Question

我有一个这样的数组：

[{Object_id: "000", label: "Avertissement"},{Object_id: "001",label: "Chapitre 1",entry: [{Object_id: "001/003",label: "Chapitre 1-1"},{Object_id: "001/004",label: "Chapitre 1-2",entry: [{Object_id: "001/004/001",label: "Chapitre 1-2-1"},{Object_id: "001/004/002",label: "Chapitre 1-2-2"}]}]}];

我想找到所有父项的Object Id，提供Object_id项因此，如果我给函数赋值“001/004/002”，它返回[“001”，“001/003”]，我的意思是，父母的所有id的数组（如面包屑）。阵列数组最多可以有五个级别。

我更喜欢基于下划线的解决方案，但纯JS或jquery都可以。

Answer 1

看起来Nina只是打败了我！无论如何，这是我的解决方案。它还使用递归来查找您要求的确切项目，但会跟踪它所经历的层次结构。

var o = [{
    Object_id: "000",
    label: "Avertissement"
}, {
    Object_id: "001",
    label: "Chapitre 1",
    entry: [{
        Object_id: "001/003",
        label: "Chapitre 1-1"
    }, {
        Object_id: "001/004",
        label: "Chapitre 1-2",
        entry: [{
            Object_id: "001/004/001",
            label: "Chapitre 1-2-1"
        }, {
            Object_id: "001/004/002",
            label: "Chapitre 1-2-2"
        }]
    }]
}];

var getParentIds = function(id, o, hierarchy) {
    var find = function(id, o) {
        for (var i = 0; i < o.length; ++i) {
            var ob = o[i];
            if (ob['Object_id'] == id) {
                return ob;
            } else {
                if (ob['entry'] != null) {
                    hierarchy.push(ob);
                    return find(id, ob['entry']);
                }
            }
        }
    };
    var theObject = find(id, o, []);

    for (var i = 0; i < hierarchy.length; ++i) {
        document.write(hierarchy[i]['Object_id'] + '<br/>');
    }
};

getParentIds('001/004/002', o, []);

Answer 2

因为你的id已经包含了路径，为什么不使用它呢？

function parentIds(id){
    for(var out = [], i = id.indexOf("/"); i > -1; i = id.indexOf("/", i+1))
        out.push(id.substr(0, index));
    return out;
}

parentIds("001/004/001")

否则

function parentIds(id, data){
    var result = [];
    function traverse(node){
        if(node){
            //if(node instanceof Array) return node.some(traverse);  //ES6
            if(node instanceof Array) return _.some(node, traverse); //underscore
            if(node.Object_id === id) return true;
            if(traverse(node.entry)){
                result.push(node.Object_id);
                return true;
            }
        }
        return false;
    }
    traverse(data);
    return result;
}

Answer 3

此提案使用iter()的递归调用获取给定字符串的所有父项。

var data = [{ Object_id: "000", label: "Avertissement" }, { Object_id: "001", label: "Chapitre 1", entry: [{ "Object_id": "001/003", "label": "Chapitre 1-1" }, { "Object_id": "001/004", "label": "Chapitre 1-2", entry: [{ "Object_id": "001/004/001", "label": "Chapitre 1-2-1" }, { "Object_id": "001/004/002", "label": "Chapitre 1-2-2" }] }] }],
    result = getParentId("001/004/002", data);

function getParentId(id, data) {

    function iter(id, data, level) {
        var key = id.split('/').filter(function (_, i) { return i <= level; }).join('/'),
            result = [];
        data.some(function (a) {
            if (a.Object_id === key) {
                result.push(a.Object_id);
                if (id !== key && Array.isArray(a.entry)) {
                    result = result.concat(iter(id, a.entry, level + 1));
                }
                return true;
            }
        });
        return result;
    }

    return iter(id.substring(0, id.lastIndexOf('/')), data, 0);
}

document.write('<pre>' + JSON.stringify(result, 0, 4) + '</pre>');

搜索多级数组中的项目

3 个答案: