Develop Reference

Replace content present in the nested brackets

Input = ABCDEF ((3) abcdef),GHIJKLMN ((4)(5) Value),OPQRSTUVW((4(5)) Value (3))
Expected Output = ABCDEF,GHIJKLMN,OPQRSTUVW
Tried so far
Output = Input.replace(/ *\([^)]*\)*/g, "");

Using a regex here probably won't work, or scale, because you expect nested parentheses in your input string. Regex works well when there is a known and fixed structure to the input. Instead, I would recommend that you approach this using a parser. In the code below, I iterate over the input string, one character at at time, and I use a counter to keep track of how many open parentheses there are. If we are inside a parenthesis term, then we don't record those characters. I also have one simple replacement at the end to remove whitespace, which is an additional step which your output implies, but you never explicitly mentioned.
var pCount = 0;
var Input = "ABCDEF ((3) abcdef),GHIJKLMN ((4)(5) Value),OPQRSTUVW((4(5)) Value (3))";
var Output = "";
for (var i=0; i < Input.length; i++) {
if (Input[i] === '(') {
pCount++;
}
else if (Input[i] === ')') {
pCount--;
}
else if (pCount == 0) {
Output += Input[i];
}
}
Output = Output.replace(/ /g,'');
console.log(Output);

If you need to remove nested parentheses, you may use a trick from Remove Nested Patterns with One Line of JavaScript.
var Input = "ABCDEF ((3) abcdef),GHIJKLMN ((4)(5) Value),OPQRSTUVW((4(5)) Value (3))";
var Output = Input;
while (Output != (Output = Output.replace(/\s*\([^()]*\)/g, "")));
console.log(Output);
Or, you could use a recursive function:
function remove_nested_parens(s) {
let new_s = s.replace(/\s*\([^()]*\)/g, "");
return new_s == s ? s : remove_nested_parens(new_s);
}
console.log(remove_nested_parens("ABCDEF ((3) abcdef),GHIJKLMN ((4)(5) Value),OPQRSTUVW((4(5)) Value (3))"));
Here, \s*\([^()]*\) matches 0+ whitespaces, (, 0+ chars other than ( and ) and then a ), and the replace operation is repeated until the string does not change.

Adding numbers within a string

I want to take a string of numbers and characters and add up the numbers.
For example: "In 2015, I want to know how much does iPhone 6+ cost?"
Output: 2021
Here is my current code:
var str = "In 2015, I want to know how much does iPhone 6+ cost?";
function sumFromString(str){
var punctuationless = str.replace(/['!"#$%&\\'()\*+,\-\.\/:;<=>?#\[\\\]\^_`{|}~']/g,"");
var finalString = punctuationless.replace(/\s{2,}/g," ");
var StringList = finalString.split(" ");
var sum = [];
for (i = 0; i < StringList.length; i++)
if (isInt(StringList[i])
sum.add(StringList[i]);
sum.reduce( (prev, curr) => prev + curr );
}
sumFromString(str);
My code takes a string and strips it of punctuation and then places each individual word/number into the array, StringList.
I can't get the next part to work.
What I tried was to iterate through each value in the array. The if statement is supposed to check if the array element is an integer. If so, it will add the integer to an empty array called sum. I then add all the values of the array, sum, together.

Much simpler:
function sumFromString(str) {
return (str.match(/\d+/g)||[]).reduce((p,c)=>+c+p);
}
Note in particular that I use +c+p - +c casting the current value from a string to a number, then adding it to p. This matches all the numbers in the string - getting an empty array if there were none - and reduces that.

For the sake of variety, here's a way to do it without regular expressions:
var myString = "5 bunnies ate 6 carrots in 2days.";
var myArray = myString.split('');
var total = 0;
for (var i = 0; i < myArray.length; i++) {
if (!isNaN(parseInt(myArray[i]))) {
total += parseInt(myArray[i]);
}
}
Fiddle Demo
note: If there's a chance myString could be null, you'd want to add a check before the split.

Split the string into an array of all characters with the split function and then run the filter function to get all numbers. Use the map function to go through all elements that include numbers, and delete characters from them that aren't digits.
Then use reduce to get the sum of all numbers. Since we're dealing with strings here, we have to perform type conversion to turn them into numbers.
string.split(' ').filter(function(word) {
return /\d+/.test(word) }
}).map(function(s) {
return s.replace(/\D/, '')
}).reduce(function(a,b) {
return Number(a) + Number(b);
});

how to retrieve a string between to same charecter

I know how to use substring() but here I have a problem, I'd like to retrieve a number between two "_" from a unknown string length. here is my string for example.
7_28_li
and I want to get the 28. How can I proceed to do so ?
Thanks.

Regex
'7_28_li'.match(/_(\d+)_/)[1]
The slashes inside match make it's contents regex.
_s are taken literally
( and ) are for retrieving the contents (the target number) later
\d is a digit character
+ is "one or more".
The [1] on the end is accesses what got matched from the first set of parens, the one or more (+) digits (\d).
Loop
var str = '7_28_li';
var state = 0; //How many underscores have gone by
var num = '';
for (var i = 0; i < str.length; i++) {
if (str[i] == '_') state++;
else if (state == 1) num += str[i];
};
num = parseInt(num);
Probably more efficient, but kind of long and ugly.
Split
'7_28_li'.split('_')[1]
Split it into an array, then get the second element.
IndexOf
var str = "7_28_li";
var num = str.substring(str.indexOf('_') + 1, str.indexOf('_', 2));
Get the start and end point. Uses the little-known second parameter of indexOf. This works better than lastIndexOf because it is guaranteed to give the first number between _s, even when there are more than 2 underscores.

First find the index of _, and then find the next position of _. Then get the substring between them.
var data = "7_28_li";
var idx = data.indexOf("_");
console.log(data.substring(idx + 1, data.indexOf("_", idx + 1)));
# 28
You can understand that better, like this
var data = "7_28_li";
var first = data.indexOf("_");
var next = data.indexOf("_", first + 1);
console.log(data.substring(first + 1, next));
# 28
Note: The second argument to indexOf is to specify where to start looking from.

Probably the easiest way to do it is to call split on your string, with your delimiter ("_" in this case) as the argument. It'll return an array with 7, 28, and li as elements, so you can select the middle one.
"7_28_li".split("_")[1]
This will work if it'll always be 3 elements. If it's more, divide the length property by 2 and floor it to get the right element.
var splitstring = "7_28_li".split("_")
console.log(splitstring[Math.floor(splitstring.length/2)]);
I'm not sure how you want to handle even length strings, but all you have to do is set up an if statement and then do whatever you want.

If you know there would be 2 underscore, you can use this
var str = "7_28_li";
var res = str.substring(str.indexOf("_") +1, str.lastIndexOf("_"));
If you want to find the string between first 2 underscores
var str = "7_28_li";
var firstIndex = str.indexOf("_");
var secondIndex = str.indexOf("_", firstIndex+1);
var res = str.substring(firstIndex+1, secondIndex);

New to javascript, how to write reverse iteration?

I'm currently taking an introduction CIS class at my university and one of the projects is javascript. It is split into two unrelated parts and I was able to do the second part but I'm stuck on the first one. My professor wants me to write an iteration that will display in a reverse order whatever name I write in the prompt screen. So if I write "John Smith" it will display "htims nhoj". The issue is that I have no idea how to write it.
<html>
<body>
<script>
var namIn = window.prompt("Enter name:" );
var namAr = namIn.split("");
var namArLen = namAr.length;
document.write(namAr + "<br /> Length: " + namArLen);
</script>
</body>
</html>

Strings in JavaScript have a function called split() which turn them in to Arrays. Arrays in JavaScript have a function called reverse() which reverse their order, and a function called join() which turn them back into Strings. You can combine these into:
"John Smith".split("").reverse().join("")
This returns:
"htimS nhoJ"
Also, and I don't know if this is a typo, but you can throw a toLowerCase() to get 100% of what your question is after:
"John Smith".split("").reverse().join("").toLowerCase()
returns:
"htims nhoj"
As for the question in your title, you can specify the direction of a for loop in the last argument like so:
var reversed = [];
var name = "John Smith".split("");
for(var i = name.length-1; i >= 0; i--) {
reversed.push(name[i]);
}
console.log(reversed.join(""));
Which will output:
"htimS nhoJ"

There's no need to split this string into an array. Just use the charAt() function and a simple for loop.
var name = window.prompt("Enter name:");
var reverse = "";
for (var i = name.length - 1; i >=0; i--) {
reverse += name.charAt(i);
}
console.log(reverse)
Instead of converting the string to an array first, you're just reading the characters out of the string directly.

You can accomplish this by iterating only half the number of characters.
DEMO: http://jsfiddle.net/vgG2P/
CODE:
var name = "Bob Dylan".split("");
// The counters will meet in the middle.
// --------------+----------------------
// first char last char | inc dec
// -------v-------------v-----------v----v----v
for(var i = 0, j = name.length-1; i < j; i++, j--) {
var temp = name[i]; // Store the `i` char
name[i] = name[j]; // Replace the `i` char with the `j` char
name[j] = temp; // Replace the `j` char with the `i` char we stored
}
console.log(name.join("")); "nalyD boB"
EXPLANATION:
What we did was split the characters into an Array, and maintain two counters, one that increments from the first character at 0, and the other that decrements from the last character at .length - 1. Then simply swap the characters.
The iteration continues while your incrementing counter is less than your decrementing counter. Since they will meet in the middle, you'll end up incrementing only half the total length.
We can also build the halves of the result without using an Array:
DEMO: http://jsfiddle.net/vgG2P/1/
var name = "Bob Dylan";
var start = "", end = ""
for(var i = 0, j = name.length-1; i < j; i++, j--) {
end = name.charAt(i) + end
start += name.charAt(j)
}
if (i === j)
start += name.charAt(i)
console.log(start + end); "nalyD boB"

I'm assuming that your professor would not be asking you how to reverse a string if he hasn't yet introduced you to the concept of arrays and loops. Basically, a string like John Smith is just an array of characters like this:
0123456789
John Smith
Again, thinking in the sense that a string is just an array of characters, you have have 10 characters that need to be reversed. So how do you go about doing this? Well, you basically need to take the last character h from the "array" you're given and make it the first character in a new "array" you're going to create. Here's an example:
var known = 'John Smith';
var reversed = ''; // I'm making an empty string aka character array
var last = known.length - 1 // This is the index of the last character
for (var i = 0; i < known.length; i++)
{
temp += known[last - i];
}
(You can see it working here)
So what's happening?
We're looping over known starting at 0 and ending at 9 (from the first character to the last)
During each iteration, i is incrementing from 0 - 9
last always has a value of 9
last - i will give us indexes in reverse order (9, 8, 7, ..., 0)
So, when i is 0, last - i is 9 and known[9] is "h"; repeat this process and you get the reversed string
Hopefully this helps explain a little better what's happening when you call reverse() on an array.

(1) A more straight forward way without built-in functions:
function reverse(str) {
let reversed_string = "";
for (let i = str.length - 1; i >= 0; i--) {
reversed_string += str[i];
}
return reversed_string;
}
(2) Using ES2015 'for' helper function:
function reverse(str) {
let reversed_string = "";
for (let character of str) {
reversed_string = character + reversed_string;
}
return reversed_string;
}
(3) Using ES6 syntax and ES5.1 reduce():
function reverse(str) {
return str.split('').reduce((reversed, char) => char + reversed, '');
}
// reduce takes in 2 arguments. (1) arrow function, and (2) empty string.
Chances are, for an interview, that you will not able to use the built-in functions, especially for "reverse()".

Javascript split array

I have a JavaScript array which contain the staff's Chinese and English names.
Example:
XXX MA Yo-Yo
Where XXX represents the Chinese name: 馬友友.
I want to split this into two parts by using the 1st space " " as an indicator.
for (i = 0; i < /* ... */)
{
w_temp = arr[i];
w_name = w_temp[1].split(' ', 1);
/* ... */
}
w_name[0] successfully returns 馬友友. But w_name[1] returns undefined, which cannot return MA Yo-Yo.
How can I split it into two parts?

Replace your existing w_name = ... line with something like this:
w_name = w_temp[1].split(' ');
w_name = [w_name[0], w_name.slice(1).join(' ')];
The ,1 you had earlier meant to stop parsing as soon as it came to the first space. This is not what you want; you want the part before the first space as one item and the part after as another. The new code parses all of the elements. After that, it sets it to an array consisting of:
The already-existing first element, and
The parts after the first element re-joined.

You have
split(' ', 1)
The 1 means to return at most one part. You should just ommit that or change it to two if thats what you need.

Because you invoke split method with limit = 1, then w_name has only 1 item.
In addition, yours input string has 2 whitespace, therefore, if you use split without limit parameter, w_name will contains 3 items ('XXX', 'CHAN', and 'Tai-Man').
I think you should use this:
var idx = str.indexOf(' ');
if (idx != -1) {
var cn_name = str.substring(0, idx);
var en_name = str.substring(idx + 1);
}

Try this
for (var i = 0; i < arr.length; i++) {
var w_temp = arr[i];
// assuming w_temp contains an array where the second item
// is the full name string
var parts = w_temp[1].match(/^(.+?) (.+)$/);
var cn_name = parts[1];
var en_name = parts[2];
}