Java Script - Count for even number of open and closed brackets - javascript

I want to make sure that a string contains an even number of closed and open brackets.
How can I count the number of open and closed brackets of a string in JavaScript and then compare the results?
Preferable without a use of regex.
Thanks you very much in advance!

You can use String.prototype.split:
str.split("(").length === str.split(")").length

Some straightforward JavaScript:
function count_chars(s) {
var i = s.length, r = {};
while (i--) r[s[i]] = r[s[i]] ? r[s[i]]+1 : 1;
return r;
}
var str = 'ab(sdf)sdf(sdf))';
var chars = count_chars(str);
console.log('( count = ' + (chars['('] || 0));
console.log(') count = ' + (chars[')'] || 0));
console.log('{ count = ' + (chars['{'] || 0));
returns:
( count = 2
) count = 3
{ count = 0

Related

How to insert - after 6th character and after 8th character in Javascript?

Hello everyone and my problem faced is when user key in Malaysian Identity Card "930402084401", I want the key in number to automatically add - to become "930402-08-4401".
When User key in 930402084401
What I want in my textbox => 930402-08-4401
I try to refer to the code where the code will automatically add space every 4 numbers, but I totally have no idea how to change the code.
document.getElementById('creditSpace').addEventListener('input', function(e) {
var target = e.target,
position = target.selectionEnd,
length = target.value.length;
target.value = target.value.replace(/[^\dA-Z]/g, '').replace(/(.{4})/g, '$1 ').trim();
target.selectionEnd = position += ((target.value.charAt(position - 1) === ' ' && target.value.charAt(length - 1) === ' ' && length !== target.value.length) ? 1 : 0);
});
<label><input id="creditSpace"> CC Details </label>
Hope someone can solve my problem. Thank you.
var s = '930402084401';
var t = s.replace( /(\d{6})(\d{2})(\d{4})/, '$1-$2-$3' ); // 930402-08-4401
You can do it with a simple for loop:
let str = '930402084401';
const myFunc = str => {
let r = '';
for (let i = 0; i < str.length; i++) {
r += str[i];
if (i == 5 || i == 7) r += '-';
}
return r;
};
myFunc(str);
Try some thing like this. Use map, slice and join
const format = (str, last = 0) =>
[6, 2, 4].map((num) => str.slice(last, (last += num))).join("-");
const str = "930402084401";
console.log(format(str));
The fastest way is also the simplest, just create separate groups of numbers and add in the - as necessary. If you want 6, 2, 4, then:
const key = '930402084401';
const text = `${key.slice(0, 6)}-${key.slice(6, 8)}-${key.slice(8)}`;
console.log(text);
Str = "930402084401";
formattedStr = Str.slice(-Str.length, 6) + "-" + Str.slice(-6,-4) + "-" + Str.slice(-4)
console.log(formattedStr);

Keep N occurrences of a single character in a string in Javascript

Let's say I have this string: "a_b_c_d_restofthestring" and I only want to keep (e.g.) 2 underscores. So,
"a_b_cdrestofthestring"
"abc_d_restofthestring"
Are both valid outputs.
My current implementation is:
let str = "___sdaj___osad$%^&*";
document.getElementById('input').innerText = str;
let u = 0;
str = str.split("").reduce((output, c) => {
if (c == "_") u++;
return u < 2 || c != "_" ? output + c : output;
});
document.getElementById('output').innerText = str;
<div id="input"></div>
<div id="output"></div>
But I'd like to know if there's a better way...
Your code seems to work fine, but here's a one-liner regular expression that replaces all but the last two underscores from the input string.
let input = "___sdaj___osad$%^&*";
let output = input.replace(/_(?=(.*_){2})/g, '');
console.log("input: " + input);
console.log("output: " + output);
This of course is not very generalized, and you'd have to modify the regular expression every time you wanted to say, replace a character other than underscore, or allow 3 occurrences. But if you're okay with that, then this solution has a bit less code to maintain.
Update: Here's an alternate version, that's fully generic and should perform a bit better:
let input = "___sdaj___osad$%^&*";
function replace(input, char = '_', max = 2, replaceWith = '') {
let result = "";
const len = input.length;
for (let i = 0, u = 0; i < len; i++) {
let c = input[i];
result += (c === char && ++u > max) ? replaceWith : c;
}
return result;
}
console.log("input: ", input);
console.log("output: ", replace(input));
See this jsPerf analysis.
You could take a regular expression which looks for an underscore and a counter of the keeping underscores and replace all others.
var string = "a_b_c_d_restofthestring",
result = string.replace(/_/g, (c => _ => c && c-- ? _ : '')(2));
console.log(result);

How to increment a string in JavaScript containing leading zeros?

I have string like:
MPG_0023
I want to find something like
MPG_0023 + 1
and I should get
MPG_0024
How to do that in JavaScript? It should take care that if there are no leading zeros, or one leading zero should still work like MPG23 should give MPG24 or MPG023 should give MPG024.
There should be no assumption that there is underscore or leading zeros, the only thing is that first part be any string or even no string and the number part may or may not have leading zeros and it is any kind of number so it should work for 0023 ( return 0024) or for gp031 ( return gp032) etc.
Here's a quick way without using regex.. as long as there's always a single underscore preceding the number and as long as the number is 4 digits, this will work.
var n = 'MPG_0023';
var a = n.split('_');
var r = a[0]+'_'+(("0000"+(++a[1])).substr(-4));
console.log(r);
Or if you do wanna do regex, the underscore won't matter.
var n = "MPG_0099";
var r = n.replace(/(\d+)/, (match)=>("0".repeat(4)+(++match)).substr(-4));
console.log(r);
You can use the regular expressions to make the changes as shown in the following code
var text = "MPG_0023";
var getPart = text.replace ( /[^\d.]/g, '' ); // returns 0023
var num = parseInt(getPart); // returns 23
var newVal = num+1; // returns 24
var reg = new RegExp(num); // create dynamic regexp
var newstring = text.replace ( reg, newVal ); // returns MPG_0024
console.log(num);
console.log(newVal);
console.log(reg);
console.log(newstring);
Using regex along with the function padStart
function add(str, n) {
return str.replace(/(\d+)/, function(match) {
var length = match.length;
var newValue = Number(match) + n;
return newValue.toString(10).padStart(length, "0");
});
}
console.log(add("MPG_023", 101));
console.log(add("MPG_0023", 101));
console.log(add("MPG_0000023", 10001));
console.log(add("MPG_0100023", 10001));
Using regular expression you can do it like this.
var text1 = 'MPG_0023';
var text2 = 'MPG_23';
var regex = /(.*_[0]*)(\d*)/;
var match1 = regex.exec(text1);
var match2 = regex.exec(text2);
var newText1 = match1[1] + (Number(match1[2]) + 1);
var newText2 = match2[1] + (Number(match2[2]) + 1);
console.log(newText1);
console.log(newText2);
Increment and pad the same value (comments inline)
var prefix = "MPG_"
var padDigit = 4; //number of total characters after prefix
var value = "MPG_0023";
console.log("currentValue ", value);
//method for padding
var fnPad = (str, padDigit) => (Array(padDigit + 1).join("0") + str).slice(-padDigit);
//method to get next value
var fnGetNextCounterValue = (value) => {
var num = value.substring(prefix.length); //extract num value
++num; //increment value
return prefix + fnPad(num, padDigit); //prepend prefix after padding
};
console.log( "Next", value = fnGetNextCounterValue(value) );
console.log( "Next", value = fnGetNextCounterValue(value) );
console.log( "Next", value = fnGetNextCounterValue(value) );
One way would e to split the string on the "_" character, increment the number and then add the zeros back to the number.
var testString = "MGP_0023";
var ary = testString.split("_");
var newNumber = Number(ary[1]) + 1;
var result = ary[0] + pad(newNumber);
// helper function to add zeros in front of the number
function pad(number) {
var str = number.toString();
while (str.length < 4) {
str = '0' + str;
}
return str;
}
You could cast to number, increment the value and cast back. Then check if you need leading zeros by looking at the length of the string.
Snippet below:
let str = "MPG_0023",
num = Number(str.substr(4)) + 1,
newStr = String(num);
function addLeading0(str) {
return str.length === 2 ? '00' + str : (str.length === 3 ? '0' + str : str);
}
console.log("MPG_" + addLeading0(newStr));

Get max value of similar items in array with Javascript

I have an array like this:
["13rq8", "13rq6", "13rq4", "13rq2", "13dl", "12dl", "13rq12", "13rq10"]
and I want to get a final array that will group similar values that changes from each other only by the last numbers of the string ("13rq8", "13rq6", "13rq4", "13rq2", "13rq12", "13rq10"), and return only the biggest values like the example below:
["13dl", "12dl", "13rq12"]
Can you help me please resolve this in Javascript?
Thank You!
Use an object (ex. tagNum) to keep track of the largest value of each prefix, and use regular expression to extract the prefix and trailing value:
var l = ["13rq8", "13rq6", "13rq4", "13rq2", "13dl", "12dl", "13rq12", "13rq10"];
var tagNum = {};
l.forEach(function(x) {
var m = x.match(/^(.*?)(\d*)$/);
var tag = m[1];
var num = parseInt("0" + m[2]);
if (tagNum[tag] === undefined || tagNum[tag] < num) tagNum[tag] = num;
});
var l2 = [];
for (var tag in tagNum) {
var num = tagNum[tag];
if (num) l2.push(tag + num);
else l2.push(tag);
}
console.log(l2);

How to make 8 digit number in javascript?

I am trying to make an auto-generator of numbers. but I'm having a problem on how to forced the number to 8 digit.
for(i=1;i<=100;i++) {
var i = x++;
var test = i.toFixed(8); // I used this but this is only for decimals
jQuery('.generated_table').append(test+'<br />');;
}
Please help.
Use toPrecision:
(10000000).toPrecision(8); //=> '10000000'
(100).toPrecision(8); //=> '100.00000'
If you meant preceding a number with leading zero's:
var i = (100).toPrecision(8).split('.').reverse().join(''); //=> '00000100'
You can also make a Number.prototype function of that:
Number.prototype.leadingZeros = function(n) {
return this.toPrecision(n).split('.').reverse().join('');
};
(100).leadinZeros(8); //=> '00000100'
Just to be complete: a more precise way to print any (number of) leading character(s) to any number may be:
Number.prototype.toWidth = function(n,chr) {
chr = chr || ' ';
var len = String(parseFloat(this)).length;
function multiply(str,nn){
var s = str;
while (--nn>0){
str+=s;
}
return str;
}
n = n<len ? 0 : Math.abs(len-n);
return (n>1 && n ? multiply(chr,n) : n<1 ? '' : chr)+this;
};
(100).toWidth(8,'0'); //=> 00000100
Whooo!!! i got anser :: Try it
for(i=1;i<=100;i++) {
//var i = x++;
var test = i.toPrecision(8).replace("\.","");
jQuery('.generated_table').append(test+'<br />');;
}
Check out this SO question for some links to various printf-style functions for Javascript: Javascript printf/string.format
var randNum = "";
var MAX_LENGTH = 8;
while(randNum.toString().length < MAX_LENGTH){
var temp = Math.floor(Math.random() * 10);
randNum += temp.toString();
}
alert(randNum);

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