How do I check if a string has all letters of alphabet in javascript

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Hi, I'm quite new to programming and started with JavaScript.
I'm curious if there is a method to check if a variable has all letters of the alphabet at least once.

What I have tried:

So I have the variable

var string = prompt("Type in message","");

Curious how to check if the message typed in has every letter a to z. if yes alert("All letters found at least once")

Posted 4-Oct-17 22:13pm

Member 13446970

Updated 5-Oct-17 0:22am

v2

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3 solutions

Solution 2

You can easily implement yourself, for instance

JavaScript

var exists =  [];

var s = "abcdefoo"; // the input string


for (var i=0; i < s.length; ++i)
{
  exists[s[i]] = 1;
}


for ( var i = 97; i<=122; ++i) // loop over all lowercase characters
{
  c = String.fromCharCode(i);
  if ( !exists[c])
  {
    alert(  c + " is missing");
    break;
  }
}

Posted 4-Oct-17 22:44pm

CPallini

Solution 3

One more solution

var Your_String= "abcdefghijklmnopqrstuvwxyz";
var Result = Your_String.replace(' ','').split('').filter(function(item, i, ar){ return ar.indexOf(item) === i; }).join('');
document.writeln(Result.length == 26 ? "All used" : "Some missing");

Made use of : javascript - Showing unique characters in a string only once - Stack Overflow[^]

Posted 5-Oct-17 0:22am

Thanks7872

Updated 5-Oct-17 0:26am

v3

Comments

Member 13446970 5-Oct-17 6:40am

Tried this but it will just pop up the message all used

Thanks7872 5-Oct-17 6:45am

Its just an example. Once you get the length, you can use it anyway you want.

Member 13446970 5-Oct-17 8:49am

oh I see. Cheers

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Karthik_Mahalingam · Accepted Answer · 2017-10-04T22:25:00

Solution 1

try

var alphabets26 = 'abcdefghijklmnopqrstuvwxyz';
       var input = prompt();
       input = input.toLowerCase();
       var icount = 0;
       for (var i = 0; i < alphabets26.length; i++) {
           var letter = alphabets26[i];
           if (input.indexOf(letter) > -1)
               icount++;
       }
       if (icount == 26)
           alert('All letters found at least once');

       else
           alert('Few letters missing');

Demo : - JSFiddle[^]

Posted 4-Oct-17 22:25pm

Karthik_Mahalingam

Updated 4-Oct-17 23:32pm

v3

Comments

Member 13446970 5-Oct-17 4:38am

Thank you. It works as expected.

Karthik_Mahalingam 5-Oct-17 4:39am

welcome

CPallini 5-Oct-17 4:45am

You have a 'z' instead of a 'x' in you alphabets26 string.

Karthik_Mahalingam 5-Oct-17 5:32am

:) typo, Hawk Eye

Member 13446970 5-Oct-17 4:57am

Could you please add some comments as in why did you use a for statement there?

Karthik_Mahalingam 5-Oct-17 5:36am

Hide Copy Code

         var alphabets26 = 'abcdefghijklmnopqrstuvwzyz'; // reference string to compare        var input = prompt(); // read the input from user        input = input.toLowerCase(); // convert to lower case to find exact match        var icount = 0; // counter to check all the letters in alphabets26 are present in input        for (var i = 0; i < alphabets26.length; i++) { // loop each letter in the alphabets26            var letter = alphabets26[i];  // assin the letter to a variable            if (input.indexOf(letter) > -1)  // check if the letter is present in the input                icount++; // if yes then increament the counter        } // the loop will run for 26 letter        if (icount == 26)   // check if the counter is equal to 26, meaning all the letters are present            alert('All letters found at least once');                    else  // else condition            alert('Few letters missing');

copy paste this code in the code editor and see

Member 13446970 5-Oct-17 6:33am

why do we have the variable icount set to 0 and how does it check the presence of the letters from alphabets26 in the input?
I get the idea of the for loop, but why isn't it for (var icount = 0 ... ) ?

Karthik_Mahalingam 5-Oct-17 6:40am

it means, the icount variable is considered as an integer,
in javascript we dont have int,float,string etc..
go through this
JavaScript Data Types[^]

Member 13446970 5-Oct-17 6:44am

Lovely, thank you for the patience!

Karthik_Mahalingam 5-Oct-17 7:05am

welcome

Member 13446970 5-Oct-17 8:40am

One last questions, why has input.indexOf(letter) to be > -1 ?

Karthik_Mahalingam 5-Oct-17 9:38am

If the letter is present in the input ,it will return the position (0 to n) of the letter in the input..
If not present then it will return -1