Develop Reference

How to check if number ends in .99

I'm attempting to do some validation a price field. I would like to check if the price entered into the price field ends in .99
I've attempted find posts about this but I can't find examples for decimal numbers only whole numbers. I tried to check by doing price % 1 but it isnt consistent as the price increases by 10, 20 etc.
Is there a quick way to check if all numbers end in .99?
const price = 9.99
console.log(price % 1)

Floating point math is inherently imprecise. The actual mathematical expression price - 9 will get those extra 0s and a 2 too.
Best you could do is convert to a string with fixed precision (rounding off any extraneous precision; for a price in dollars, you'd only need two digits, but you might go to three or more to verify the price entered didn't end with nonsense fractions of a cent) and perform a string test, e.g.
price.toFixed(2).endsWith('.99')
which doesn't try to perform math on price at all, it just rounds off to two digits after the decimal place to produce a string, then checks if the string ends with .99.

You can try regular expression as well. See following code for example:
function testRegex() {
var re = /^[0-9]*[.](99)$/;
var val = document.getElementById("inputValue").value;
if(re.exec(val)) {
document.getElementById("result").innerText = "Found match!!!";
} else {
document.getElementById("result").innerText = "Found no match!!!";
}
}
<input type="text" id="inputValue" value="" onkeyup="testRegex()" />
<div id="result"></div>

You can perform that validation using the following regular expression:
/\.99$/
First, we try to match an explicit . by escaping it with a backlash. Then, the next two characters must be 99, and then the string end must occur for the successful match. We check that with $. For example,
prices = [0.99, 0.0099, 1.99, 2.99, 3.98, 4.01];
for (const price of prices) {
if (/\.99$/.test(price.toString())) {
console.log(`${price} ends with .99`);
}
}
will print:
0.99 ends with .99
1.99 ends with .99
2.99 ends with .99

How to make regular expression only accept special formula?

I'm making html page for special formula using angularJS.
<input ng-model="expression" type="text" ng-blur="checkFormula()" />
function checkFormula() {
let regex;
if (scope.formulaType === "sum") {
regex = "need sum regular expression here"; // input only like as 1, 2, 5:6, 8,9
} else {
regex = "need arithmetic regular expression here"; // input only like as 3 + 4 + 6 - 9
}
if (!regex.test(scope.expression)) {
// show notification error
Notification.error("Please input expression correctly");
return;
}
// success case
if (scope.formulaType === "sum") {
let fields = expression.split(',');
let result = fields.reduce((acc, cur) => { return acc + Number(cur) }, 0);
// processing result
} else {
// need to get fields with + and - sign.
// TODO: need coding more...
let result = 0;
// processing result
}
}
So I want to make inputbox only accept my formula.
Formulas are two cases.
1,2,3:7,9
or
4-3+1+5
First case, means sum(1,2,3,4,5,6,7,9) and second case means (4-3+1+5).
But I don't know regular expression how to process it.
I searched google, but I didn't get result for my case.
So I want to need 2 regex match.

1,2,3:7,9
Fot this pattern, you can try this one:
^\d+(?::\d+)?(?:,\d+(?::\d+)?)*$
^\d+(?::\d+)?
matches string starts with a number(e.g. 1) or two numbers separated by a column (e.g. 1:2)
(?:,\d+(?::\d+)?)*$
repeats the previous pattern with a comma in front of it as many time as possible until meets the end of the string (e.g. ,2:3,4:5,6)
4-3+1+5
Fot this pattern, you can try this one:
^\d+(?:[+-]\d+)*$
Like the previous one, this is much simpler
^\d+
starts with a number(e.g. 12)
(?:[+-]\d+)*$
repeats the previous pattern with a - or + in front of it as many time as possible until meets the end of the string (e.g. +2-3+14)
Also, if you need at least one pair of numbers.
Such as 1,2 is allowed but just 1 is not. You can just change the * before $ to +:
^\d+(?::\d+)?(?:,\d+(?::\d+)?)+$
^\d+(?:[+-]\d+)+$
And if you allow white spaces in between them:
^\d+(?:\s*:\s*\d+)?(?:\s*,\s*\d+(?:\s*:\s*\d+)?)+$
^\d+(?:\s*[+-]\s*\d+)+$

Searching keywords in JavaScript

Here's an example of the customer codes:
C000000123
C000000456
If I input C123 in the search box, "C000000123" will automatically display.
9 numbers are fixed.
Please help me, a short sample was shown to me but I don't get it.
function test(key, num, digit) {
let retStr;
xxxx (condition)
retun retStr;
}
here's an elaboration:
**
input:123
output:A00000123
input:1
output:A00000001
input:99999
output:A00099999
**
here's the detailed demand:
Since it takes time and effort to enter the management number “alphabet + numeric value 9 digits” on the search screen, when the alphabetic number and the number excluding the leading 0 are entered, it is automatically complemented so that it becomes 9 padded with zeros.
sorry i'm very very new to programming in javascript

Try this:
May be what you want...
Please test it and tell if its what you want.
function getOutput(input){
var str=input.substring(1,input.length);
var padd0=9-str.length;
var zr="000000000";
var zrsub=zr.substring(0,padd0);
var output=input[0]+zrsub+""+str;
return output;
}
//Example: Call it like (NB any letter can be used):
getOutput("C123"); //or
getOutput("D123");

You can use .endsWith in js which takes a string and a search string and returns true if the specified string ends with the search string.
This function takes an array of customer ids and a search string and returns the matching customer id
function searchCustomer(customers, searchString) {
return customers.find(customer => customer.endsWith(searchString));
}
searchCustomer(['C000000123', 'C000000456'], 123); // "C000000123"
searchCustomer(['C000000123', 'C000000456'], 456); // "C000000456"
searchCustomer(['C000000123', 'C000000456', 'A00000001'], 1); //"A00000001"

JavaScript regex: capturing optional groups while having a strict matching

I am trying to extract some data from user input that should follow this format: 1d 5h 30m, which means the user is entering an amount of time of 1 day, 5 hours and 30 minutes.
I am trying to extract the value of each part of the input. However, each group is optional, meaning that 2h 20m is a valid input.
I am trying to be flexible in the input (in the sense that not all parts need to be input) but at the same time I don't watch my regex to match some random imput like asdfasdf20m. This one should be rejected (no match).
So first I am getting rid of any separator the user might have used (their input can look like 4h, 10m and that's ok):
input = input.replace(/[\s.,;_|#-]+/g, '');
Then I am capturing each part, which I indicate as optional using ?:
var match = /^((\d+)d)?((\d+)h)?((\d+)m)?$/.exec(input);
It is kind of messy capturing an entire group including the letter when I only want the actual value, but I cannot say that cluster is optional without wrapping it with parentheses right?
Then, when an empty group is captured its value in match is undefined. Is there any function to default undefined values to a particular value? For example, 0 would be handy here.
An example where input is "4d, 20h, 55m", and the match result is:
["4d20h55m", "4d", "4", "20h", "20", "55m", "55", index: 0, input: "4d20h55m"]
My main issues are:
How can I indicate a group as optional but avoid capturing it?
How can I deal with input that can potentially match, like abcdefg6d8m?
How can I deal with an altered order? For example, the user could input 20m 10h.
When I'm asking "how to deal with x" I mean I'd like to be able to reject those matches.

As variant:
HTML:
<input type="text">
<button>Check</button>
<div id="res"></div>
JS:
var r = [];
document.querySelector('button').addEventListener('click', function(){
var v = document.querySelector('input').value;
v.replace(/(\d+d)|(\d+h)|(\d+m)/ig, replacer);
document.querySelector('#res').innerText = r;
}, false);
function trim(s, mask) {
while (~mask.indexOf(s[0])) {
s = s.slice(1);
}
while (~mask.indexOf(s[s.length - 1])) {
s = s.slice(0, -1);
}
return s;
}
function replacer(str){
if(/d$/gi.test(str)){
r[0] = str;
}
else if(/h$/gi.test(str)){
r[1] = str;
}
else if(/m$/gi.test(str)){
r[2] = str;
}
return trim(r.join(', '), ',');
}
See here.

How to get jQuery animateNumber to pull value dynamically?

I am trying to implement jQuery animateNumber, and I have it working like the demo just fine. However, I want to modify it so that it pulls the number from the HTML, rather than setting it in the script. I tried the following, but it just shows "NAN." What am I doing wrong?
<div class="stat-title animate-number">$16,309</div>
<script>
$('.animate-number').each(function(){
var value = new Number;
// Grab contents of element and turn it into a number
value = $(this).text();
value = parseInt(value);
// Set the starting text to 0
$(this).text('0');
$(this).animateNumber(
{
number: value,
},
1000
)
});
</script>

ParseInt is failing because of the $ character. Also, the comma is messing with parseInt and giving you the value 16 instead of 16309. After that, the animation seems to be working. Also, you don't necessarily have to replace it with 0 since animateNumber automatically starts from 0. Here's my work so far on JSfiddle:
http://jsfiddle.net/0m9828kn/1
$('.animate-number').each(function(){
var value = $(this).text();
value = value.substring(1, value.length);//Remove $ sign
value = value.replace(/,/g, "");//Remove all commas
value = parseInt(value);
//$(this).text("0"); //This part isn't actually necessary
$(this).animateNumber(
{
number: value
},
1000
)
});

Looks like the problem is that ParseInt is choking on the dollar sign. Then when you remove that, it only reads up to comma and animates to 16. Take them both out, and it works.
value = $(this).text().replace(/[\,\$]/g,'');
But, you'll need to include this option to get your comma back:
var comma_separator_number_step = $.animateNumber.numberStepFactories.separator(',')
$(this).animateNumber(
{
number: value,
numberStep: comma_separator_number_step
},
1000
)
To get your dollar sign back, you could change your html to something like this:
<div>$<span class="stat-title animate-number">16,309</span></div>