How to Add Leading Zero When Number Is Less Than 10

Show a leading zero if a number is less than 10

There's no built-in JavaScript function to do this, but you can write your own fairly easily:

function pad(n) {
    return (n < 10) ? ("0" + n) : n;
}

EDIT:

Meanwhile there is a native JS function that does that. See String#padStart

console.log(String(5).padStart(2, '0'));

less than 10 add 0 to number

You can always do

('0' + deg).slice(-2)

See slice():

You can also use negative numbers to select from the end of an array

Hence

('0' + 11).slice(-2) // '11'
('0' + 4).slice(-2)  // '04'

For ease of access, you could of course extract it to a function, or even extend Number with it:

Number.prototype.pad = function(n) {
    return new Array(n).join('0').slice((n || 2) * -1) + this;
}

Which will allow you to write:

c += deg.pad() + '° '; // "04° "

The above function pad accepts an argument specifying the length of the desired string. If no such argument is used, it defaults to 2. You could write:

deg.pad(4) // "0045"

Note the obvious drawback that the value of n cannot be higher than 11, as the string of 0's is currently just 10 characters long. This could of course be given a technical solution, but I did not want to introduce complexity in such a simple function. (Should you elect to, see alex's answer for an excellent approach to that).

Note also that you would not be able to write 2.pad(). It only works with variables. But then, if it's not a variable, you'll always know beforehand how many digits the number consists of.

How to concatenate with a leading zero in number less than 10

Try this:

=A1&"."&TEXT(B1,"00")

Or this also works:

=TEXT(TIME(A1,B1,0),"h.mm")

As required:

enter image description here

adding leading zero if number is < 10

You can use the slice method

diff.sec = tmp % 60;
if( diff.sec < 10 ){
    diff.sec = ("0" + diff.sec).slice(-2);
}

JSFiddle with sample value

var test = 9;if( test < 10 ){  test = ("0" + test).slice(-2);}console.log(test);

Adding a 0 if number is less than 10

Your problem is i is still an integer, it needs to be assigned to a string

  for (int i = 1; i <= 36; i++)
    {
        var iString = i.ToString();

        if(iString.Length == 1)
        {
            iString = iString.PadLeft(2,'0'); //RIGHT HERE!!!
        }
        Response.Write("Test: " + iString);
    }

However, much of this code is superflous, the if statement is not needed. Pad will only ped with zeroes up to the length (2) given. If it's already 2 or more characters long, it won't pad anything. All you need is this

    for (int i = 1; i <= 36; i++)
    {
        var iString = i.ToString().PadLeft(2,'0');
        Response.Write("Test: " + iString);
    }

For that matter, the variable is no longer needed.

    for (int i = 1; i <= 36; i++)
    {
        Response.Write("Test: " + i.ToString().PadLeft(2,'0'));
    }

And if you'll be padding with zeroes all the time, and not some other character, you could just do this

    for (int i = 1; i <= 36; i++)
    {
        Response.Write("Test: " + i.ToString("00"));
    }

And you should get into the habit of using string.Format

    for (int i = 1; i <= 36; i++)
    {
        Response.Write(string.Format("Test: {0}", i.ToString("00")));
    }

And to simplify the string.Format even further:

    for (int i = 1; i <= 36; i++)
    {
        Response.Write(string.Format("Test: {0:00}", i));
    }

R-- Add leading zero to string, with no fixed string format

A one-liner using regular expressions:

my_strings <- c("9453", "55489", "4588", 
      "18893", "4457", "2339", "45489HQ", "7833HQ")

gsub("^([0-9]{1,4})(HQ|$)", "0\\1\\2",my_strings)

[1] "09453"   "55489"   "04588"   "18893"   
    "04457"   "02339"   "45489HQ" "07833HQ"

Explanation:

^ start of string
[0-9]{1,4} one to four numbers in a row
(HQ|$) the string "HQ" or the end of the string

Parentheses represent capture groups in order. So 0\\1\\2 means 0 followed by the first capture group [0-9]{1,4} and the second capture group HQ|$.

Of course if there is 5 numbers, then the regex isn't matched, so it doesn't change.