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Check if first character of a string is a letter in JavaScript

I'm trying to write a function that checks a string for multiple conditions. However, I have reached a wall when trying to figure out how to check if the first character in a string is a letter only.
function SearchingChallenge(str) {
// code goes here
let onlyLetters = /^[a-zA-Z]+$/;
if (str.length > 4 && str.length < 25){
if (onlyLetters.test(str)){
return true;
} else {
return false;
}
} else {
return false;
}
}
"u__adced_123" should return true but it's returning false. I've tried str[0]==onlyLetters but still the same.

onlyLetters.test(str) checks the whole string. To get the first character, use str.charAt(0).
function SearchingChallenge(str) {
let onlyLetters = /^[a-zA-Z]+$/;
if (str.length > 4 && str.length < 25) {
if (onlyLetters.test(str.charAt(0))) {
return true;
} else {
return false;
}
} else {
return false;
}
}
console.log(SearchingChallenge('Hello World!'));
console.log(SearchingChallenge('!dlroW olleH'));
console.log(SearchingChallenge('u__adced_123'));

const SearchingChallenge = str => (
!!str[4] && // > 4
!str[24] && // < 25
(/^[a-z]+$/i).test(str) // alpha
);
// Tests...
// true: greater than four characters
console.log(SearchingChallenge('Fiver'));
// true: less than twenty-five characters
console.log(SearchingChallenge('TwentyFouroooooooooooooo'));
// false: twenty-five or more characters
console.log(SearchingChallenge('TwentyFiveooooooooooooooo'));
// false: contains numbers
console.log(SearchingChallenge('abcd1234'));
// false: less than five characters
console.log(SearchingChallenge('Four'));

connect 3 different true statements to an IF ELSE

I want return true if the number is an integer with 4 or 6 digits. no decimals or letter allowed
The thing is not working is the if its really a number check and if its got a decimal.
i think i got already the right functions applied to it but i just cant connect them properly to my if statement.
so i want to check if 3 different things a true then return true but didnt figured out
Please if possible only answer with a hint or a link or SUDO Code or stuff i can look up.
gonna answer the question myself when i figured it out
JS
function validatePIN (pin) {
//return true or false
var result = (pin - Math.floor(pin)) !== 0;
if( pin.length === 4 || isNaN(pin) || result) {
return true
} else if ( pin.length === 6 || isNaN(pin) || result) {
return true
} else return false
}
Thanks

A simple regular expression can be used to test that is is 4 or 6 numbers.
function isValidPin (pin) {
return /^(\d{4}|\d{6})$/.test(pin.toString());
}
console.log(isValidPin(123));
console.log(isValidPin("1234"));
console.log(isValidPin("12345"));
console.log(isValidPin("123456"));
console.log(isValidPin("1234567"));
console.log(isValidPin("12.45"));
console.log(isValidPin("12e45"));

You can check the conditions with the AND operator (&&).
function validatePIN (pin) {
//return true or false
var result = (pin - Math.floor(pin)) !== 0;
if( pin.length === 4 && isNaN(pin) && result)
{ return true} else if ( pin.length === 6 && isNaN(pin) && result) {
return true
} else return false
}

You need to change your or to and
function validatePIN (pin) {
//return true or false
var result = (pin - Math.floor(pin)) !== 0;
if( pin.length === 4 && isNaN(pin) && result)
{ return true}
else if ( pin.length === 6 && isNaN(pin) && result) {
return true
} else return false
}

Ty this:
function validatePIN(pin) {
var parsed = Math.parseInt(pin, 10);
// if it's not an integer
if(pin !== parsed.toString()) return false;
return pin.length === 4 || pin.length === 6;
}

perhaps I'm mistaken - but you could just check if the length is 4 OR 6, and continue your other two checks:
function validatePIN (pin) {
//return true or false
var result = (pin - Math.floor(pin)) !== 0;
if(!isNaN(pin) && (pin.length === 4 || pin.length === 6) && result) {
return true
} else return false
}
I've also edited your code, as it seemed illogical returning true for NaN.

I think you want this. You should use regular expression for simplicity.
console.log(445584, validatePin(445584));
console.log("445584", validatePin("445584"));
console.log("alj454", validatePin("alj454"));
console.log(4455.84, validatePin(4455.84));
function validatePin(pin){
return /^(\d{4}|\d{6})$/.test(pin);
}

You might want to look into using isNaN() to detect if there are any characters that aren't numbers.
Also, using .toString() so you can check the .length.

Why does function return undefined when it explicitly is told to return true?

I was trying to solve the following coding exercise.
We have two special characters. The first character can be represented
by one bit 0. The second character can be represented by two bits (10
or 11).
Now given a string represented by several bits. Return whether the
last character must be a one-bit character or not. The given string
will always end with a zero.
example:
Input: bits = [1, 0, 0] Output: True
Below is my solution for the above challenge. Why is this returning undefined? If I use [1,0,1,0] as input, it should return true but I am getting undefined. I am explicitly writing true in the return statement and not the results of a variable.
var isOneBitCharacter = function(bits) {
console.log(bits);
var length = bits.length;
if (length == 1) {
return true;
}
if (length == 0) {return false;}
if (length == 2 && bits[0] === 1) {
return false;
}
if (bits[0] === 1) {
isOneBitCharacter(bits.slice(1));
} else {
isOneBitCharacter(bits.slice(2));
}
};
isOneBitCharacter([1,0,1,0]);

I guess you are missing returns. Here is adjusted code:
var isOneBitCharacter = function(bits) {
console.log(bits);
var length = bits.length;
if (length == 1) {
return true;
}
if (length == 0) {return false;}
// added return here and next statements
if (length == 2 && bits[0] === 1) {
return false;
}
if (bits[0] === 1) {
return isOneBitCharacter(bits.slice(1));
} else {
return isOneBitCharacter(bits.slice(2));
}
};
isOneBitCharacter([1,0,1,0]);

How to print a value of parameter inside of function in javascript?

I am trying to create a function pluralizeParam(n, word, pluralWord) with these requirements:
If n is 1, return the non-plural word (parameter word);otherwise, add an “s” to the plural word;
If the pluralWord parameter is provided, instead of adding an “s,” return the pluralWord.
What I have done so far is following:
function returnPluralWord(n, word, pluralWord) {
if (n === 1) {
return word;
} else if (n === 1 && (word.lastIndexOf("s")) || (word.lastIndexOf("ess"))) {
return word + "s";
} else if (n !== 1 && word.length - 2 === "es") {
return word + "s";
} else if (pluralWord !== 'undefined') {
return pluralWord;
}
}
var result = returnPluralWord(2, "lioness", "lionesses");
console.log(result);
My problem is: it is not printing pluralWord. How do I do that?
Thanks

word.length - 2 can't never be equal to "es". You need also to rearrange your statements, the 2nd is already catched by 1.
When you use the word.lastIndexOf('s') ( which is wrong logic I think ), it returns the last index of the s character, not if it ends on s.
You can check String#endsWith and String#startsWith methods, these ones check if the string starts or ends with the given part
const str = 'less';
console.log(str.endsWith('s'))

Determining the case (upper/lower) of the first letter in a string

In a web application, how do I determine whether the first letter in a given string is upper- or lower-case using JavaScript?

You can use toUpperCase:
if(yourString.charAt(0) === yourString.charAt(0).toUpperCase()) {
//Uppercase!
}
If you're going to be using this on a regular basis, I would suggest putting it in a function on the String prototype, something like this:
String.prototype.isFirstCapital = function() {
return this.charAt(0) === this.charAt(0).toUpperCase();
}
if(yourString.isFirstCapital()) {
//Uppercase!
}
Update (based on comments)
I don't know what you actually want to do in the case that the string does not being with a letter, but a simple solution would be to add a quick check to see if it does or not, and return false if not:
String.prototype.isFirstCapital = function() {
return /^[a-z]/i.test(this) && this.charAt(0) === this.charAt(0).toUpperCase();
}

This will work only with English alphabet.
var ch = myStr.chatAt(0);
if (ch >= 'a' && ch <= 'z') {
// small
} else if (ch >= 'A' && ch <= 'Z') {
// capital
} else {
// not english alphabet char
}

var mystring = "Test string";
var first= "";
if (mystring )
{
first= mystring[1];
}
if (first)
{
$('p').each(function()
{
if ($(this).text().charAt(0).toUpperCase() === $(this).text().charAt(0))
{
alert("Uppercase");
}
});
}

This will be called recursively until a first letter in a string is approached, otherwise returns 'no letters'.
function getFirstCase(string) {
if (string === '') return 'no letters';
var firstChar = string.charAt(0);
/*
* If both lowercase and uppercase
* are equal, it is not a letter
*/
if (firstChar.toLowerCase() === firstChar.toUpperCase()) {
return getFirstCase(string.substr(1));
} else {
return firstChar.toLowerCase() === firstChar ? 'lowercase' : 'uppercase';
}
}
Testing:
console.log(getFirstCase('alphabet'),
getFirstCase('Sunshine'),
getFirstCase('123123'),
getFirstCase('#Hi'),
getFirstCase('\nHAHA'));

I'm surprised no one's offered a regex solution to this - it seems like the easiest by far:
function getFirstCase(s) {
return (/^[\d\W]*[A-Z]/).test(s) ? 'upper' :
(/^[\d\W]*[a-z]/).test(s) ? 'lower' :
'none';
}
Blatantly stealing #Lapple's test cases:
console.log(getFirstCase('alphabet'),
getFirstCase('Sunshine'),
getFirstCase('123123'),
getFirstCase('#Hi'),
getFirstCase('\nHAHA'));
// lower upper none upper upper
See http://jsfiddle.net/nrabinowitz/a5cQa/