We Keep Coding

How to find time stamp in the string and then convert to number of hours/minutes?

In the column where the hours/minutes are stored for some of the business facilities time stamp(s) are presented in this format 0000-0000. The first two digits represent hours and the other two minutes. Here is example of some business hours:
0700-2300 M-F 0700-1700 S&S
0600-2200
0600-2300 Mon-Fri 0700-2200 Sat&Sun
Local 1 0000-2359 Local 2 0630-2230
0700-2100
0600-2345
The original solution that I had was to convert the values in JavaScript and that it was pretty simple. The problem I have is when there is more than one set of time hours/minutes in the string. In the example above that is the case where hours/minutes are different during the weekend or for the different locations. The JS code example is here:
var time = calcDifference("0600-2345");
function calcDifference(time) {
var arr = time.split('-').map(function(str) {
var hours = parseInt(str.substr(0, 2), 10),
minutes = parseInt(str.substr(2, 4), 10);
var result = (hours * 60 + minutes) / 60;
return result.toFixed(2);
});
return arr[1] - arr[0];
}
console.log(time);
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
The code above works just fine if I pass the argument with one time stamp. I'm wondering how to handle situation where I have two time stamps? What would be a good solution to search if string has more than one hours/minutes values and then convert each of them and return to the user.

Assuming the HHMM-HHMM format is consistent in the input, and you don't care about discarding the remaining information in the string, regex is probably the simplest approach (and much safer than your current approach of splitting on hyphens, which might easily occur in other parts of the string you don't care about.)
Note that you won't be able to distinguish between "weekend" and "weekday" times, because that information isn't in a consistent format in your input. (This looks like human input, which pretty much guarantees that your HHMM-HHMM format also won't be strictly consistent; consider allowing for optional whitespace around the hyphen for example, and logging strings which show no match so you can check them manually.)
var testinputs = [
"0700-2300 M-F 0700-1700 S&S",
"0600-2200",
"0600-2300 Mon-Fri 0700-2200 Sat&Sun",
"Local 1 0000-2359 Local 2 0630-2230",
"0700-2100",
"0600-2345"
]
var reg = /(\d\d)(\d\d)\-(\d\d)(\d\d)/g; // \d means any digit 0-9; \- matches a literal "-", parens capture the group for easier access later
for (input of testinputs) {
console.log("Input: ", input)
var matches;
while ((matches = reg.exec(input)) !== null) { // loop through every matching instance in the string
// matches[0] is the full HHMM-HHMM string; the remainder is
// the HH and MM for each parenthetical in the regexp:
console.log(matches)
}
}

There are plenty of ways to do this ( based on your point of view ), but this is my favourite one. you can manipulate the text then pass numbers individually.
var date = '0700-2300 M-F 0700-1700 S&S'.match( /[-\d]+/gi ).filter( e => ~e.search( /\d+/gi ) )
now you have an array of multiple timestamps saved on your database and you pass them individually to your function.
date.forEach( each => calcDifference( each ) );

You can use a regex like /\d{4}\-\d{4}/g to extract all of the digits from the string and map them to their time differences or replace text in the original.
const calcDifference = range => {
const time = range.split`-`
.map(e => (+e.substr(0, 2) * 60 + (+e.substr(2))) / 60)
return time[1] - time[0];
};
const diffs = `0700-2300 M-F 0700-1700 S&S
0600-2200
0600-2300 Mon-Fri 0700-2200 Sat&Sun
Local 1 0000-2359 Local 2 0630-2230
0700-2100
0600-2345`.replace(/\d{4}\-\d{4}/g, calcDifference);
console.log(diffs);

How to format numeric values with pattern?

I'm finding an equivalent to Java's DecimalFormat in JavaScript. I would like to format given numeric values with given decimal pattern like
"#,###.00"
"###.##"
"#,###.##"
"$#,##0.00"
"###,###.00"
"$###,###.00"
"###,###.###"
"####,####.000"
Has there any way to achieve it?

Part 1 - formatting
I would recommend using the Intl.NumberFormat natively supported in javascript, although it may not be supported by older browsers. Looks like IE11 has it, but not android.
So if you wanted to support US Dollars you would simply use something like this
var dollarFormat = new Intl.NumberFormat('en-US', { style: 'currency', currency: 'USD' });
var amountInDollars = dollarFormat.format(123456.123);
// amountInDollars = "$123,456.12"
But this also rounds up for you, for example
var roundedDollars = dollarFormat.format(555.555);
// roundedDollars = "$555.56";
For the numeric cases just use a different formatter. The default 'en-US' adds commas, a decimal before fractional numbers, and limits to 3 fractional numbers. But this is configurable.
var numberFormat = new Intl.NumberFormat('en-US');
var formatted = numberFormat.format(123456.123456);
// formatted = "123,456.123"
var numberFormat2decimals = new Intl.NumberFormat('en-US', { maximumFractionDigits: 2 });
var formatted2 = numberFormat2decimals.format(123456.123456);
// formatted2 = "123,456.12"
You can set maximum and minimum for fraction, integer, and significant digits, and this also supports international formats. Since it's native javascript, I think it's a good way to go if your platforms support it.
MDN is an excellent reference: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/NumberFormat
Part 2 - the 0's
To achieve the 0's in your formats you'll have to modify the value before passing to the formatter. If you require a minimum fractional amount, you're fine for things like .00, the currency formatter will do that by default. If you've got fractal numbers you don't want, just use Math.trun() to truncate the values.
var num = Math.trun(1234.1234);
// num = 1234
Now to change something like 12345 to 12340 we'll have to remove some of the numeric value. We can find out much by converting to a string, pulling the last character, and converting back.
var num = 123456.12345;
var wholeNum = Math.trunc(num);
// wholeNum = 123456;
var toRemove = Number.parseInt(wholeNum.toString().slice(-1), 10);
// toRemove = 6
// slice(-1) gives us the right-most character of a string.
// Notice the ', 10' at the end, this is important to indicate which number base to use for parseInt.
var wholeNumEnding0 = wholeNum - toRemove;
// wholeNumEnding0 = 123450
Hopefully that's what you're looking to accomplish? I didn't perform any rounding here.

Note: I typed this at speed, excuse any mistakes, there might be better ways to do it too.
If you don't want to rely on a library, you could do something like the following:
var number = 100000.00000000000012422;
function FormatNumber(no){
no = no.toFixed(2);
no = no.toString().split('.');
var p1 = no[0];
p1 = p1.substring(0, p1.length - 3) + ',' + p1.substring(p1.length - 3);
no = p1 + '.' + no[1];
console.log(no);
}
FormatNumber(number);
The FormatNumber function takes a number as a parameter (you would probably want to expand that to include e.g. decimal places). It converts the number to the required decimal places, the turns it into a string and splits it by the decimal separator '.'.
The next step is to add a thousands separator three characters from the back, then it's just a matter of joining the remaining characters back together.
JSFiddle
If you wanted to get a ',' every 3 characters you could write a little more 'complex' formatter, something along the lines of the following:
no = no.toFixed(2);
no = no.toString().split('.');
var p1 = no[0];
var arr = [];
arr = p1.split("").reverse().join("").match(/[\s\S]{1,3}/g) || [];
arr = arr.reverse();
p1 = "";
for(var i = 0; i < arr.length; i++){
p1 += arr[i].split("").reverse().join("");
if(i != arr.length - 1){
p1 += ',';
}
}
no = p1 + '.' + no[1];
This method splits the number into an array by each number, reverses the array as we need to start from the end of the string to get accurate result.
Then we iterate the array of strings with 3 or less values by splitting the number into an array again, reversing it and joining back together and then appending to p1. If it's the last item, it doesn't add a comma.
Lastly we take the decimal and append to built string.
JSFiddle

Javascript use split on date range

I get a date range from some API (not in my control) which has the following possibilities of date format
dd-Mmm-yy
dd-Mmm-yyyy
dd/mm/yy
dd/mm/yyyy
mm/dd/yy
mm/dd/yyyy
yyyy-mm-dd
So I get the date range as
01-Dec-16-06-Dec-16 (as per dd-Mmm-yy) OR
01/12/16-06/12/16 (as per dd/mm/yy)
So hyphen (-) is the from/to date separator which the API uses (not in my control) & I get this single combined value
Now, in my code, I want to get the from & to dates separately.
So I use
range.split("-")
However, this does not work properly for 01-Dec-16-06-Dec-16
Not sure what is the best way to account for all the possibilities (considering that the from/to date separator would always be -)

Since this is a case of an ugly API, the only way to do this is by using "hacky" solutions.
Use #Rory McCrossan's suggestion: count the number of - in the string. If 1, split. If 5, split by the third.
Since the API uses the same format of date for both the left side and the right side, the total length of the string will always be ("left side" + "-" + "right side"). You can split the text on the middle character.
e.g.
let date = "01-Dec-16-06-Dec-16";
let idx = date.length / 2;
let start = date.substr(0, idx);
let end = date.substr(idx + 1);
Use regex.

From the formats provided, it looks like the from and to dates will always be the same length split by -. In that case, just do this:
var len = (yourDateRange.length - 1) / 2
var dateFrom = yourDateRange.substring(0, len)
var dateTo = yourDateRange.substring(len + 1)
If you have any format where the length is variable (Such as full name for month), this obviously won't work

It's a bit hacky, but gets the job done.
I used a chained indexOf call with the previous call as the fromIndex parameter (which is the 2nd parameter of indexOf). And seeing as there is either / in the string (then split by -) or not (then split by 3rd -), there was no need for any special checks.
function splitDates(date) {
// ~ is just a fancy way to turn indexOf into a
// boolean-equivalent check (-1 is 0, 0 is 1, etc)
if (~date.indexOf('/')) {
return date.split('-');
} else {
var idx = date.indexOf('-', 1 + date.indexOf('-', 1 + date.indexOf('-')));
return [date.slice(0, idx), date.slice(idx + 1)];
}
}
var dates = ['01-Dec-16-06-Dec-16', '01/12/16-06/12/16'];
dates.forEach((date) => {
alert(splitDates(date))
});

not giving value after decimal point

This method gives me wrong value, why? Result must be 5.088100 but it gives me 5.o88100 in any value. If the first number after decmal point is zero then it removes it, why?
var precision = descriptor.digits ? descriptor.digits[1] : 2;
var int_part = Math.floor(value);
var dec_part = Math.abs(Math.floor((value % 1) * Math.pow(10, precision)));

Because you're turning the part after the decimal point into an integer, so 08 becomes 8 — just like var dec_part = 08 would give you 8.
If you want the decimal part use string manipulation:
var tmp = value.toFixed(precision).split("."),
int_part = tmp[0],
dec_part = tmp[1];

if you treat the dec_part as a number, if absolutely clear that the leading zeroes are droped off. numbers don't and can't have leading zeros.
what you're really trying to do is getting the two parts as strings, wich is a lot easier using split():
var parts = value.split(".");
var int_part = parts[0],
var dec_part = parts[1];
but at least, if you just want to round your value to a specific precision, you can just drop that complicated stuff out and use toFixed():
value = value.toFixed(precision);

How do I select the nth digit in a large integer inside javascript?

When I want to select the nth character, I use the charAt() method, but what's the equivalent I can use when dealing with integers instead of string values?

Use String():
var number = 132943154134;
// convert number to a string, then extract the first digit
var one = String(number).charAt(0);
// convert the first digit back to an integer
var one_as_number = Number(one);

It's a stupid solution but seems to work without converting to string.
var number = 123456789;
var pos = 4;
var digit = ~~(number/Math.pow(10,pos))- ~~(number/Math.pow(10,pos+1))*10;

You could convert the number to a string and do the same thing:
parseInt((number + '').charAt(0))

If you want an existing method, convert it to a string and use charAt.
If you want a method that avoids converting it to a string, you could play games with dividing it by 10 repeatedly to strip off enough digits from the right -- say for 123456789, if you want the 3rd-from-right digit (6), divide by 10 3 times yielding 123456, then take the result mod 10 yielding 6. If you want to start counting digits from the left, which you probably do, then you need to know how many digits (base 10) are in the entire number, which you could deduce from the log base 10 of the number... All this is unlikely to be any more efficient than just converting it to a string.

function digitAt(val, index) {
return Math.floor(
(
val / Math.pow(10, Math.floor(Math.log(Math.abs(val)) / Math.LN10)-index)
)
% 10
);
};
digitAt(123456789, 0) // => 1
digitAt(123456789, 3) // => 4
A bit messy.
Math.floor(Math.log(Math.abs(val)) / Math.LN10)
Calculates the number of digits (-1) in the number.

var num = 123456;
var secondChar = num.toString()[1]; //get the second character

var number = 123456789
function return_digit(n){
r = number.toString().split('')[n-1]*1;
return r;
}
return_digit(3); /* returns 3 */
return_digit(6); /* returns 6 */