How do I search a string in JavaScript array using jQuery?

Question

I have a JavaScript array:

var j_array = new Array();
j_arry=["class:1","division:a","class:5","class:3","division:b","division:c","division:d","class:10"];

I need to find how many times the class is coming and its array key, so I use:

found = $.inArray('class', j_array); ` But it returns `-1`;

Then I use:

var search = 'class';
$.each([j_array], function(index, value){
    $.each(value, function(key, cell){
        if (search.indexOf(cell) !== -1)
            console.log('found in array '+index, cell);
    });
});

But that is also wrong. How do I solve this?

From this array I want to get the following:

Class coming 4 times, at key 0, 2, 3, and 7

I want to make a separate array of class only, that is,

new_array = ["class:1", "class:2", "class:3", "class:10"];

Currently there are four classes in j_array. How can I get the Nth class value

That is, 1st class value ="class:1", 2nd class value="class:5", etc.

1 . dont use nested loop 2. do reverse e.g `item.indexOf(searchTerm);` — DivineCoder, Dec 24 '16 at 13:47
This question is definitely too broad—there are many possible ways to do it. — Michał Perłakowski, Dec 24 '16 at 21:14

kevin ternet · Answer 1 · 2016-12-24T13:50:41.293

13

You could filter elements which match in a new array and just return the length of this new array

var j_arry = ["class:1","division:a","class:5","class:3","division:b","division:c","division:d","class:10"];
var res = j_arry.filter(x => x.includes("class"));
var key = res.map(x => x.split(":")[1]);
console.log("Class coming " + res.length + " , at key " + key.join(","));
console.log("new array = ", res);

edited Dec 24 '16 at 13:50

answered Dec 24 '16 at 13:37

kevin ternet

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kukkuz · Accepted Answer · 2016-12-24T13:50:59.197

8

Use Array.prototype.filter to filter out the elements of the array that contains the string class - see demo below:

var j_array =["class:1","division:a","class:5","class:3","division:b","division:c","division:d","class:10"];

var result = j_array.filter(function(e){
  return e.indexOf('class')!==-1;
});

console.log(result);

EDIT:

To get the list of indexes too, you can try this:

var j_array =["class:1","division:a","class:5","class:3","division:b","division:c","division:d","class:10"];

var filteredIndices = []

var filtered = j_array.filter(function(e,i){
  if(e.indexOf('class')!==-1) {
    filteredIndices.push(i);
    return true;
  } else {
    return false;
  }
});

console.log(filtered);
console.log(filteredIndices);

// Nth class value
console.log(filtered[2]); // this prints the 3rd one

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

edited Dec 24 '16 at 13:50

answered Dec 24 '16 at 13:32

kukkuz

41,512
6
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1

Thank you friend . But i have 3 question . Could you please answer for this separate 3 question . – Kamal Dec 24 '16 at 13:34
question (1) and (2) would be clear from the `result` in the demo above... For (3) you can do `result[N]`... – kukkuz Dec 24 '16 at 13:36
2

No friend . Actually you give the answer for question 2, and result[N] is based on new array . But i want result[N] based on J_array. And i cannot find answer for 1st question . – Kamal Dec 24 '16 at 13:39
for (1), you need a array like [0,2,3,7]? – kukkuz Dec 24 '16 at 13:45
Yes. Please update this in your answer . – Kamal Dec 24 '16 at 13:46
@Kamal see updated answer, thanks! – kukkuz Dec 24 '16 at 13:52

score 3 · Answer 3 · answered Dec 24 '16 at 13:58

Here is the answer to your questions 1 + 2. It is also 'n' proof so answers your part 3 also. This works by old-fashioned hard graft rather than funky functions. The original array entries are split and filtered then if qualifying we store in an associative array (results) using a pointer array (list) to make it easier to give a sorted result and pull the values from the associative array. The max variable is probably not necessary but included for clarity - could have used list.length instead. Note that the list[] array will be sparse (missing steps) so we test each entry before use in the output steps.

  var j_array = new Array();
  j_arry=["class:1","division:a","class:5","class:3","division:b","division:c","division:d","class:10","class:1"];

  var a, result = [], list=[], max = -1    
  for (var i =0; i < j_arry.length; i = i + 1) {

   var a = j_arry[i].split(":")
    if ( a[0] === "class") {
    
    var key = "c" + a[1]
     if ( !result[key] ) { result[key] = {pos:[]}}
    result[key].cnt = result[key].cnt ? result[key].cnt + 1 : 1; 
     result[key].pos.push(i)
     list[parseInt(a[1])] = "c" + a[1]  
     
     max = parseInt(a[1]) > max ? a[1] : max;
    }
  }

  // say locations
  for (var i = 0; i < max; i = i + 1) {
   if (list[i]) {
   key = "c" + i 
    console.log("Class " + i + " occurs  at " + result[key].pos.toString()  )
    }
  }

  // make new array
  var newArray=[]
  for (var i = 0; i < max; i = i + 1) {
   if (list[i]) {
    newArray.push("Class:" + i)
    }
  }
  console.log("New array=" + newArray.toString()  )

Results are:

Class 1 occurs at 0,8 Class 3 occurs at 3 Class 5 occurs at 2 New array=Class:1,Class:3,Class:5

score 3 · Answer 4 · answered Dec 24 '16 at 15:13

Single reduce is sufficient here.

var arr = ["class:1","division:a","class:5","class:3","division:b","division:c","division:d","class:10"],
    res = arr.reduce((p,c) => c.includes("class") ? (p.count++, p.keys.push(c.split(":")[1]), p)
                                                  : p ,{count:0, keys:[]});
console.log(res);

score 3 · Answer 5 · answered Dec 24 '16 at 15:21

You can use the filter and map functions to filter your array to have only elements that match the text 'class', and use array index notation to access the nth element in the array. Check the below code snippet I hope it will be of help to you.

The below code snippet uses ES6 arrow syntax.

var arr = ["class:1", "division:a", "class:5", "class:3", "division:b", "division:c", "division:d", "class:10"];
var result = arr.filter(x => x.indexOf('class') !== -1);
var indices = result.map(x => arr.indexOf(x));
console.log(indices);
console.log(result);

var nValue = window.prompt('Enter n value');
console.log(result[nValue]);

score 3 · Answer 6 · edited May 23 '17 at 12:24

If you're using jQuery to support some really old browser that still don't implement the new Array functions, and you don't want to polyfill those because you're already using jQuery, then you can use the jQuery equivalents:

var arr = ["class:1", "division:a", "class:5", "class:3", "division:b", "division:c", "division:d", "class:10"]
var result = $.grep(arr, function (x) { return x.indexOf('class') !== -1 })
var indices = $.map(result, function (x) { return arr.indexOf(x) })

This is the same code as this answer, but using jQuery.

score 1 · Answer 7 · edited Dec 24 '16 at 14:13

1

You have to do map first then filter.

var j_array = ["class:1", "division:a", "class:5", "class:3", "division:b", "division:c", "division:d", "class:10"];

var result = j_array.map(function(e, i) {
  return e.indexOf('class') > -1 ? '' + i : false;
}).filter(function(e) {
  return !!e;
});
console.log(result);

edited Dec 24 '16 at 14:13

kukkuz

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answered Dec 24 '16 at 13:59

Anil Talla

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How do I search a string in JavaScript array using jQuery?

7 Answers7