重新格式化分层JSON数据

Question

我有以下使用PHP从PostgreSQL数据库中检索到的分层JSON数据：

[{"treelevel":"1","app":"Front","lrflag":null,"ic":null,"price":null,"parentlevel":"0","seq":"27", "indexlistid":439755},
{"treelevel":"2","app":"V-Series","lrflag":null,"ic":null,"price":null,"parentlevel":"1","seq":"28", "indexlistid":439755},
{"treelevel":"3","app":"opt J56","lrflag":null,"ic":null,"price":null,"parentlevel":"2","seq":"29", "indexlistid":439755},
{"treelevel":"4","app":"R.","lrflag":"R","ic":"536-01132AR","price":"693.00","parentlevel":"3","seq":"30", "indexlistid":439755},
{"treelevel":"4","app":"L.","lrflag":"L","ic":"536-01133AL","price":"693.00","parentlevel":"3","seq":"31", "indexlistid":439755},
{"treelevel":"3","app":"opt J63","lrflag":null,"ic":null,"price":null,"parentlevel":"2","seq":"32", "indexlistid":439755},
{"treelevel":"4","app":"R.","lrflag":"R","ic":"536-01130R","price":null,"parentlevel":"3","seq":"33", "indexlistid":439755},
{"treelevel":"4","app":"L.","lrflag":"L","ic":"536-01131L","price":null,"parentlevel":"3","seq":"34", "indexlistid":439755}]

我需要一些重新格式化数据的方法：

[{"app": "Front-V-Series-opt J56-R. R", "price": "$693", "ic": "536-01132AR"},
{"app": "Front-V-Series-opt J56-L. L", "price": "$693", "ic": "536-01132AL"},
{"app": "Front-V-Series-opt J63-R. R", "price": null, "ic": "536-01130R"},
{"app": "Front-V-Series-opt J63-L. L", "price": null, "ic": "536-01131L"}]

父级和树级是构成树关系的两个键。

新格式的应用程序值是一个树级别中所有节点的应用程序和lrflag值加上该级别中最深叶节点的价格和ic的串联。 这是可视化为树的数据：

[
    {
        "app": "Front",
        "children": [
            {
                "app": "V-Series",
                "children": [
                    {
                        "app": "opt J56",
                        "children": [
                            {
                                "app": "R. ,
                                "lrflag": "R",
                                "ic": "536-01132AR",
                                "price": "$693"
                            },
                            {
                                "app": "L. ,
                                "lrflag": "L",
                                "ic": "536-01132AL",
                                "price": "$693"
                            }
                        ]
                    },
                    {
                        "app": "opt J63",
                         "children": [
                            {
                                "app": "R. ,
                                "lrflag": "R",
                                "ic": "536-01130R"
                            },
                            {
                                "app": "L. ,
                                "lrflag": "L",
                                "ic": "536-01131L"
                            }
                        ]
                    }
                ]
            }
        ]
    }
]

我尝试了几种不同的方法来做到这一点，但被困住了。 这是我尝试修改并使用的一些功能，以免碰运气。 我什至无法从这些树中正确构建树。

function buildTree(list) {
    var map = {}, node, roots = [], i;
    for (i = 0; i < list.length; i += 1) {
        map[list[i].treelevel] = i; // initialize the map
        list[i].children = []; // initialize the children
    }
    for (i = 0; i < list.length; i += 1) {
        node = list[i];
        if (node.parentlevel !== "0") {
            // if you have dangling branches check that map[node.parentId] exists
            list[map[node.parentlevel]].children.push(node);
        } else {
            roots.push(node);
        }
    }
    return roots;
}

function listToTree(data, options) {
    options = options || {};
    var ID_KEY = options.idKey || 'treelevel';
    var PARENT_KEY = options.parentKey || 'parentlevel';
    var CHILDREN_KEY = options.childrenKey || 'children';

    var tree = [],
        childrenOf = {};
    var item, id, parentId;

    for (var i = 0, length = data.length; i < length; i++) {
        item = data[i];
        id = item[ID_KEY];
        parentId = item[PARENT_KEY] || 0;
        // every item may have children
        childrenOf[id] = childrenOf[id] || [];
        // init its children
        item[CHILDREN_KEY] = childrenOf[id];
        if (parentId != 0) {
            // init its parent's children object
            childrenOf[parentId] = childrenOf[parentId] || [];
            // push it into its parent's children object
            childrenOf[parentId].push(item);
        } else {
            tree.push(item);
        }
    };

    return tree;
}

unflattenToObject = function(array, parent) {
    var tree = {};
    parent = typeof parent !== 'undefined' ? parent : {id: 0};

    var childrenArray = array.filter(function(child) {
        return child.treelevel == parent.parentlevel;
    });

    if (childrenArray.length > 0) {
        var childrenObject = {};
        // Transform children into a hash/object keyed on token
        childrenArray.forEach(function(child) {
            childrenObject[child.treelevel] = child;
        });
        if (parent.treelevel == 0) {
            tree = childrenObject;
        } else {
            parent['children'] = childrenObject;
        }
        childrenArray.forEach(function(child) {
            unflattenToObject(array, child);
        })
    }
    return tree;
};

Answer 1

这个想法是在最大深度treelevel : 4获取第一个元素的索引treelevel : 4然后从那里开始并向后循环，确保您没有两次遍历父级，并在到达顶部时连接您需要的内容级别，则删除该元素

将其包装在函数中，并递归调用它，直到没有treelevel : 4还剩treelevel : 4

 var data = [{ "treelevel": "1", "app": "Front", "lrflag": null, "ic": null, "price": null, "parentlevel": "0", "seq": "27", "indexlistid": 439755 }, { "treelevel": "2", "app": "V-Series", "lrflag": null, "ic": null, "price": null, "parentlevel": "1", "seq": "28", "indexlistid": 439755 }, { "treelevel": "3", "app": "opt J56", "lrflag": null, "ic": null, "price": null, "parentlevel": "2", "seq": "29", "indexlistid": 439755 }, { "treelevel": "4", "app": "R.", "lrflag": "R", "ic": "536-01132AR", "price": "693.00", "parentlevel": "3", "seq": "30", "indexlistid": 439755 }, { "treelevel": "4", "app": "L.", "lrflag": "L", "ic": "536-01133AL", "price": "693.00", "parentlevel": "3", "seq": "31", "indexlistid": 439755 }, { "treelevel": "3", "app": "opt J63", "lrflag": null, "ic": null, "price": null, "parentlevel": "2", "seq": "32", "indexlistid": 439755 }, { "treelevel": "4", "app": "R.", "lrflag": "R", "ic": "536-01130R", "price": null, "parentlevel": "3", "seq": "33", "indexlistid": 439755 }, { "treelevel": "4", "app": "L.", "lrflag": "L", "ic": "536-01131L", "price": null, "parentlevel": "3", "seq": "34", "indexlistid": 439755 } ]; // first go look for the max depth in the tree, if it's always four, skip this part and put var max = 4 var max = 0; data.forEach(function(elem) { if (elem.treelevel >= max) max = elem.treelevel; }); // implement a function to get the index of the first matched element with treelevel = max depth function getIndex() { return data.indexOf(data.find((elem) => { return elem.treelevel == max; })); } var myTree = []; function formObject(ndx) { var myObj = {}; myObj.app = []; var nextTreeLevel = 3; // start looping from the index of the element in max depth backwards for (var i = ndx; i >= 0; i--) { // if the next element ( backwards ) is a parent , this is to avoid going through parent's siblings if (nextTreeLevel == data[i].parentlevel) { if (data[i].ic != null) myObj.ic = data[i].ic; if (data[i].price != null) myObj.price = data[i].price; if (data[i].lrflag != null) myObj.app.push(data[i].lrflag); myObj.app.push(data[i].app); nextTreeLevel = data[i].parentlevel - 1; // prent's level } } data.splice(ndx, 1); // remove the lement once you're done with it // glue the "app" together, would be better to use array.join but it's not the same join everywhere myObj.app = myObj.app[4] + '-' + myObj.app[3] + '-' + myObj.app[2] + '-' + myObj.app[1] + ' ' + myObj.app[0]; // just to fill the price with null if there is none if (myObj.price == undefined) myObj.price = null; // push the object to the result's array myTree.push(myObj); // get the index of the next element in max depth ( 4 ) var nextIndex = getIndex(); // if there's still another element in the max depth, recall the same function with it's index if (nextIndex > -1) formObject(nextIndex) } formObject(getIndex()) console.log(myTree);