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I'm new in StackOverflow and JavaScript, I'm trying to get the first letter that repeats from a string considering both uppercase and lowercase letters and counting and obtaining results using the for statement. The problem is that the form I used is too long Analyzing the situation reaches such a point that maybe you can only use a "For" statement for this exercise, which I get to iterate, but not with a cleaner and reduced code has me completely blocked, this is the reason why I request help to understand and continue with the understanding and use of this sentence. In this case, the result was tested in a JavaScript script inside a function and 3 "For" sentences obtaining quite positive results, but I can not create it in 1 only For (Sorry for my bad english google translate)
I making in HTML with JavasScript
var letter = "SYAHSVCXCyXSssssssyBxAVMZsXhZV";
var contendor = [];
var calc = [];
var mycalc = 0;
letter = letter.toUpperCase()
console.log(letter)
function repeats(){
for (var i = 0; i < letter.length; i++) {
if (contendor.includes(letter[i])) {
}else{
contendor.push(letter[i])
calc.push(0)
}
}
for (var p = 0; p < letter.length; p++) {
for (var l = 0; l < contendor.length; l++) {
if (letter[p] == contendor[l]) {
calc [l]= calc [l]+1
}
}
}
for (var f = 0; f < calc.length; f++) {
if ( calc[f] > calc[mycalc]) {
mycalc = f
}
}
}
repeats()
console.log("The most repeated letter its: " + contendor[mycalc]);
I Expected: A result with concise code
It would probably be a lot more concise to use a regular expression: match a character, then lookahead for more characters until you can match that first character again:
var letter = "SYAHSVCXCyXSssssssyBxAVMZsXhZV";
const firstRepeatedRegex = /(.)(?=.*\1)/;
console.log(letter.match(firstRepeatedRegex)[1]);
Of course, if you aren't sure whether a given string contains a repeated character, check that the match isn't null before trying to extract the character:
const input = 'abcde';
const firstRepeatedRegex = /(.)(?=.*\1)/;
const match = input.match(firstRepeatedRegex);
if (match) {
console.log(match[0]);
} else {
console.log('No repeated characters');
}
You could also turn the input into an array and use .find to find the first character whose lastIndexOf is not the same as the index of the character being iterated over:
const getFirstRepeatedCharacter = (str) => {
const chars = [...str];
const char = chars.find((char, i) => chars.lastIndexOf(char) !== i);
return char || 'No repeated characters';
};
console.log(getFirstRepeatedCharacter('abcde'));
console.log(getFirstRepeatedCharacter('SYAHSVCXCyXSssssssyBxAVMZsXhZV'));
If what you're actually looking for is the character that occurs most often, case-insensitive, use reduce to transform the string into an object indexed by character, whose values are the number of occurrences of that character, then identify the largest value:
const getMostRepeatedCharacter = (str) => {
const charsByCount = [...str.toUpperCase()].reduce((a, char) => {
a[char] = (a[char] || 0) + 1;
return a;
}, {});
const mostRepeatedEntry = Object.entries(charsByCount).reduce((a, b) => a[1] >= b[1] ? a : b);
return mostRepeatedEntry[0];
};
console.log(getMostRepeatedCharacter('abcde'));
console.log(getMostRepeatedCharacter('SYAHSVCXCyXSssssssyBxAVMZsXhZV'));
If the first repeated character is what you want, you can push it into an array and check if the character already exists
function getFirstRepeating( str ){
chars = []
for ( var i = 0; i < str.length; i++){
var char = str.charAt(i);
if ( chars.includes( char ) ){
return char;
} else {
chars.push( char );
}
}
return -1;
}
This will return the first repeating character if it exists, or will return -1.
Working
function getFirstRepeating( str ){
chars = []
for ( var i = 0; i < str.length; i++){
var char = str.charAt(i);
if ( chars.includes( char ) ){
return char;
} else {
chars.push( char );
}
}
return -1;
}
console.log(getFirstRepeating("SYAHSVCXCyXSssssssyBxAVMZsXhZV"))
Have you worked with JavaScript objects yet?
You should look into it.
When you loop through your string
let characters = "hemdhdksksbbd";
let charCount = {};
let max = { count: 0, ch: ""}; // will contain max
// rep letter
//Turn string into an array of letters and for
// each letter create a key in the charcount
// object , set it to 1 (meaning that's the first of
// that letter you've found) and any other time
// you see the letter, increment by 1.
characters.split("").forEach(function(character)
{
if(!charCount[character])
charCount[character] = 1;
else
charCount[character]++;
}
//charCount should now contain letters and
// their counts.
//Get the letters from charCount and find the
// max count
Object.keys(charCount). forEach (function(ch){
if(max.count < charCount[ch])
max = { count: charCount[ch], ch: ch};
}
console.log("most reps is: " , max.ch);
This is a pretty terrible solution. It takes 2 loops (reduce) and doesn't handle ties, but it's short and complicated.
Basically keep turning the results into arrays and use array methods split and reduce to find the answer. The first reduce is wrapped in Object.entries() to turn the object back into an array.
let letter = Object.entries(
"SYAHSVCXCyXSssssssyBxAVMZsXhZV".
toUpperCase().
split('').
reduce((p, c) => {
p[c] = isNaN(++p[c]) ? 1 : p[c];
return p;
}, {})
).
reduce((p, c) => p = c[1] > p[1] ? c : p);
console.log(`The most repeated letter is ${letter[0]}, ${letter[1]} times.`);
I have a page with a grid where user's numbers get saved. It has a following pattern - every number ends with 3 digits after comma. It doesn't look nice, when for example user's input is
123,450
123,670
123,890
It's much better to have just 2 numbers after comma, because last 0 is absolutely meaningless and redundant.
The way it still should have 3 digits is only if at least one element in an array doesn't end up with 0
For example:
123,455
123,450
123,560
In this case 1st element of the array has the last digit not equal to 0 and hence all the elements should have 3 digits. The same story with 2 or 1 zeros
Zeros are redundant:
123,30
123,40
123,50
Zeros are necessary:
123,35
123,40
123,50
The question is how can I implement it programatically? I've started like this:
var zeros2Remove = 0;
numInArray.forEach(function(item, index, numInArray)
{
var threeDigitsAfterComma = item.substring(item.indexOf(',') + 1);
for(var j = 2; j <= 0; j--)
{
if(threeDigitsAfterComma[j] == 0)
{
zeros2Remove =+ 1;
}
else //have no idea what to do..
}
})
Well in my implementation I don't know how to do it since I have to iterate through every element but break it if at least 1 number has a last digit equal to zero.. In order to do that I have to break outer loop, but don't know how and I'm absolutely sure that I don't have to...
I think the following code what you are looking for exactly , please manipulate numbers and see the changes :
var arr = ["111.3030", "2232.0022", "3.001000", "4","558.0200","55.00003000000"];
var map = arr.map(function(a) {
if (a % 1 === 0) {
var res = "1";
} else {
var lastNumman = a.toString().split('').pop();
if (lastNumman == 0) {
var m = parseFloat(a);
var res = (m + "").split(".")[1].length;
} else {
var m = a.split(".")[1].length;
var res = m;
}
}
return res;
})
var maxNum = map.reduce(function(a, b) {
return Math.max(a, b);
});
arr.forEach(function(el) {
console.log(Number.parseFloat(el).toFixed(maxNum));
});
According to MDN,
There is no way to stop or break a forEach() loop other than by throwing an exception. If you need such behavior, the forEach() method is the wrong tool. Use a plain loop or for...of instead.
If you convert your forEach loop to a for loop, you can break out of it with a label and break statement:
// unrelated example
let i;
let j;
outerLoop:
for (i = 2; i < 100; ++i) {
innerLoop:
for (j = 2; j < 100; ++j) {
// brute-force prime factorization
if (i * j === 2183) { break outerLoop; }
}
}
console.log(i, j);
I gave you an unrelated example because your problem doesn't need nested loops at all. You can find the number of trailing zeroes in a string with a regular expression:
function getTrailingZeroes (str) {
return str.match(/0{0,2}$/)[0].length;
}
str.match(/0{0,2}$/) finds between 0 and 2 zeroes at the end of str and returns them as a string in a one-element array. The length of that string is the number of characters you can remove from str. You can make one pass over your array of number-strings, breaking out when necessary, and use Array.map as a separate truncation loop:
function getShortenedNumbers (numInArray) {
let zeroesToRemove = Infinity;
for (const str of numInArray) {
let candidate = getTrailingZeroes(str);
zeroesToRemove = Math.min(zeroesToRemove, candidate);
if (zeroesToRemove === 0) break;
}
return numInArray.map(str => str.substring(0, str.length - zeroesToRemove);
}
All together:
function getTrailingZeroes (str) {
return str.match(/0{0,2}$/)[0].length;
}
function getShortenedNumbers (numInArray) {
let zeroesToRemove = Infinity;
for (const str of numInArray) {
let candidate = getTrailingZeroes(str);
zeroesToRemove = Math.min(zeroesToRemove, candidate);
if (zeroesToRemove === 0) break;
}
return numInArray.map(str => str.substring(0, str.length - zeroesToRemove));
}
console.log(getShortenedNumbers(['123,450', '123,670', '123,890']));
console.log(getShortenedNumbers(['123,455', '123,450', '123,560']));
This solution might seem a little cumbersome but it should work for all possible scenarios. It should be easy enough to make always return a minimal number of decimals places/leading zeros.
I hope it helps.
// Define any array
const firstArray = [
'123,4350',
'123,64470',
'123,8112390',
]
const oneOfOfYourArrays = [
'123,30',
'123,40',
'123,50',
]
// Converts 123,45 to 123.45
function stringNumberToFloat(stringNumber) {
return parseFloat(stringNumber.replace(',', '.'))
}
// For 123.45 you get 2
function getNumberOfDecimals(number) {
return number.split('.')[1].length;
}
// This is a hacky way how to remove traling zeros
function removeTralingZeros(stringNumber) {
return stringNumberToFloat(stringNumber).toString()
}
// Sorts numbers in array by number of their decimals
function byNumberOfValidDecimals(a, b) {
const decimalsA = getNumberOfDecimals(a)
const decimalsB = getNumberOfDecimals(b)
return decimalsB - decimalsA
}
// THIS IS THE FINAL SOLUTION
function normalizeDecimalPlaces(targetArray) {
const processedArray = targetArray
.map(removeTralingZeros) // We want to remove trailing zeros
.sort(byNumberOfValidDecimals) // Sort from highest to lowest by number of valid decimals
const maxNumberOfDecimals = processedArray[0].split('.')[1].length
return targetArray.map((stringNumber) => stringNumberToFloat(stringNumber).toFixed(maxNumberOfDecimals))
}
console.log('normalizedFirstArray', normalizeDecimalPlaces(firstArray))
console.log('normalizedOneOfOfYourArrays', normalizeDecimalPlaces(oneOfOfYourArrays))
Try this
function removeZeros(group) {
var maxLength = 0;
var newGroup = [];
for(var x in group) {
var str = group[x].toString().split('.')[1];
if(str.length > maxLength) maxLength = str.length;
}
for(var y in group) {
var str = group[y].toString();
var substr = str.split('.')[1];
if(substr.length < maxLength) {
for(var i = 0; i < (maxLength - substr.length); i++)
str += '0';
}
newGroup.push(str);
}
return newGroup;
}
Try it on jsfiddle: https://jsfiddle.net/32sdvzn1/1/
My script checks the length of every number decimal part, remember that JavaScript removes the last zeros in a decimal number, so 3.10 would be 3.1, so the length is less when there is a number with zeros in the end, in this case we just add a zero to the number.
Update
I've updated the script, the new version adds as much zeros as the different between the max decimal length and the decimal length of the analyzed number.
Example
We have: 3.11, 3.1423, 3.1
The max length would be: 4 (1423)
maxLenght (4) - length of .11 (2) = 2
We add 2 zeros to 3.11, that will become 3.1100
I think you can start out assuming you will remove two extra zeros, and loop through your array looking for digits in the last two places. With the commas, I'm assuming your numArray elements are strings, all starting with the same length.
var numArray = ['123,000', '456,100', '789,110'];
var removeTwo = true, removeOne = true;
for (var i = 0; i < numArray.length; i++) {
if (numArray[i][6] !== '0') { removeTwo = false; removeOne = false; }
if (numArray[i][5] !== '0') { removeTwo = false; }
}
// now loop to do the actual removal
for (var i = 0; i < numArray.length; i++) {
if (removeTwo) {
numArray[i] = numArray[i].substr(0, 5);
} else if (removeOne) {
numArray[i] = numArray[i].substr(0, 6);
}
}
I'm sorry it's a stupid question but I'm try to list a sequence of numbers from 1 to a limit and they should be separated by comma. I have a problem. I don't know how to stop the comma. I should have 1,2,3 but I have ,1,2,3. Can you help me? Here is my code.
function getNumberSequence(number) {
var result = ""
if(number <= 0){
return result
}
else{
if(number == 1){
result = result + 1
} else {
for(i = 1; i <= number; i++){
result = result + ',' + i;
}
}
}
return result
}
Thanks for all
You can simply add a flag (in the example "first") and check if this is your first iteration - if so, don't add a comma but set it to false ... see here: http://jsfiddle.net/fdfxc5zq/
var first = true; //have a flag that tells you if this is your first iteration - don't add a comma the first time around
for (i = 0; i <= number; i++) {
if (first) {
first = false;
} else {
result += ", ";
}
result += i;
}
Using your code, simply declare result with an initial value of 1 and start the for loop with i=2. Solved. (And in case of number being 1, just return result)
A one-liner for you:
return Object.keys(Array.apply(null,new Array(number))).map(function(n) {return +n+1;}).join(",");
Basically, creates an array of length number, "applies" it (basically ends up giving undefined values to each index), then uses Object.keys to get the indices of the array, increments all the items, then joins the whole thing with commas.
Two easy methods:
1) Just add a comma after each number and return a substring of length-1 to get rid of the last comma
if (i === 0) return ""; //to avoid substring() issues on empty strings
var result = "";
for (i = 0; i <= number; i++) {
result = i + ",";
}
return result.substr(0, result.length-1);
2) Using array and join
var result = [];
for (i = 0; i <= number; i++) {
result.push(i)
}
return result.join(); //default join separator is the comma
YourCommaSeparatedString.split(',');
I am trying to find the Rank of the given string in the list of possible permutations. I tried to come up with a solution that tries to find all possible permutations, assign a rank to them and then display it.
But this drastically reduces the performance when the length of the string keeps increasing. So was wondering if someone can think of an efficient solution for this problem..
function permute(str) {
// Sort the string
var arr = [];
for (var i = 0; i < str.length; i++) arr.push(str[i]);
var sortedString = arr.sort().join(''),
// Length of the string
length = str.length,
used = [];
// Create a boolean array for length of the string
while (length--) used.push(false);
// String buffer that holds the current string
var out = '';
// Call the function
doPermute(sortedString, str, out, used, str.length, 0);
}
var count = 0;
function doPermute(inp, givenString, out, used, length, level) {
// Only if length of the string equal to current level print it
// That is permutation length is eqaul to string length
if (level == length) {
count++;
//console.log('Perm :: ' + out + ' -- ' + count);
if (out === givenString) {
var pp = 'Rank of :: ' + out + ' -- ' + count;
$('div').append('<p>' + pp + '</p>');
}
return;
}
for (var i = 0; i < length; ++i) {
// If variable used continue
if (used[i]) continue;
// Append the current char in loop
out += inp[i];
// set variable to true
used[i] = true;
// Call the function again
doPermute(inp, givenString, out, used, length, level + 1);
// Set it to false as the variable can be reused
used[i] = false;
// remove the last character of the buffer
out = out.slice(0, out.length - 1)
}
}
permute('dbcarf')
Fiddle
Sure: if input string is "cab".
What is the lowest rank that a string starting with c could get?
c
Note the strings that come before it.
abc
acb
bac
bca
So a string starting with c has minimum rank 5.This is just number of characters in input string that come lexicographically before c.(in order a,b,c,d,e,f...)So we have 2.Each word starting with a letter can have 2 words.
Next letter is "a"?
What is minimum rank that a word starting with "ca" can get?
5
Why?
"a" is the best way we can fill the second spot with the remaining letters.
And the same goes for third element "b".
So rank of "cab" is 5.
In general.(Assuming no duplicates, though this is not much harder)
var W; //input string
var C[26];
var rank = 1;
for (var i = 0; i < W.length; i++) C[W[i] - 'a']++;
for (var i = 0; i < W.length; i++) {
//How many characters which are not used, that come before current character
var count = 0;
for (var j = 0; j < 26; j++) {
if (j == (W[i] - 'a')) break;
if (C[j] > 0) count++;
}
C[W[i] - 'a'] = 0;
rank += count * fact(W.length - i - 1);
}
There is an explanation in https://en.wikipedia.org/wiki/Permutation#Numbering_permutations of how to convert a permutation on n objects to a number in the range 0..n!-1 and it goes on to say that "Converting successive natural numbers to the factorial number system produces those sequences in lexicographic order (as is the case with any mixed radix number system), and further converting them to permutations preserves the lexicographic ordering, provided the Lehmer code interpretation is used" So I would try doing this number conversion and see if it produces something related to the rank that you need, by your definition.
What's the shortest way (within reason) to generate a random alpha-numeric (uppercase, lowercase, and numbers) string in JavaScript to use as a probably-unique identifier?
I just came across this as a really nice and elegant solution:
Math.random().toString(36).slice(2)
Notes on this implementation:
This will produce a string anywhere between zero and 12 characters long, usually 11 characters, due to the fact that floating point stringification removes trailing zeros.
It won't generate capital letters, only lower-case and numbers.
Because the randomness comes from Math.random(), the output may be predictable and therefore not necessarily unique.
Even assuming an ideal implementation, the output has at most 52 bits of entropy, which means you can expect a duplicate after around 70M strings generated.
If you only want to allow specific characters, you could also do it like this:
function randomString(length, chars) {
var result = '';
for (var i = length; i > 0; --i) result += chars[Math.floor(Math.random() * chars.length)];
return result;
}
var rString = randomString(32, '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ');
Here's a jsfiddle to demonstrate: http://jsfiddle.net/wSQBx/
Another way to do it could be to use a special string that tells the function what types of characters to use. You could do that like this:
function randomString(length, chars) {
var mask = '';
if (chars.indexOf('a') > -1) mask += 'abcdefghijklmnopqrstuvwxyz';
if (chars.indexOf('A') > -1) mask += 'ABCDEFGHIJKLMNOPQRSTUVWXYZ';
if (chars.indexOf('#') > -1) mask += '0123456789';
if (chars.indexOf('!') > -1) mask += '~`!##$%^&*()_+-={}[]:";\'<>?,./|\\';
var result = '';
for (var i = length; i > 0; --i) result += mask[Math.floor(Math.random() * mask.length)];
return result;
}
console.log(randomString(16, 'aA'));
console.log(randomString(32, '#aA'));
console.log(randomString(64, '#A!'));
Fiddle: http://jsfiddle.net/wSQBx/2/
Alternatively, to use the base36 method as described below you could do something like this:
function randomString(length) {
return Math.round((Math.pow(36, length + 1) - Math.random() * Math.pow(36, length))).toString(36).slice(1);
}
UPDATED:
One-liner solution, for random 20 characters (alphanumeric lowercase):
Array.from(Array(20), () => Math.floor(Math.random() * 36).toString(36)).join('');
Or shorter with lodash:
_.times(20, () => _.random(35).toString(36)).join('');
Another variation of answer suggested by JAR.JAR.beans
(Math.random()*1e32).toString(36)
By changing multiplicator 1e32 you can change length of random string.
Or to build upon what Jar Jar suggested, this is what I used on a recent project (to overcome length restrictions):
var randomString = function (len, bits)
{
bits = bits || 36;
var outStr = "", newStr;
while (outStr.length < len)
{
newStr = Math.random().toString(bits).slice(2);
outStr += newStr.slice(0, Math.min(newStr.length, (len - outStr.length)));
}
return outStr.toUpperCase();
};
Use:
randomString(12, 16); // 12 hexadecimal characters
randomString(200); // 200 alphanumeric characters
This is cleaner
Math.random().toString(36).substr(2, length)
Example
Math.random().toString(36).substr(2, 5)
function randomString(len) {
var p = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789";
return [...Array(len)].reduce(a=>a+p[~~(Math.random()*p.length)],'');
}
Summary:
Create an array of the size we want (because there's no range(len) equivalent in javascript.
For each element in the array: pick a random character from p and add it to a string
Return the generated string.
Some explanation:
[...Array(len)]
Array(len) or new Array(len) creates an array with undefined pointer(s). One-liners are going to be harder to pull off. The Spread syntax conveniently defines the pointers (now they point to undefined objects!).
.reduce(
Reduce the array to, in this case, a single string. The reduce functionality is common in most languages and worth learning.
a=>a+...
We're using an arrow function.
a is the accumulator. In this case it's the end-result string we're going to return when we're done (you know it's a string because the second argument to the reduce function, the initialValue is an empty string: ''). So basically: convert each element in the array with p[~~(Math.random()*p.length)], append the result to the a string and give me a when you're done.
p[...]
p is the string of characters we're selecting from. You can access chars in a string like an index (E.g., "abcdefg"[3] gives us "d")
~~(Math.random()*p.length)
Math.random() returns a floating point between [0, 1) Math.floor(Math.random()*max) is the de facto standard for getting a random integer in javascript. ~ is the bitwise NOT operator in javascript.
~~ is a shorter, arguably sometimes faster, and definitely funner way to say Math.floor( Here's some info
I think the following is the simplest solution which allows for a given length:
Array(myLength).fill(0).map(x => Math.random().toString(36).charAt(2)).join('')
It depends on the arrow function syntax.
for 32 characters:
for(var c = ''; c.length < 32;) c += Math.random().toString(36).substr(2, 1)
Random character:
String.fromCharCode(i); //where is an int
Random int:
Math.floor(Math.random()*100);
Put it all together:
function randomNum(hi){
return Math.floor(Math.random()*hi);
}
function randomChar(){
return String.fromCharCode(randomNum(100));
}
function randomString(length){
var str = "";
for(var i = 0; i < length; ++i){
str += randomChar();
}
return str;
}
var RandomString = randomString(32); //32 length string
Fiddle: http://jsfiddle.net/maniator/QZ9J2/
Using lodash:
function createRandomString(length) {
var chars = "abcdefghijklmnopqrstufwxyzABCDEFGHIJKLMNOPQRSTUFWXYZ1234567890"
var pwd = _.sampleSize(chars, length || 12) // lodash v4: use _.sampleSize
return pwd.join("")
}
document.write(createRandomString(8))
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>
Random Key Generator
keyLength argument is the character length you want for the key
function keyGen(keyLength) {
var i, key = "", characters = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789";
var charactersLength = characters.length;
for (i = 0; i < keyLength; i++) {
key += characters.substr(Math.floor((Math.random() * charactersLength) + 1), 1);
}
return key;
}
keyGen(12)
"QEt9mYBiTpYD"
var randomString = function(length) {
var str = '';
var chars ='0123456789ABCDEFGHIJKLMNOPQRSTUVWXTZabcdefghiklmnopqrstuvwxyz'.split(
'');
var charsLen = chars.length;
if (!length) {
length = ~~(Math.random() * charsLen);
}
for (var i = 0; i < length; i++) {
str += chars[~~(Math.random() * charsLen)];
}
return str;
};
When I saw this question I thought of when I had to generate UUIDs. I can't take credit for the code, as I am sure I found it here on stackoverflow. If you dont want the dashes in your string then take out the dashes. Here is the function:
function generateUUID() {
var d = new Date().getTime();
var uuid = 'xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx'.replace(/[xy]/g,function(c) {
var r = (d + Math.random()*16)%16 | 0;
d = Math.floor(d/16);
return (c=='x' ? r : (r&0x7|0x8)).toString(16);
});
return uuid.toUpperCase();
}
Fiddle: http://jsfiddle.net/nlviands/fNPvf/11227/
This function should give a random string in any length.
function randString(length) {
var l = length > 25 ? 25 : length;
var str = Math.random().toString(36).substr(2, l);
if(str.length >= length){
return str;
}
return str.concat(this.randString(length - str.length));
}
I've tested it with the following test that succeeded.
function test(){
for(var x = 0; x < 300000; x++){
if(randString(x).length != x){
throw new Error('invalid result for len ' + x);
}
}
}
The reason i have chosen 25 is since that in practice the length of the string returned from Math.random().toString(36).substr(2, 25) has length 25. This number can be changed as you wish.
This function is recursive and hence calling the function with very large values can result with Maximum call stack size exceeded. From my testing i was able to get string in the length of 300,000 characters.
This function can be converted to a tail recursion by sending the string to the function as a second parameter. I'm not sure if JS uses Tail call optimization
A simple function that takes the length
getRandomToken(len: number): string {
return Math.random().toString(36).substr(2, len);
}
Ff you pass 6 it will generate 6 digit alphanumeric number
Nice and simple, and not limited to a certain number of characters:
let len = 20, str = "";
while(str.length < len) str += Math.random().toString(36).substr(2);
str = str.substr(0, len);
One could just use lodash uniqueId:
_.uniqueId([prefix=''])
Generates a unique ID. If prefix is given, the ID is appended to it.
Here's a simple code to generate random string alphabet.
Have a look how this code works.
go(lenthOfStringToPrint); - Use this function to generate the final string.
var letters = {
1: ["q","w","e","r","t","y","u","i","o","p","a","s","d","f","g","h","j","k","l","z","x","c","v","b","n","m"],
2: ["Q","W","E","R","T","Y","U","I","O","P","A","S","D","F","G","H","J","K","L","Z","X","C","V","B","N","M"]
},i,letter,final="";
random = (max,min) => {
return Math.floor(Math.random()*(max-min+1)+min);
}
function go(length) {
final="",letter="";
for (i=1; i<=length; i++){
letter = letters[random(0,3)][random(0,25)];
final+=letter;
}
return final;
}
I used #Nimphious excellent second approach and found that occasionally the string returned was numeric - not alphanumeric.
The solution I used was to test using !isNaN and use recursion to call the function again.
Why bother? I was using this function to create object keys, if all the keys are alphanumeric everything sorts properly but if you use
numbers as keys mixed with alphanumeric (strings) looping through the object will produce a different order to original order.
function newRandomString(length, chars) {
var mask = '';
if (chars.indexOf('a') > -1) mask += 'abcdefghijklmnopqrstuvwxyz';
if (chars.indexOf('A') > -1) mask += 'ABCDEFGHIJKLMNOPQRSTUVWXYZ';
if (chars.indexOf('#') > -1) mask += '0123456789';
if (chars.indexOf('$') > -1) mask += '0123456789';
var result = '';
for (var i = length; i > 0; --i) result += mask[Math.floor(Math.random() *
mask.length)];
/*
we need a string not a number !isNaN(result)) will return true if '1234' or '3E77'
because if we're looping through object keys (created by newRandomString()) and
a number is used and all the other keys are strings then the number will
be first even if it was the 2nd or third key in object
*/
//use recursion to try again
if(!isNaN(result)){
console.log('found a number....:'+result);
return newRandomString(length, chars)
}else{
return result;
}
};
var i=0;
while (i < 1000) {
var a = newRandomString(4, '#$aA');
console.log(i+' - '+a);
//now we're using recursion this won't occur
if(!isNaN(a)){
console.log('=============='+i+' - '+a);
}
i++;
}
console.log('3E77:'+!isNaN('3E77'));//true
console.log('1234:'+!isNaN('1234'));//true
console.log('ab34:'+!isNaN('ab34'));//false
After looking at solutions in answers to this question and other sources, this is the solution that is simplest while allowing for modification of the included characters and selection in the length of the returned result.
// generate random string of n characters
function randomString(length) {
const characters = '0123456789abcdefghijklmnopqrstuvwxyz'; // characters used in string
let result = ''; // initialize the result variable passed out of the function
for (let i = length; i > 0; i--) {
result += characters[Math.floor(Math.random() * characters.length)];
}
return result;
}
console.log(randomString(6));
Use md5 library: https://github.com/blueimp/JavaScript-MD5
The shortest way:
md5(Math.random())
If you want to limit the size to 5:
md5(Math.random()).substr(0, 5)