以递归方式修剪Object键和值中的空格

时间:2015-11-03 22:52:37

标签: javascript object recursion trim

如何以递归方式修剪JavaScript对象中的键和值中的空格?

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

我们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"字符串"。但是,在这种情况下的问题是键和值不如所希望的那样干净。说一个{" key_with_leading_n_trailing_spaces":" my_value_with_leading_spaces" }。

在这种情况下,它很容易导致你出色的JavaScript程序试图利用这些数据(或者我们应该把它称为脏数据?),因为当你的代码试图从这个JSON对象中获取值时,不仅关键不匹配而且价值也无法匹配。我已经浏览了谷歌并找到了一些提示,但没有一种方法可以治愈这一切。

鉴于这个JSON在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   "}
    }
  ]

};

所以这就是我提出的(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));

请注意:我是在1,2步中做到这一点,因为我无法找到更好的一击来解决所有问题。如果我的代码中存在问题或更好,请与我们分享。

谢谢!

8 个答案:

答案 0 :(得分:30)

您可以对其进行字符串化,字符串替换和重新分析

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

答案 1 :(得分:18)

您可以使用 Object.keys 清除属性名称和属性以获取密钥数组,然后使用 Array.prototype.reduce 来迭代密钥和使用修剪的键和值创建一个新对象。该函数需要递归,以便它还修剪嵌套的对象和数组。

请注意,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 的调用需要更复杂才能确定对象的类型(例如适当聪明的 new obj.constructor())。

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)? []:{});
}

答案 2 :(得分:3)

epascarello的上述答案加上一些单元测试(仅供我确定):

{{1}}

答案 3 :(得分:1)

我认为通用的map函数可以很好地解决这一问题。它将深层对象的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,以轻松应用其他转换-

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编写为函数-

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现在必须接受两个功能参数-

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是分开的。字符串值获取简单的.trim,而键获取.trim().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"]}

展开以下代码段,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"]}

答案 4 :(得分:0)

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 }"));

答案 5 :(得分:0)

我尝试了上面的解决方案JSON.stringify解决方案,但是它不适用于''this is \'my \'test''这样的字符串。您可以使用stringify的replacer函数来解决它,而只需修剪输入的值即可。

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

答案 6 :(得分:0)

@RobG谢谢您的解决方案。再添加一个条件将不会创建更多的嵌套对象

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)? []:{});
    }

答案 7 :(得分:0)

我使用的最佳解决方案是这个。查看replacer function上的文档。

var obj = {"data": {"address": {"city": "\n \r     New York", "country": "      USA     \n\n\r"}}};

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

console.log(JSON.parse(trimmed));