归遍历对象的任何目的。
let findValuesDeepByKey = (obj, key, res = []) => (
_.cloneDeepWith(obj, (v,k) => {k==key && res.push(v)}) && res
)
因此,您作为_.cloneDeepWith
的第二个参数传递的回调将以递归方式遍历所有key/value
对,您要做的就是对每个参数进行操作。上面的代码只是您的情况的一个示例。这是一个工作示例:
var object = {
prop1: 'ABC1',
prop2: 'ABC2',
prop3: {
prop4: 'ABC3',
prop5Arr: [{
prop5: 'XYZ'
},
{
prop5: 'ABC4'
},
{
prop6: {
prop6NestedArr: [{
prop1: 'XYZ Nested Arr'
},
{
propFurtherNested: {key100: '100 Value'}
}
]
}
}
]
}
}
let findValuesDeepByKey = (obj, key, res = []) => (
_.cloneDeepWith(obj, (v,k) => {k==key && res.push(v)}) && res
)
console.log(findValuesDeepByKey(object, 'prop1'));
console.log(findValuesDeepByKey(object, 'prop5'));
console.log(findValuesDeepByKey(object, 'key100'));
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.10/lodash.min.js"></script>
答案 5 :(得分:0)
Array.prototype.findpath = function(item,path) {
return this.find(function(f){return item==eval('f.'+path)});
}
答案 6 :(得分:0)
我是这样做的:
function _find( obj, field, results )
{
var tokens = field.split( '.' );
// if this is an array, recursively call for each row in the array
if( obj instanceof Array )
{
obj.forEach( function( row )
{
_find( row, field, results );
} );
}
else
{
// if obj contains the field
if( obj[ tokens[ 0 ] ] !== undefined )
{
// if we're at the end of the dot path
if( tokens.length === 1 )
{
results.push( obj[ tokens[ 0 ] ] );
}
else
{
// keep going down the dot path
_find( obj[ tokens[ 0 ] ], field.substr( field.indexOf( '.' ) + 1 ), results );
}
}
}
}
用以下方法测试:
var obj = {
document: {
payload: {
items:[
{field1: 123},
{field1: 456}
]
}
}
};
var results = [];
_find(obj.document,'payload.items.field1', results);
console.log(results);
输出
[ 123, 456 ]
答案 7 :(得分:0)
使用Deepdash,您可以pickDeep,然后从中获取paths,或indexate(构建路径->值对象)
var obj = { 'aa': 1, 'bb': 2, 'cc': {'d':{'x':9}}, dd:{'d':{'y':9}}}
var cherry = _.pickDeep(obj,"d");
console.log(JSON.stringify(cherry));
// {"cc":{"d":{}},"dd":{"d":{}}}
var paths = _.paths(cherry);
console.log(paths);
// ["cc.d", "dd.d"]
paths = _.paths(cherry,{pathFormat:'array'});
console.log(JSON.stringify(paths));
// [["cc","d"],["dd","d"]]
var index = _.indexate(cherry);
console.log(JSON.stringify(index));
// {"cc.d":{},"dd.d":{}}
这里是Codepen demo
答案 8 :(得分:0)
如果您没有看到来自@eugene的最新答案,此调整可让您传递要搜索的键列表!
// Method that will find any "message" in the Apex errors that come back after insert attempts
// Could be a validation rule, or duplicate record, or pagemessage.. who knows!
// Use in your next error toast from a wire or imperative catch path!
// message: JSON.stringify(this.findNested(error, ['message', 'stackTrace'])),
// Testing multiple keys: this.findNested({thing: 0, list: [{message: 'm'}, {stackTrace: 'st'}], message: 'm2'}, ['message', 'stackTrace'])
findNested(obj, keys, memo) {
let i,
proto = Object.prototype,
ts = proto.toString,
hasOwn = proto.hasOwnProperty.bind(obj);
if ('[object Array]' !== ts.call(memo)) memo = [];
for (i in obj) {
if (hasOwn(i)) {
if (keys.includes(i)) {
memo.push(obj[i]);
} else if ('[object Array]' === ts.call(obj[i]) || '[object Object]' === ts.call(obj[i])) {
this.findNested(obj[i], keys, memo);
}
}
}
return memo.length == 0 ? null : memo;
}
答案 9 :(得分:0)
我们将对象扫描用于数据处理任务。一旦把头缠在如何使用上,它就会很棒。
const objectScan = require('object-scan');
const haystack = { a: { b: { c: 'd' }, e: { f: 'g' } } };
const r = objectScan(['a.*.*'], { joined: true, rtn: 'entry' })(haystack);
console.log(r);
// => [ [ 'a.e.f', 'g' ], [ 'a.b.c', 'd' ] ]
网站上还有更多示例。