Trim 对象键和值中递归的空格

Question

如何在 JavaScript 对象的键和值中递归地 trim 空格？

我遇到了一个问题，我试图 "clean" 用户提供 JSON 字符串并将其发送到我的其他代码中以供进一步处理。

假设我们有一个用户提供的 JSON 字符串，其属性键和值的类型为 "string"。但是，在这种情况下，问题在于键和值并不像预期的那样干净。说 { " key_with_leading_n_trailing_spaces ": " my_value_with_leading_spaces" }.

在这种情况下，它很容易导致试图使用此类数据（或者我们应该称之为脏数据？）的编写出色的 JavaScript 程序出现问题，因为当您的代码试图获取value out this JSON object，不仅键不匹配，而且值也无法匹配。我环顾四周 google 并发现了一些技巧，但没有一种方法可以包治百病。

鉴于此 JSON 键和值中有很多空格。

var badJson = {
  "  some-key   ": "    let it go    ",
  "  mypuppy     ": "    donrio   ",
  "   age  ": "   12.3",
  "  children      ": [
    { 
      "   color": " yellow",
      "name    ": "    alice"
    },    { 
      "   color": " silver        ",
      "name    ": "    bruce"
    },    { 
      "   color": " brown       ",
      "     name    ": "    francis"
    },    { 
      "   color": " red",
      "      name    ": "    york"
    },

  ],
  "     house": [
    {
      "   name": "    mylovelyhouse     ",
      " address      " : { "number" : 2343, "road    "  : "   boardway", "city      " : "   Lexiton   "}
    }
  ]

};

这就是我想出的（在使用 lodash.js 的帮助下）：

//I made this function to "recursively" hunt down keys that may 
//contain leading and trailing white spaces
function trimKeys(targetObj) {

  _.forEach(targetObj, function(value, key) {

      if(_.isString(key)){
        var newKey = key.trim();
        if (newKey !== key) {
            targetObj[newKey] = value;
            delete targetObj[key];
        }

        if(_.isArray(targetObj[newKey]) || _.isObject(targetObj[newKey])){
            trimKeys(targetObj[newKey]);
        }
      }else{

        if(_.isArray(targetObj[key]) || _.isObject(targetObj[key])){
            trimKeys(targetObj[key]);
        }
      }
   });

}

//I stringify this is just to show it in a bad state
var badJson = JSON.stringify(badJson);

console.log(badJson);

//now it is partially fixed with value of string type trimed
badJson = JSON.parse(badJson,function(key,value){
    if(typeof value === 'string'){
        return value.trim();
    }
    return value;
});

trimKeys(badJson);

console.log(JSON.stringify(badJson));

此处注意：我分 1、2 步完成此操作，因为我找不到更好的一次性解决方案。如果我的代码有问题或有更好的地方，请与我们分享。

谢谢！

Answer 1

您可以对其进行字符串化、字符串替换和重新解析

JSON.parse(JSON.stringify(badJson).replace(/"\s+|\s+"/g,'"'))

Answer 2

您可以使用 Object.keys 清理属性名称和属性以获取键数组，然后 Array.prototype.reduce 迭代键并创建一个具有修剪键和值的新对象。该函数需要递归，以便它还可以修剪嵌套的对象和数组。

注意它只处理普通数组和对象，如果你想处理其他类型的对象，对reduce的调用需要更复杂来确定类型对象（例如 new obj.constructor() 的一个合适的聪明版本）。

function trimObj(obj) {
  if (!Array.isArray(obj) && typeof obj != 'object') return obj;
  return Object.keys(obj).reduce(function(acc, key) {
    acc[key.trim()] = typeof obj[key] == 'string'? obj[key].trim() : trimObj(obj[key]);
    return acc;
  }, Array.isArray(obj)? []:{});
}

Answer 3

上面 epascarello 的回答加上一些单元测试（只是为了我确定）：

function trimAllFieldsInObjectAndChildren(o: any) {
  return JSON.parse(JSON.stringify(o).replace(/"\s+|\s+"/g, '"'));
}

import * as _ from 'lodash';
assert.true(_.isEqual(trimAllFieldsInObjectAndChildren(' bob '), 'bob'));
assert.true(_.isEqual(trimAllFieldsInObjectAndChildren('2 '), '2'));
assert.true(_.isEqual(trimAllFieldsInObjectAndChildren(['2 ', ' bob ']), ['2', 'bob']));
assert.true(_.isEqual(trimAllFieldsInObjectAndChildren({'b ': ' bob '}), {'b': 'bob'}));
assert.true(_.isEqual(trimAllFieldsInObjectAndChildren({'b ': ' bob ', 'c': 5, d: true }), {'b': 'bob', 'c': 5, d: true}));
assert.true(_.isEqual(trimAllFieldsInObjectAndChildren({'b ': ' bob ', 'c': {' d': 'alica c c '}}), {'b': 'bob', 'c': {'d': 'alica c c'}}));
assert.true(_.isEqual(trimAllFieldsInObjectAndChildren({'a ': ' bob ', 'b': {'c ': {'d': 'e '}}}), {'a': 'bob', 'b': {'c': {'d': 'e'}}}));
assert.true(_.isEqual(trimAllFieldsInObjectAndChildren({'a ': ' bob ', 'b': [{'c ': {'d': 'e '}}, {' f ': ' g ' }]}), {'a': 'bob', 'b': [{'c': {'d': 'e'}}, {'f': 'g' }]}));

Answer 4

类似于 epascarello 的回答。这就是我所做的：

import java.util.regex.Matcher;
import java.util.regex.Pattern;

........

public String trimWhiteSpaceAroundBoundary(String inputJson) {
    String result;
    final String regex = "\"\s+|\s+\"";
    final Pattern pattern = Pattern.compile(regex);
    final Matcher matcher = pattern.matcher(inputJson.trim());
    // replacing the pattern twice to cover the edge case of extra white space around ','
    result = pattern.matcher(matcher.replaceAll("\"")).replaceAll("\"");
    return result;
}

测试用例

assertEquals("\"2\"", trimWhiteSpace("\" 2 \""));
assertEquals("2", trimWhiteSpace(" 2 "));
assertEquals("{   }", trimWhiteSpace("   {   }   "));
assertEquals("\"bob\"", trimWhiteSpace("\" bob \""));
assertEquals("[\"2\",\"bob\"]", trimWhiteSpace("[\"  2  \",  \"  bob  \"]"));
assertEquals("{\"b\":\"bob\",\"c c\": 5,\"d\": true }",
              trimWhiteSpace("{\"b \": \" bob \", \"c c\": 5, \"d\": true }"));

Answer 5

我尝试了上面的 JSON.stringify 解决方案，但它不适用于像“"this is \'my\' test"”这样的字符串。您可以使用 stringify 的替换函数绕过它，只需 trim 输入的值。

JSON.parse(JSON.stringify(obj, (key, value) => (typeof value === 'string' ? value.trim() : value)))

Answer 6

我认为通用 map 函数可以很好地处理这个问题。它将深度对象遍历和转换与我们希望执行的特定操作分开 -

const identity = x =>
  x

const map = (f = identity, x = null) =>
  Array.isArray(x)
    ? x.map(v => map(f, v))
: Object(x) === x
    ? Object.fromEntries(Object.entries(x).map(([ k, v ]) => [ map(f, k), map(f, v) ]))
    : f(x)

const dirty = 
` { "  a  ": "  one "
  , " b": [ null,  { "c ": 2, " d ": { "e": "  three" }}, 4 ]
  , "  f": { "  g" : [ "  five", 6] }
  , "h " : [[ [" seven  ", 8 ], null, { " i": " nine " } ]]
  , " keep  space  ": [ " betweeen   words.  only  trim  ends   " ]
  }
`
  
const result =
  map
   ( x => String(x) === x ? x.trim() : x // x.trim() only if x is a String
   , JSON.parse(dirty)
   )
   
console.log(JSON.stringify(result))
// {"a":"one","b":[null,{"c":2,"d":{"e":"three"}},4],"f":{"g":["five",6]},"h":[[["seven",8],null,{"i":"nine"}]],"keep  space":["betweeen   words.  only  trim  ends"]}

map 可以重复使用以轻松应用不同的转换 -

const result =
  map
   ( x => String(x) === x ? x.trim().toUpperCase() : x
   , JSON.parse(dirty)
   )

console.log(JSON.stringify(result))
// {"A":"ONE","B":[null,{"C":2,"D":{"E":"THREE"}},4],"F":{"G":["FIVE",6]},"H":[[["SEVEN",8],null,{"I":"NINE"}]],"KEEP  SPACE":["BETWEEEN   WORDS.  ONLY  TRIM  ENDS"]}

使map实用

感谢 Scott 的评论，我们为 map 添加了一些人体工程学。在此示例中，我们将 trim 写为函数 -

const trim = (dirty = "") =>
  map
   ( k => k.trim().toUpperCase()          // transform keys
   , v => String(v) === v ? v.trim() : v  // transform values
   , JSON.parse(dirty)                    // init
   )

这意味着 map 现在必须接受两个函数参数 -

const map = (fk = identity, fv = identity, x = null) =>
  Array.isArray(x)
    ? x.map(v => map(fk, fv, v)) // recur into arrays
: Object(x) === x
    ? Object.fromEntries(
        Object.entries(x).map(([ k, v ]) =>
          [ fk(k)           // call fk on keys
          , map(fk, fv, v)  // recur into objects
          ] 
        )
      )
: fv(x) // call fv on values

现在我们可以看到键转换与值转换分开工作。字符串值得到一个简单的 .trim 而键得到 .trim() 和 .toUpperCase() -

console.log(JSON.stringify(trim(dirty)))
// {"A":"one","B":[null,{"C":2,"D":{"E":"three"}},4],"F":{"G":["five",6]},"H":[[["seven",8],null,{"I":"nine"}]],"KEEP  SPACES":["betweeen   words.  only  trim  ends"]}

展开下面的代码片段以在您自己的浏览器中验证结果 -

const identity = x =>
  x

const map = (fk = identity, fv = identity, x = null) =>
  Array.isArray(x)
    ? x.map(v => map(fk, fv, v))
: Object(x) === x
    ? Object.fromEntries(
        Object.entries(x).map(([ k, v ]) =>
          [ fk(k), map(fk, fv, v) ]
        )
      )
: fv(x)

const dirty = 
` { "  a  ": "  one "
  , " b": [ null,  { "c ": 2, " d ": { "e": "  three" }}, 4 ]
  , "  f": { "  g" : [ "  five", 6] }
  , "h " : [[ [" seven  ", 8 ], null, { " i": " nine " } ]]
  , " keep  spaces  ": [ " betweeen   words.  only  trim  ends   " ]
  }
`

const trim = (dirty = "") =>
  map
   ( k => k.trim().toUpperCase()
   , v => String(v) === v ? v.trim() : v
   , JSON.parse(dirty)
   )
   
console.log(JSON.stringify(trim(dirty)))
// {"A":"one","B":[null,{"C":2,"D":{"E":"three"}},4],"F":{"G":["five",6]},"H":[[["seven",8],null,{"I":"nine"}]],"KEEP  SPACES":["betweeen   words.  only  trim  ends"]}

Answer 7

@RobG 感谢您提供解决方案。再添加一个条件不会创建更多嵌套对象

function trimObj(obj) {
      if (obj === null && !Array.isArray(obj) && typeof obj != 'object') return obj;
      return Object.keys(obj).reduce(function(acc, key) { 
        acc[key.trim()] = typeof obj[key] === 'string' ? 
          obj[key].trim() : typeof obj[key] === 'object' ?  trimObj(obj[key]) : obj[key];
        return acc;
      }, Array.isArray(obj)? []:{});
    }

Answer 8

我使用的最好的解决方案是这个。查看 replacer function.

上的文档

function trimObject(obj){
  var trimmed = JSON.stringify(obj, (key, value) => {
    if (typeof value === 'string') {
      return value.trim();
    }
    return value;
  });
  return JSON.parse(trimmed);
}

var obj = {"data": {"address": {"city": "\n \r     New York", "country": "      USA     \n\n\r"}}};
console.log(trimObject(obj));

Trim 对象键和值中递归的空格

Trim white spaces in both Object key and value recursively

javascript

recursion

trim

object