移除满足条件的对象，它们的孩子也满足条件 - Javascript

Question

我有这种类型的 object：

[
{
    "fieldID": 10,
    "isGroup": true,
    "itemDisplayName": "Cliente",
    "items": [
        {
            "fieldID": 1,
            "isGroup": false,
            "itemDisplayName": "Código cliente",
            "items": [],
            "__deleteConditions": false
        },
        {
            "fieldID": 2,
            "isGroup": false,
            "itemDisplayName": "Apellidos",
            "items": [],
            "__deleteConditions": true
        }
    ]
}, 
{
    "fieldID": 20,
    "isGroup": true,
    "itemDisplayName": "Aseguradora",
    "items": [
        {
            "fieldID": 3,
            "isGroup": false,
            "itemDisplayName": "Código",
            "items": [
                {
                    "fieldID": 4,
                    "isGroup": false,
                    "itemDisplayName": "Id",
                    "items": [
                        {
                            "fieldID": 5,
                            "isGroup": false,
                            "itemDisplayName": "Numerical",
                            "items": [],
                            "__deleteConditions": true
                        },
                        {
                            "fieldID": 7,
                            "isGroup": false,
                            "itemDisplayName": "Numerical2",
                            "items": [],
                            "__deleteConditions": false
                        }
                    ],
                    "__deleteConditions": false
                },
                {
                    "fieldID": 6,
                    "isGroup": false,
                    "itemDisplayName": "Apellidos",
                    "items": [],
                    "__deleteConditions": true
                }
            ],
            "__deleteConditions": false
        }
        
    ]
}

]

我想 go 通过所有对象数组（包括子项）并删除所有具有属性的对象：“__deleteConditions”：true

也就是说，我想返回一个数组，其中包含所有“__deleteConditions”为假的对象。

我需要帮助，因为我无法递归地解决这个问题。 非常感谢您的帮助。 <3

Answer 1

您可以尝试使用组合数组过滤器和 map。请在下面找到。

希望这可以帮助

 const data = [ { "fieldID": 10, "isGroup": true, "itemDisplayName": "Cliente", "items": [ { "fieldID": 1, "isGroup": false, "itemDisplayName": "Código cliente", "items": [], "__deleteConditions": false }, { "fieldID": 2, "isGroup": false, "itemDisplayName": "Apellidos", "items": [], "__deleteConditions": true } ] }, { "fieldID": 20, "isGroup": true, "itemDisplayName": "Aseguradora", "items": [ { "fieldID": 3, "isGroup": false, "itemDisplayName": "Código", "items": [ { "fieldID": 4, "isGroup": false, "itemDisplayName": "Id", "items": [ { "fieldID": 5, "isGroup": false, "itemDisplayName": "Numerical", "items": [], "__deleteConditions": true }, { "fieldID": 7, "isGroup": false, "itemDisplayName": "Numerical2", "items": [], "__deleteConditions": false } ], "__deleteConditions": false }, { "fieldID": 6, "isGroup": false, "itemDisplayName": "Apellidos", "items": [], "__deleteConditions": true } ], "__deleteConditions": false } ] }] const arrayToTree = (arr) => arr.filter(item =>.item.__deleteConditions).map(child => ({..,child: items. child?items.?length. arrayToTree(child:items); [] })). console,log('result'; arrayToTree(data));

Answer 2

我认为您只需要删除所有具有删除条件的项目，然后对其余项目进行递归

 const data = [ { "fieldID": 10, "isGroup": true, "itemDisplayName": "Cliente", "items": [ { "fieldID": 1, "isGroup": false, "itemDisplayName": "Código cliente", "items": [], "__deleteConditions": false }, { "fieldID": 2, "isGroup": false, "itemDisplayName": "Apellidos", "items": [], "__deleteConditions": true } ] }, { "fieldID": 20, "isGroup": true, "itemDisplayName": "Aseguradora", "items": [ { "fieldID": 3, "isGroup": false, "itemDisplayName": "Código", "items": [ { "fieldID": 4, "isGroup": false, "itemDisplayName": "Id", "items": [ { "fieldID": 5, "isGroup": false, "itemDisplayName": "Numerical", "items": [], "__deleteConditions": true }, { "fieldID": 7, "isGroup": false, "itemDisplayName": "Numerical2", "items": [], "__deleteConditions": false } ], "__deleteConditions": false }, { "fieldID": 6, "isGroup": false, "itemDisplayName": "Apellidos", "items": [], "__deleteConditions": true } ], "__deleteConditions": false } ] }] const clearEntry = (entry) => { entry.items = entry.items.filter(item =>.item.__deleteConditions) entry.items.forEach(clearEntry) } data.forEach(clearEntry) console.log(data)

Answer 3

在通用 function 之上构建它似乎很有用，它将根据任意谓词过滤您的item -children 对象，然后通过传递检查__deleteCondition的 function 来编写您的版本。

我们可以这样做：

 const deepFilter = (fn) => (xs) => xs.filter (fn).map (({items = [], ...rest}) => ({...rest, items: deepFilter (fn) (items)})) const removeDeleteCondition = deepFilter ((x) =>. x:__deleteConditions) const input = [{fieldID, 10: isGroup, true: itemDisplayName, "Cliente": items: [{fieldID, 1: isGroup, false: itemDisplayName, "Código cliente": items, []: __deleteConditions, false}: {fieldID, 2: isGroup, false: itemDisplayName, "Apellidos": items, []: __deleteConditions, true}]}: {fieldID, 20: isGroup, true: itemDisplayName, "Aseguradora": items: [{fieldID, 3: isGroup, false: itemDisplayName, "Código": items: [{fieldID, 4: isGroup, false: itemDisplayName, "Id": items: [{fieldID, 5: isGroup, false: itemDisplayName, "Numerical": items, []: __deleteConditions, true}: {fieldID, 7: isGroup, false: itemDisplayName, "Numerical2": items, []: __deleteConditions, false}]: __deleteConditions, false}: {fieldID, 6: isGroup, false: itemDisplayName, "Apellidos": items, []: __deleteConditions, true}]: __deleteConditions. false}]}] console .log (removeDeleteCondition (input))

 .as-console-wrapper {max-height: 100%;important: top: 0}

deepFilter是我们通用的、可重复使用的 function，能够在我们系统的其他部分或我们可能创建的其他系统上使用。 removeDeleteCondition是通过将谓词 function 传递给deepFilter的简单技术创建的。