在 JavaScript 中递归遍历数组并修改对象属性的值

Question

这篇文章可能看起来很长，但很容易理解，如果不是，我会添加更多细节

我有 criteria 数组，看起来像：

   let criteria = [
        "and",
        {
          "Collection": "persons",
          "Property": "phone",
          "operator": "eq",
          "operatorValue": "$p:phone"  
        },
        {
          "Collection": "persondetails",
          "Property": "country",
          "operator": "eq",
          "operatorValue": "$p:country" 
        },
        ["or",
        {
          "Collection": "persons",
          "Property": "city",
          "operator": "eq",
          "operatorValue": "$p:city"  
        }]
      ]

criteria的特征：

1. 它可以有嵌套数组。
2. 第一项数组（或嵌套数组）总是“与”或“或”
3. 数组中的第二个项目，项目可以是具有此特定结构的对象

{ "Collection": "persons", "Property": "phone", "operator": "eq", "operatorValue": "$p:phone" }

或者它可以是一个数组，例如：

["or", { "Collection": "persons", "Property": "city", "operator": "eq", "operatorValue": "$p:city" }]

4. 对象永远不会是嵌套对象

还有一个parameters对象：

let parameters = {phone:"23138213", "country": "Russia", "city":"york"}

目的是递归遍历criteria 数组中的所有operatorValue 属性，如果遇到诸如$p:phone 之类的值，则将其替换为parameters["phone"] 求值的任何值。

预期输出：

[
    "and",
    {
      "Collection": "persons",
      "Property": "phone",
      "operator": "eq",
      "operatorValue": "23138213"  
    },
    {
      "Collection": "persondetails",
      "Property": "country",
      "operator": "eq",
      "operatorValue": "Russia" 
    },
    ["or",
    {
      "Collection": "persons",
      "Property": "city",
      "operator": "eq",
      "operatorValue": "york"  
    }]
  ]

我能够递归遍历数组。唯一的问题是我不知道如何修改原来的criteria变量。

REPL

请参阅 repl 中的第 43 行。 item[1]=parameters[item[1].split('$p:')[1]] 我理解为什么它不会修改标准，因为这里的 item 是完全不同范围内的不同变量。

尝试失败：

  function traverse(obj,parameters){
  
    obj.forEach((item,index)=>{
      
      
     if( typeof item == 'string' ){
       //do nothing
     }
     else if( !(item instanceof Array)){
         
       Object.entries(item).forEach((item,index)=>{
         
         if( item[1] instanceof Array){ 
                      
           traverse(item,parameters);
         }else{
           if(item[1].startsWith('$p:')){
               item[1]=parameters[item[1].split('$p:')[1]] //values dont get replaced for obvious reason
               console.log(item[1])
           } 
         }
       })        
     }
     else if( item  instanceof Array){           
           traverse(item,parameters);
      }  
    })
  }

  traverse(criteria,parameters)
  console.log(criteria)

我该如何解决这个问题？

Answer 1

您可以简化您的功能。您不需要遍历对象的条目。你也不需要splitoperationValue。 parameters 的映射键存在于 Property 键中。

• 遍历数组中的每一项并检查该项是否为Array。
• 如果是，则递归调用该项目上的traverse。
• 如果是对象，则将其operatorValue 属性更新为parameters[val.Property]

function traverse(arr, parameters) {
  for (const item of arr) {
    if (Array.isArray(item))
        traverse(item, parameters)
    else if (typeof item === 'object')
        item.operatorValue = parameters[item.Property]
  }
  return arr
}

let criteria=["and",{Collection:"persons",Property:"phone",operator:"eq",operatorValue:"$p:phone"},{Collection:"persondetails",Property:"country",operator:"eq",operatorValue:"$p:country"},["or",{Collection:"persons",Property:"city",operator:"eq",operatorValue:"$p:city"}]],
    parameters = {phone:"23138213", "country": "Russia", "city":"york"};

console.log(traverse(criteria, parameters))

Answer 2

您需要获取对象的键来分配值。

function traverse(obj, parameters) {
    obj.forEach(item => {
        if (item instanceof Array) return traverse(item, parameters);
        if (item?.operatorValue?.startsWith('$p:')) {
            item.operatorValue = parameters[item.operatorValue.split('$p:')[1]];
        }
    });
}

const
    criteria = ["and", { Collection: "persons", Property: "phone", operator: "eq", operatorValue: "$p:phone" }, { Collection: "persondetails", Property: "country", operator: "eq", operatorValue: "$p:country" }, ["or", { Collection: "persons", Property: "city", operator: "eq", operatorValue: "$p:city" }]],
    parameters = { phone: "23138213", "country": "Russia", "city": "york" };

traverse(criteria, parameters);
console.log(criteria);

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

Answer 3

一个干净的递归版本：

const traverse = ([conj, ...nodes], params) => [
  conj, 
  ... nodes .map (node => Array .isArray (node) 
        ? traverse (node, params)
        : {... node, operatorValue: params [nodes .Property]}
      )
]

const criteria = ["and", {Collection: "persons", Property: "phone", operator: "eq",operatorValue: "$p:phone"}, {Collection: "persondetails", Property: "country", operator: "eq", operatorValue: "$p:country"}, ["or", {Collection: "persons", Property: "city", operator: "eq", operatorValue: "$p:city"}]]
const parameters = {phone:"23138213", "country": "Russia", "city":"york"}


console .log (traverse (criteria, parameters))

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

这仅适用于 adiga 的简化证明对象的 Property 节点与 operatorValue 节点中的后缀具有相同的值。如果没有，我可能会使用正则表达式替换来处理它，并将其分离成一个辅助函数：

const replaceOpValue = (params) => (obj) => ({
  ... obj, 
  operatorValue: obj .operatorValue .replace (/\$p:(.+)/, ((_, key) => params [key]))
})
  

const traverse = ([conj, ...nodes], params) => [
  conj, 
  ... nodes .map (node => Array .isArray (node) 
        ? traverse (node, params)
        : replaceOpValue (params) (node)
      )
]

Answer 4

这是使用object-scan 的解决方案。根据您的要求，可能会更加灵活和可维护

// const objectScan = require('object-scan');

const criteria = [ 'and', { Collection: 'persons', Property: 'phone', operator: 'eq', operatorValue: '$p:phone' }, { Collection: 'persondetails', Property: 'country', operator: 'eq', operatorValue: '$p:country' }, [ 'or', { Collection: 'persons', Property: 'city', operator: 'eq', operatorValue: '$p:city' } ] ];

const substitute = (obj, params) => objectScan(['**.operatorValue'], {
  rtn: 'count',
  filterFn: ({ value, parent, property }) => {
    if (value.startsWith('$p:') && value.slice(3) in params) {
      parent[property] = params[value.slice(3)];
      return true;
    }
    return false;
  }
})(obj);

console.log(substitute(criteria, { phone: '23138213', country: 'Russia', city: 'york' })); // returns number of substitutions
// => 3

console.log(criteria);
// => [ 'and', { Collection: 'persons', Property: 'phone', operator: 'eq', operatorValue: '23138213' }, { Collection: 'persondetails', Property: 'country', operator: 'eq', operatorValue: 'Russia' }, [ 'or', { Collection: 'persons', Property: 'city', operator: 'eq', operatorValue: 'york' } ] ]

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

<script src="https://bundle.run/object-scan@13.8.0"></script>

免责声明：我是object-scan的作者