check multiple string contains to each other using javascript

Question

I have three strings which contain comma separated numbers.

var str1 = "1,5,3";
var str2 = "2,5,1";
var str3 = "3,2,1,5";

I want to check these three strings with each other (To get the common elements between all)
Expected output

var result = 1,5;

If I have only two strings, this is the logic I have already used

var array = str2.split(',');
for(var item in array) {
var contains = (str1.indexOf(array[item]) > -1);
if(contains == 1) {
    var result = array[item]+',';
    getele2 += result;
    geteleresult = getele2.replace(/\,$/, '');
  }
}
alert(geteleresult);

but when multiple strings are checked i dont know how to apply sort logic for these. Any idea then please help thanks...

Ian · Accepted Answer · 2013-05-09 04:56:19Z

You can try this Function and logic:

function findCommon() {
    var nums = {};
    var args_length = arguments.length;
    for (var i = 0; i < args_length; i++) {
        var cur_arg = arguments[i];
        var cur_found = {};

        var cur_arg_split = cur_arg.split(",");
        for (var j = 0; j < cur_arg_split.length; j++) {
            var cur_val = cur_arg_split[j];
            if (!(cur_val in cur_found)) {
                cur_found[cur_val] = 1;
                if (!(cur_val in nums)) {
                    nums[cur_val] = 0;
                }
                nums[cur_val]++;
            }
        }
    }

    var ret = [];
    for (var key in nums) {
        if (nums[key] === args_length) {
            ret.push(key);
        }
    }
    return ret;
}

DEMO: http://jsfiddle.net/LkEyj/1/

It allows you to pass any number of variables (strings that contain comma-separated numbers) to the function.

dfsq · Accepted Answer · 2013-05-09 05:13:45Z

1

What about this simple approach:

function common() {
    var arg = [].slice.call(arguments),
        com = [], i;

    for (i = 1; i < arg.length; i++) {
        com = arg[i].match(RegExp((com.length ? com : arg[0].split(',')).join('|'), 'g'));
    }

    return com;
}

http://jsfiddle.net/tUvWM/

It's very simple, and supports any number of strings. E.g.

var arr = common(str1, str2, str3, str4);

answered May 9, 2013 at 5:13

dfsq

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Comments

Prasath K · Accepted Answer · 2013-05-09 04:52:27Z

0

You can use indexOf() Method

var str1 = "1,5,3";
var str2 = "2,5,1";
var str3 = "3,2,1,5";

var array1 = str1.split(',');
var array2 = str2.split(',');
var array3 = str3.split(',');
var output = [];
for(var i=0;i<array1.length;i++)
{
    if(array2.indexOf(array1[i]) >= 0 && array3.indexOf(array1[i]) >= 0)
       output.push(array1[i]);
}
alert(output);

JS Fiddle

answered May 9, 2013 at 4:52

Prasath K

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4 Comments

HaBo Over a year ago

@prashanth this looks good to me. Not sure why this is down voted .

Prasath K Over a year ago

@HaBo i have some haters here who used to downvote for all of my answers

HaBo Over a year ago

We all experience that at times :)

Prasath K Over a year ago

@HaBo i think they are jealous of us :)

Vaibhav Jain · Accepted Answer · 2013-05-09 04:53:09Z

0

you can store the value inside an array and then apply .sort() method to sort the values.

var str1 = "1,5,3";
var str2 = "2,5,1";
var str3 = "3,2,1,5";
var getele2="";
var rarray = new Array();
var array = str2.split(',');
for(var item in array) {
var contains = (str1.indexOf(array[item]) > -1);
if(contains == 1){
var result = array[item];
rarray.push(result);
}
}
alert(rarray.sort());

Here is the fiddle : Demo

http://jsfiddle.net/jvaibhav/BsFPW/

answered May 9, 2013 at 4:53

Vaibhav Jain

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3 Comments

HaBo Over a year ago

I doubt if this works for strings like '11,2,3' and '1,22,33,4'

Vaibhav Jain Over a year ago

@HaBo : Agree. but OP spacify the string and expected output so i posted this.

Vaibhav Jain Over a year ago

@HaBo : any ways thanks. For universal situation indexOf() method is fine as prashant answered.

gnomical · Accepted Answer · 2013-05-09 05:01:31Z

It would be easiest if you could store the strings in an array to start. If you can do so the following should do the trick. This also scales to check more than three if need be with no changes.

str[0] = "1,5,3";
str[1] = "2,5,1";
str[2] = "3,2,1,5"; 

var matches = '';
for(var i=0;i<str.length-1;i++) {
    if(i==0)
        var str1 = str[i].split(',');
    else
        var str1 = matches.split(',');
    var str2 = str[i+1].split();
    var newMatches = '';
    for(var item in str2) {
        var contains = (str1.indexOf(array[item]) > -1);
        if(contains == 1) {
            var result = array[item]+',';
            newMatches += result;
        }
    }
    matches = newMatches.replace(/\,$/, '');
}

return matches;

kennebec · Accepted Answer · 2013-05-09 05:35:24Z

You can call a function recursively-

after you find the elements common to the first two strings,

compare the commons to the next string, and so on,

for as many arguments as you pass to the function.

This example will return two '5's if all the strings contain two comma separated '5's. (And they needn't be numbers, strings will work the same.)

function inCommon(s1, s2){
    var L1= s1.length, L2= s2.length, rx, 
    A1,  next, next1, next2, least, common= [];
    A1= (L1<= L2)? s1.split(','):s2.split(',');

    for(var i= 0, L= A1.length; i<L; i++){
        next= A1[i];
        rx= RegExp('(^|,)'+next+'(,|$)', 'g');
        next1= (s1.match(rx) || []).length;
        next2= (s2.match(rx) || []).length;
        least= Math.min(next1, next2);
        while(least--) common.push(next);
    }
    if(arguments.length>2 && common.length){
        var a= common.slice.call(arguments, 2);
        a.unshift(common.join(','));
        return inCommon.apply(window, a);
    }
    return common;
}
var str1= "1,5,3";
var str2= "2,5,1";
var str3= "3,2,1,5";

inCommon(str1, str2, str3)

/*  returned value: (Array)
1,5
*/

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