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How can I exclude punctuation in "pig latin" function?

I apologize if this question has been answered somewhere - please point me in the right direction if so. I have read through a bunch of solutions and have not yet cracked it!
Sooo...basically, I need to:
Move the first letter of each word to the end of it, then add "ay" to the end of the word. Leave punctuation marks untouched.
This is my code so far:
function pigIt(str) {
var newStr = str.split(" ");
var changed = newStr.map(function(input) {
return input.substring(1) + input.charAt(0) + "ay";
});
changed = changed.join(" ");
return changed;
}
console.log(pigIt('Pig latin is cool'));
As you can see, the code will work for any input that doesn't include punctuation. Great. Now I need to maybe add a Regex somewhere to exclude punctuation but I don't know where to put it! Please help!!

You could split by the word boundary /(\W+)/ while capturing separator. Transform words only. And then join back.
function pigIt(str) {
var newStr = str.split(/(\W+)/); // ['Pig', ' ', 'latin', ',- ',..]
var changed = newStr.map(function(input) {
if (!/\w/.test(input)) return input // keep non word elements as is
return input.substring(1) + input.charAt(0) + "ay";
});
return changed.join("");
}
console.log(pigIt('Pig latin,- is cool!'));

I think because you are going to want to put the punctuation back in the same place after processing, then you will probably be better of doing it all manually.
Loop the input 1 char at a time and build a 'word buffer', every time you hit a non-letter character then process the word buffer and append the non-letter character too.
function pigIt(str) {
var process = function(s) {
if (s.length < 2) {
return s;
}
return s.substring(1) + s.charAt(0) + "ay";
};
var result = '';
var buffer = '';
for (var i = 0; i < str.length; i++) {
var c = str[i];
if (c.match(/[a-zA-Z]/i)) {
buffer += c;
} else {
if (buffer.length) {
result += process(buffer);
buffer = '';
}
result += c;
}
}
result += process(buffer);
buffer = '';
return result;
}
var output = pigIt('Pig latin is cool.');
console.log(output);

Reversing words in a string JS without using built in function

Without using the split reverse and join functions, how would one do such a thing?
The Problem Given: Reverse the words in a string
Sample Input: "Hello World"
Sample Output: "World Hello"
<script>
var newString = "";
var theString = prompt("Enter a Phrase that you would like to reverse (Ex. Hello world)");
newString = theString.split(" ").reverse().join(" ")
document.write(newString);
</script>

Arrays can be used like stacks out of the box. And stacks are LIFO, which is what you need.
function reverseWords(str) {
var word, words, reverse;
words = str.match(/(?:\w+)/g);
reverse = '';
while(word = words.pop()) {
reverse += word + ' ';
}
return reverse.trim();
}
reverseWords('hello world');
Or use the call stack as your stack:
function reverseWords(str) {
var result = '';
(function readWord(i = 0) {
var word = '';
if(i > str.length) {
return '';
}
while(str[i] !== ' ' && i < str.length) {
word += str[i];
i++;
}
readWord(++i); // Skip over delimiter.
result += word + ' ';
}());
return result.trim();
}
reverseWords('hello world');

Another idea for reversing the words in a String is using a Stack data structure. Like so:
var newString = "";
var theString = prompt("Enter a Phrase that you would like to reverse (Ex. Hello world)");
var word = "";
var c;
var stack = [];
for (var i = 0, len = theString.length; i < len; i++) {
c = theString[i];
word += c;
if (c == " " || i == (len-1)) {
word = word.trim();
stack.push(word);
word = "";
}
}
while (s = stack.pop()) {
newString += s + " ";
}
console.log(newString);

You could also go fancy and try something like this:
I couldn't come up with a shorter solution.
var newString = "";
var theString = prompt("Enter a Phrase that you would like to reverse (Ex. Hello world)");
theString.replace(/[^\s]*/g, function (value) {
newString = value + ' ' + newString;
});
document.write(newString);

Of the millions of different solutions, the least amount of typing I could come up with involves using lastIndexOf and substring.
var str = "The quick brown fox",
reversed = "",
idx;
while(true) {
idx = str.lastIndexOf(" ")
reversed = reversed + str.substring(idx).trim() + " "
if (idx < 0) break;
str = str.substring(0, idx)
}
reversed.trim() # Oh, yes, trim too
Output:
"fox brown quick The"

The simplest way to do in javascript. Here replace() have /,/g it will replace all comma from the string to space.
var msg = 'Hello world I am Programmer';
var newstr = msg.split(" ").reverse().join().replace(/,/g, ' ');
console.log(newstr)
;

Javascript Return the location of the first letter of each word in a string using a loop

Having trouble coming up with code doing this.
So for example here is my string.
var str = "Hello how are you today?";
How would I manipulate this string to return the position of the first letter of each word using a loop?

this will give you the result with less complicated code and a single loop
function foo(str) {
var pos = [];
var words = str.split(' ');
pos.push(1);
var prevWordPos;
for (var i = 1; i < words.length; i++) {
prevWordPos = pos[i - 1] + words[i - 1].length;
pos.push((str.indexOf(words[i], prevWordPos) + 1));
}
return pos;
}

You should search for a question before asking it in case it's already been asked and answered.
Get first letter of each word in a string, in Javascript

You can use a regexp replace passing a function instead of a replacement string, this will call the function for each match:
str.replace(/[^ ]+/g, function(match, pos) {
console.log("Word " + match + " starts at position " + pos);
});
The regexp meaning is:
[^ ]: anything excluding space
+: one or more times
"g" option: not only first match, but each of them
in other words the function will be called with sequences of non-spaces. Of course you can define what you consider a "word" differently.

Here is a Solution with two Loops, i hope that is close enough ;)
var starts = [];
var str = "How are you doing today?";
//var count = 0;
var orgStr = str;
while (str.indexOf(" ") > 0) {
if (starts.length > 0) {
starts.push(starts[starts.length - 1] + str.indexOf(" ") +1);
} else {
starts.push(1);
starts.push(str.indexOf(" ") +2);
//alert(str);
}
str = str.substring(str.indexOf(" ") + 1);
}
for (var i = 0; i < starts.length; i++) {
alert(starts[i] + ": " + orgStr.substring(starts[i]-1,starts[i]))
}

Easiest would be to search a regular expression \b\w and collect match.start() match.index for each match. Loop while there's matches.
EDIT: wrong language. lol.

Replace letters in string with the next letter, and capitalize vowels in the changed string

function LetterChanges(str) {
var alphabet = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";
for (var i = 0; i < str.length; i++) {
var index = alphabet.indexOf(str[i])
if (/[a-zA-Z]/.test(str[i])) {
str = str.replace(str[i], alphabet.charAt(index + 1));
}
if (/[aeiou]/.test(str[i])) {
str = str.replace(str[i], alphabet.charAt(index + 26));
}
}
return str;
}
When I call LetterChanges("hello"), it returns 'Ifmmp' which is correct, but when "sent" is passed it returns 'ufOt' instead of 'tfOu'. Why is that?

str.replace() replaces the first occurrence of the match in the string with the replacement. LetterChanges("sent") does the following:
i = 0 : str.replace("s", "t"), now str = "tent"
i = 1 : str.replace("e", "f"), now str = "tfnt"
i = 2 : str.replace("n", "o"), now str = "tfot", then
str.replace("o", "O"), now str = "tfOt"
i = 3 : str.replace("t", "u"), now str = "ufOt"
return str

There are several issues. The main one is that you could inadvertently change the same letter several times.
Let's see what happens to the s in sent. You first change it to t. However, when it comes to changing the final letter, which is also t, you change the first letter again, this time from t to u.
Another, smaller, issue is the handling of the letter z.
Finally, your indexing in the second if is off by one: d becomes D and not E.

You can use String.replace to avoid that:
function LetterChanges(str) {
return str.replace(/[a-zA-Z]/g, function(c){
return String.fromCharCode(c.charCodeAt(0)+1);
}).replace(/[aeiou]/g, function(c){
return c.toUpperCase();
});
}
But there is still a bug: LetterChanges('Zebra') will return '[fcsb'. I assume that is not your intention. You will have to handle the shift.

Try this one:
function LetterChanges(str) {
var alphabet = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";
var result = '';
var temp;
for (var i = 0; i < str.length; i++) {
var index = alphabet.indexOf(str[i])
if (/[a-zA-Z]/.test(str[i])) {
//str = str.replace(str[i], alphabet.charAt(index + 1));
temp= alphabet.charAt(index + 1);
index = index+1;
}
else if(str[i] == ' '){
temp = ' ';
}
if (/[aeiou]/.test(temp)) {
temp = alphabet.charAt(index + 26);
}
result += temp;
}
return result;
}

var str = 'bcd12';
str = str.replace(/[a-z]/gi, function(char) { //call replace method
char = String.fromCharCode(char.charCodeAt(0)+1);//increment ascii code of char variable by 1 .FromCharCode() method will convert Unicode values into character
if (char=='{' || char=='[') char = 'a'; //if char values goes to "[" or"{" on incrementing by one as "[ ascii value is 91 just after Z" and "{ ascii value is 123 just after "z" so assign "a" to char variable..
if (/[aeiuo]/.test(char)) char = char.toUpperCase();//convert vowels to uppercase
return char;
});
console.log(str);

Check this code sample. There is no bug in it. Not pretty straight forward but Works like a charm. Cheers!
function LetterChanges(str) {
var temp = str;
var tempArr = temp.split("");//Split the input to convert it to an Array
tempArr.forEach(changeLetter);/*Not many use this but this is the referred way of using loops in javascript*/
str = tempArr.join("");
// code goes here
return str;
}
function changeLetter(ele,index,arr) {
var lowerLetters ="abcdefghijklmnopqrstuvwxyza";
var upperLetters ="ABCDEFGHIJKLMNOPQRSTUVWXYZA";
var lowLetterArr = lowerLetters.split("");
var upLetterArr = upperLetters.split("");
var i =0;
for(i;i<lowLetterArr.length;i++){
if(arr[index] === lowLetterArr[i]){
arr[index] = lowLetterArr[i+1];
arr[index]=arr[index].replace(/[aeiou]/g,arr[index].toUpperCase());
return false;
}
if(arr[index] === upLetterArr[i]){
arr[index] = upLetterArr[i+1];
arr[index]=arr[index].replace(/[aeiou]/g,arr[index].toUpperCase());
return false;
}
}
}
// keep this function call here
// to see how to enter arguments in JavaScript scroll down
LetterChanges(readline());

Javascript word-count for any given DOM element

I'm wondering if there's a way to count the words inside a div for example. Say we have a div like so:
<div id="content">
hello how are you?
</div>
Then have the JS function return an integer of 4.
Is this possible? I have done this with form elements but can't seem to do it for non-form ones.
Any ideas?
g

If you know that the DIV is only going to have text in it, you can KISS:
var count = document.getElementById('content').innerHTML.split(' ').length;
If the div can have HTML tags in it, you're going to have to traverse its children looking for text nodes:
function get_text(el) {
ret = "";
var length = el.childNodes.length;
for(var i = 0; i < length; i++) {
var node = el.childNodes[i];
if(node.nodeType != 8) {
ret += node.nodeType != 1 ? node.nodeValue : get_text(node);
}
}
return ret;
}
var words = get_text(document.getElementById('content'));
var count = words.split(' ').length;
This is the same logic that the jQuery library uses to achieve the effect of its text() function. jQuery is a pretty awesome library that in this case is not necessary. However, if you find yourself doing a lot of DOM manipulation or AJAX then you might want to check it out.
EDIT:
As noted by Gumbo in the comments, the way we are splitting the strings above would count two consecutive spaces as a word. If you expect that sort of thing (and even if you don't) it's probably best to avoid it by splitting on a regular expression instead of on a simple space character. Keeping that in mind, instead of doing the above split, you should do something like this:
var count = words.split(/\s+/).length;
The only difference being on what we're passing to the split function.

Paolo Bergantino's second solution is incorrect for empty strings or strings that begin or end with whitespaces. Here's the fix:
var count = !s ? 0 : (s.split(/^\s+$/).length === 2 ? 0 : 2 +
s.split(/\s+/).length - s.split(/^\s+/).length - s.split(/\s+$/).length);
Explanation: If the string is empty, there are zero words; If the string has only whitespaces, there are zero words; Else, count the number of whitespace groups without the ones from the beginning and the end of the string.

string_var.match(/[^\s]+/g).length
seems like it's a better method than
string_var.split(/\s+/).length
At least it won't count "word " as 2 words -- ['word'] rather than ['word', '']. And it doesn't really require any funny add-on logic.

Or just use Countable.js to do the hard job ;)

document.deepText= function(hoo){
var A= [];
if(hoo){
hoo= hoo.firstChild;
while(hoo!= null){
if(hoo.nodeType== 3){
A[A.length]= hoo.data;
}
else A= A.concat(arguments.callee(hoo));
hoo= hoo.nextSibling;
}
}
return A;
}
I'd be fairly strict about what a word is-
function countwords(hoo){
var text= document.deepText(hoo).join(' ');
return text.match(/[A-Za-z\'\-]+/g).length;
}
alert(countwords(document.body))

Or you can do this:
function CountWords (this_field, show_word_count, show_char_count) {
if (show_word_count == null) {
show_word_count = true;
}
if (show_char_count == null) {
show_char_count = false;
}
var char_count = this_field.value.length;
var fullStr = this_field.value + " ";
var initial_whitespace_rExp = /^[^A-Za-z0-9]+/gi;
var left_trimmedStr = fullStr.replace(initial_whitespace_rExp, "");
var non_alphanumerics_rExp = rExp = /[^A-Za-z0-9]+/gi;
var cleanedStr = left_trimmedStr.replace(non_alphanumerics_rExp, " ");
var splitString = cleanedStr.split(" ");
var word_count = splitString.length -1;
if (fullStr.length <2) {
word_count = 0;
}
if (word_count == 1) {
wordOrWords = " word";
} else {
wordOrWords = " words";
}
if (char_count == 1) {
charOrChars = " character";
} else {
charOrChars = " characters";
}
if (show_word_count & show_char_count) {
alert ("Word Count:\n" + " " + word_count + wordOrWords + "\n" + " " + char_count + charOrChars);
} else {
if (show_word_count) {
alert ("Word Count: " + word_count + wordOrWords);
} else {
if (show_char_count) {
alert ("Character Count: " + char_count + charOrChars);
}
}
}
return word_count;
}

The get_text function in Paolo Bergantino's answer didn't work properly for me when two child nodes have no space between them. eg <h1>heading</h1><p>paragraph</p> would be returned as headingparagraph (notice lack of space between the words). So prepending a space to the nodeValue fixes this. But it introduces a space at the front of the text but I found a word count function that trims it off (plus it uses several regexps to ensure it counts words only). Word count and edited get_text functions below:
function get_text(el) {
ret = "";
var length = el.childNodes.length;
for(var i = 0; i < length; i++) {
var node = el.childNodes[i];
if(node.nodeType != 8) {
ret += node.nodeType != 1 ? ' '+node.nodeValue : get_text(node);
}
}
return ret;
}
function wordCount(fullStr) {
if (fullStr.length == 0) {
return 0;
} else {
fullStr = fullStr.replace(/\r+/g, " ");
fullStr = fullStr.replace(/\n+/g, " ");
fullStr = fullStr.replace(/[^A-Za-z0-9 ]+/gi, "");
fullStr = fullStr.replace(/^\s+/, "");
fullStr = fullStr.replace(/\s+$/, "");
fullStr = fullStr.replace(/\s+/gi, " ");
var splitString = fullStr.split(" ");
return splitString.length;
}
}
EDIT
kennebec's word counter is really good. But the one I've found includes a number as a word which is what I needed. Still, that's easy to add to kennebec's. But kennebec's text retrieval function will have the same problem.

This should account for preceding & trailing whitespaces
const wordCount = document.querySelector('#content').innerText.trim().split(/\s+/).length;

string_var.match(/[^\s]+/g).length - 1;