Javascript как найти элемент в объекте

Something like this:

var objects = [
  {
    "foo" : "bar",
    "bar" : "sit"
  },
  {
    "foo" : "lorem",
    "bar" : "ipsum"
  },
  {
    "foo" : "dolor",
    "bar" : "amet"
  }
];

var results = [];

var toSearch = "lo";

for(var i=0; i<objects.length; i++) {
  for(key in objects[i]) {
    if(objects[i][key].indexOf(toSearch)!=-1) {
      results.push(objects[i]);
    }
  }
}

The results array will contain all matched objects.

If you search for ‘lo’, the result will be like:

[{ foo="lorem", bar="ipsum"}, { foo="dolor", bar="amet"}]

NEW VERSION — Added trim code, code to ensure no duplicates in result set.

function trimString(s) {
  var l=0, r=s.length -1;
  while(l < s.length && s[l] == ' ') l++;
  while(r > l && s[r] == ' ') r-=1;
  return s.substring(l, r+1);
}

function compareObjects(o1, o2) {
  var k = '';
  for(k in o1) if(o1[k] != o2[k]) return false;
  for(k in o2) if(o1[k] != o2[k]) return false;
  return true;
}

function itemExists(haystack, needle) {
  for(var i=0; i<haystack.length; i++) if(compareObjects(haystack[i], needle)) return true;
  return false;
}

var objects = [
  {
    "foo" : "bar",
    "bar" : "sit"
  },
  {
    "foo" : "lorem",
    "bar" : "ipsum"
  },
  {
    "foo" : "dolor blor",
    "bar" : "amet blo"
  }
];

function searchFor(toSearch) {
  var results = [];
  toSearch = trimString(toSearch); // trim it
  for(var i=0; i<objects.length; i++) {
    for(var key in objects[i]) {
      if(objects[i][key].indexOf(toSearch)!=-1) {
        if(!itemExists(results, objects[i])) results.push(objects[i]);
      }
    }
  }
  return results;
}

console.log(searchFor('lo '));

answered Dec 15, 2011 at 8:32

All the other old answers use a for in loop, modern JavaScript has Object.keys. Combine that with some, includes, and filter and it is a bit nicer.

var a = [{
  name: 'xyz',
  grade: 'x'
}, {
  name: 'yaya',
  grade: 'x'
}, {
  name: 'x',
  frade: 'd'
}, {
  name: 'a',
  grade: 'b'
}];

function filterIt(arr, searchKey) {
  return arr.filter(function(obj) {
    return Object.keys(obj).some(function(key) {
      return obj[key].includes(searchKey);
    })
  });
}

console.log("find 'x'", filterIt(a,"x"));
console.log("find 'a'", filterIt(a,"a"));
console.log("find 'z'", filterIt(a,"z"));

Or with ES6

function filterIt(arr, searchKey) {
  return arr.filter(obj => Object.keys(obj).some(key => obj[key].includes(searchKey)));
}

answered Nov 30, 2016 at 14:18

This is a cool solution that works perfectly

const array = [{"title":"tile hgfgfgfh"},{"title":"Wise cool"},{"title":"titlr DEytfd ftgftgfgtgtf gtftftft"},{"title":"This is the title"},{"title":"yeah this is cool"},{"title":"tile hfyf"},{"title":"tile ehey"}];

var item = array.filter(item=>item.title.toLowerCase().includes('this'));

 alert(JSON.stringify(item))

EDITED

const array = [{"title":"tile hgfgfgfh"},{"title":"Wise cool"},{"title":"titlr DEytfd ftgftgfgtgtf gtftftft"},{"title":"This is the title"},{"title":"yeah this is cool"},{"title":"tile hfyf"},{"title":"tile ehey"}];


// array.filter loops through your array and create a new array returned as Boolean value given out "true" from eachIndex(item) function 

var item = array.filter((item)=>eachIndex(item));

//var item = array.filter();



function eachIndex(e){
console.log("Looping each index element ", e)
return e.title.toLowerCase().includes("this".toLowerCase())
}

console.log("New created array that returns "true" value by eachIndex ", item)

answered May 4, 2018 at 18:31

This is a proposal which uses the key if given, or all properties of the object for searching a value.

function filter(array, value, key) {
    return array.filter(key
        ? a => a[key] === value
        : a => Object.keys(a).some(k => a[k] === value)
    );
}

var a = [{ name: 'xyz', grade: 'x' }, { name: 'yaya', grade: 'x' }, { name: 'x', frade: 'd' }, { name: 'a', grade: 'b' }];


console.log(filter(a, 'x'));
console.log(filter(a, 'x', 'name'));

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

The search function will return all objects which contain a value which has contains the search query

function search(arr, s){
    var matches = [], i, key;
    
    for( i = arr.length; i--; )
        for( key in arr[i] )
            if( arr[i].hasOwnProperty(key) && arr[i][key].indexOf(s) > -1 )
                matches.push( arr[i] );  // <-- This can be changed to anything

    return matches;
};

// dummy data
var items = [
      {
        "foo" : "bar",
        "bar" : "sit"
      },
      {
        "foo" : "lorem",
        "bar" : "ipsum"
      },
      {
        "foo" : "dolor",
        "bar" : "amet"
      }
];
    
var result = search(items, 'lo'); // search "items" for a query value
console.log(result); // print the result

answered Dec 15, 2011 at 8:41

AsafAsaf

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search(searchText) {
  let arrayOfMatchedObjects = arrayOfAllObjects.filter(object => {
    return JSON.stringify(object)
      .toString()
      .toLowerCase()
      .includes(searchText);
  });
  return arrayOfMatchedObjects;
}

This could be very simple, easy, fast and understandable Search function for some of you just like me.

answered Dec 25, 2019 at 19:56

var search(subject, objects) {

    var matches = [];
    var regexp = new RegExp(subject, 'g');

    for (var i = 0; i < objects.length; i++) {
        for (key in objects[i]) {
            if (objects[i][key].match(regexp)) matches.push(objects[i][key]);
        }
    }
    return matches;
};

var items = [
  {
    "foo" : "bar",
    "bar" : "sit"
  },
  {
    "foo" : "lorem",
    "bar" : "ipsum"
  },
  {
    "foo" : "dolor",
    "bar" : "amet"
  }
];

search('r', items);    // ["bar", "lorem", "dolor"]

answered Dec 15, 2011 at 8:27

As a Javascripter Lv. 1 I just learned to search for strings in objects with this:

function isThere( a_string, in_this_object )
{
    if( typeof a_string != 'string' )
    {
        return false;
    }

    for( var key in in_this_object )
    {
        if( typeof in_this_object[key] == 'object' || typeof in_this_object[key] == 'array' )
        {
            if ( isThere( a_string, in_this_object[key] ) )
            {
                return true;
            }
        }
        else if( typeof in_this_object[key] == 'string' )
        {
            if( a_string == in_this_object[key] )
            {
                return true;
            }
        }
    }

    return false;
}

I know is far from perfect but it is useful.

Feel free to comment in order to improve this.

Nadeem JamaliNadeem Jamali

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Hakan BilginHakan Bilgin

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//Search function
var locateInObject = function(obj, key, find, result, currentLocation){
  if(obj === null) return;
  result = result||{done:[],found:{}};
  if(typeof obj == 'object'){
    result.done.push(obj);
  }
  currentLocation = currentLocation||key;
  var keys = Object.keys(obj);
  for(var k=0; k<keys.length; ++k){
    var done = false;
    for(var d=0; d<result.done.length; ++d){
      if(result.done[d] === obj[keys[k]]){
        done = true;
        break;
      }
    }
    if(!done){
      var location = currentLocation+'.'+keys[k];
      if(typeof obj[keys[k]] == 'object'){
        locateInObject(obj[keys[k]], keys[k], find, result, location)
      }else if((typeof find == 'string' && obj[keys[k]].toString().indexOf(find) > -1) || (typeof find == 'function' && find(obj[keys[k]], keys[k]))){
        result.found[location] = obj[keys[k]];
      }
    }
  }
  return result.found;
}

//Test data
var test = {
  key1: {
    keyA: 123,
    keyB: "string"
  },
  key2: {
    keyC: [
      {
        keyI: "string123",
        keyII: 2.3
      },
      "string"
    ],
    keyD: null
  },
  key3: [
    1,
    2,
    123,
    "testString"
  ],
  key4: "123string"
}
//Add a circular reference
test.key5 = test;

//Tests
console.log(locateInObject(test, 'test', 'string'))
console.log(locateInObject(test, 'test', '123'))
console.log(locateInObject(test, 'test', function(val, key){ return key.match(/keyd/) && val.indexOf('string') > -1}))

JstnPwllJstnPwll

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Came across this problem today and using a modified version of the provided code by epascarello in this thread did the trick, because that version had trouble when the object contained some values others than strings (like a number of booleans for example).

console.log('find: ', findIn(arrayOfObjects, searchKey));

const findIn = (arr, searchKey) => {
 return arr.filter(obj => 
  Object.keys(obj).some(key => {
   if (typeof obj[key] === 'string') {
    return obj[key].includes(searchKey);
   }
  })
 );
};

answered Nov 22, 2019 at 2:34

Here is the answer in 100% PURE JavaScript:

<html>

<head>
  <meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
  <title></title>
  <script type="text/javascript">
    var mySet = [{
        "foo": "bar",
        "bar": "sit"
      },
      {
        "foo": "lorem",
        "bar": "ipsum"
      },
      {
        "foo": "dolor",
        "bar": "amet"
      }
    ];

    function queryObject(needle, set) {
      var results = new Array();
      for (index = 0; index < set.length; index++) {
        for (key in set[index]) {
          if (set[index][key].indexOf(needle) > -1) {
            results.push(set[index]);
          }
        }
      }

      if (results.length) {
        return JSON.stringify(results);
      } else {
        return "No match!";
      }
    }
  </script>
</head>

<body>
  <form>
    <input type="text" id="prompt" onFocus="this.value='';" value="Type your query HERE" size="20" onKeyDown="document.getElementById('submit').disabled = false;">
    <input id="submit" type="button" value="Find in Object" onClick="var prompt=document.getElementById('prompt'); if(prompt.value){document.getElementById('output').innerHTML = queryObject(prompt.value, mySet);}else{prompt.value='Type your query HERE';}"
      disabled="disabled">
    <div id="output"></div>
  </form>
</body>

</html>

There are, of course, more fancy ways to traverse your object using JQuery, but this is the basic concept.

Cheers!

*EDIT: Sorry, I didn’t read your question carefully enough, and modified the code to return an array of objects as you requested.

var students=[{name:"ali",family:"romandeh",age:18,curse:[
   {lesson1:"arabic"},
   {lesson2:"english"},
   {lesson3:"history"}
   ]},
   {name:"hadi",family:"porkar",age:48,curse:[
   {lesson1:"arabic"},
   {lesson2:"english"},
   {lesson3:"history"}
   ]},
   {name:"majid",family:"porkar",age:30,curse:[
   {lesson1:"arabic"},
   {lesson2:"english"},
   {lesson3:"history"}
   ]}
   ];
  
    function searchInChild(objects, toSearch){
        var _finded=false;
        for(var i=0; i<objects.length; i++) {
            for(key in objects[i]) {
                if(objects[i][key]!=null && typeof(objects[i][key] )!="boolean" && typeof(objects[i][key] )!="number"){
                    if (typeof objects[i][key] == 'object') {
                        _finded= searchInChild(objects[i][key],toSearch);

                    }
                    else if(objects[i][key].indexOf(toSearch)!=-1) {
                        _finded=true;
                    }
                }
            }
        }
        return _finded;
    }
    function findNested(objects, toSearch) {
        var _results=[];
        for(var i=0; i<objects.length; i++) {
            for(key in objects[i]) {
                if(objects[i][key]!=null && typeof(objects[i][key] )!="boolean" && typeof(objects[i][key] )!="number"){
                    if (typeof objects[i][key] == 'object') {
                        if(searchInChild(objects[i][key],toSearch)){
                            _results.push(objects[i]);
                        }
                    }
                    else if(objects[i][key].indexOf(toSearch)!=-1) {
                        _results.push(objects[i]);
                    }
                }
            }
        }

        return _results;

    }
    $('.quickSearch').on('click',function(){
          var _inputSeach=$('#evertingSearch').val();
          if(_inputSeach!=''){
          var _finded=findNested(students,_inputSeach);
          $('.count').html(_finded.length);}
    
    });

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<html>
<head>
</head>
<body>
<span>
<pre><code>
       var students=[{name:"ali",family:"romandeh",age:18,curse:[
   {lesson1:"arabic"},
   {lesson2:"english"},
   {lesson3:"history"}
   ]},
   {name:"hadi",family:"porkar",age:48,curse:[
   {lesson1:"arabic"},
   {lesson2:"english"},
   {lesson3:"history"}
   ]},
   {name:"majid",family:"rezaeiye",age:30,curse:[
   {lesson1:"arabic"},
   {lesson2:"english"},
   {lesson3:"history"}
   ]}
   ];
</code></pre>

<span>

<input id="evertingSearch" placeholder="Search on students" />
<input type="button" class="quickSearch" value="search" />
<lable>count:</lable><span class="count"></span>


</body>



</html>

Yann ThibodeauYann Thibodeau

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asked Mar 3, 2011 at 13:45

If you have an array such as

var people = [
  { "name": "bob", "dinner": "pizza" },
  { "name": "john", "dinner": "sushi" },
  { "name": "larry", "dinner": "hummus" }
];

You can use the filter method of an Array object:

people.filter(function (person) { return person.dinner == "sushi" });
  // => [{ "name": "john", "dinner": "sushi" }]

In newer JavaScript implementations you can use a function expression:

people.filter(p => p.dinner == "sushi")
  // => [{ "name": "john", "dinner": "sushi" }]

You can search for people who have "dinner": "sushi" using a map

people.map(function (person) {
  if (person.dinner == "sushi") {
    return person
  } else {
    return null
  }
}); // => [null, { "name": "john", "dinner": "sushi" }, null]

or a reduce

people.reduce(function (sushiPeople, person) {
  if (person.dinner == "sushi") {
    return sushiPeople.concat(person);
  } else {
    return sushiPeople
  }
}, []); // => [{ "name": "john", "dinner": "sushi" }]

I’m sure you are able to generalize this to arbitrary keys and values!

answered Mar 3, 2011 at 14:05

answered Mar 25, 2013 at 18:32

var getKeyByDinner = function(obj, dinner) {
    var returnKey = -1;

    $.each(obj, function(key, info) {
        if (info.dinner == dinner) {
           returnKey = key;
           return false; 
        };   
    });

    return returnKey;       

}

jsFiddle.

So long as -1 isn’t ever a valid key.

answered Mar 3, 2011 at 13:56

If you’re going to be doing this search frequently, consider changing the format of your object so dinner actually is a key. This is kind of like assigning a primary clustered key in a database table. So, for example:

Obj = { 'pizza' : { 'name' : 'bob' }, 'sushi' : { 'name' : 'john' } }

You can now easily access it like this: Object['sushi']['name']

Or if the object really is this simple (just ‘name’ in the object), you could just change it to:

Obj = { 'pizza' : 'bob', 'sushi' : 'john' }

And then access it like: Object['sushi'].

It’s obviously not always possible or to your advantage to restructure your data object like this, but the point is, sometimes the best answer is to consider whether your data object is structured the best way. Creating a key like this can be faster and create cleaner code.

answered Mar 8, 2013 at 17:12

You can find the object in array with Alasql library:

var data = [ { name : "bob" , dinner : "pizza" }, { name : "john" , dinner : "sushi" },
     { name : "larry", dinner : "hummus" } ];

var res = alasql('SELECT * FROM ? WHERE dinner="sushi"',[data]);

Try this example in jsFiddle.

answered Dec 21, 2014 at 20:48

RotaretiRotareti

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TomDotTomTomDotTom

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MarcMarc

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

const findDinner = (dinner, data) => objectScan(['*'], {
  abort: true,
  rtn: 'value',
  filterFn: ({ value }) => value.dinner === dinner
})(data);

const data = { 1: { name: 'bob', dinner: 'pizza' }, 2: { name: 'john', dinner: 'sushi' }, 3: { name: 'larry', dinner: 'hummus' } };

console.log(findDinner('sushi', data));
// => { name: 'john', dinner: 'sushi' }

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

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

vincentvincent

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