We Keep Coding

Does the input string have a . or ? or ' or space?

Im writing a program that will tell you if the input string contains a period, question mark, colon or space. If it doesn't it'll return "false". If it does, it'll return all of the characters before the found punctuation. Ex. str= "grey cat", program will return "grey" because it was stopped by the space. Can I use "or" in Javascript? EDIT Im trying to do this without any built-in functions. Im not looking for an efficient way
update - now it's just printing the str. How do I get it to just print what comes before the punctuation (if any)
function find_punctuation(str){
let result = "";
for (let i = 0; i < str.length; i ++ )
if (str[i] === "." || str[i] === "?"|| str[i] === "" || str[i] === ","){
}
else result += str[i]
return result
}
console.log(find_punctuation('he. y'));

function find_punctuation(str) {
let result = "";
for (let i = 0; i < str.length; i++) {
// Determine if the current character is punctuation.
const is_punctuation = (
str[i] === "." ||
str[i] === "?" ||
str[i] === " " ||
str[i] === ","
);
if (is_punctuation) {
// If the current character is punctuation, then exit the loop.
break;
} else {
// Otherwise, keep adding to the result string.
result += str[i];
}
}
// The result string will now contain all characters until the first
// punctuation character, because the loop that added characters to
// this string was exited as soon as the first punctuation character
// was detected.
return result;
}
console.log(find_punctuation('he. y'));
console.log(find_punctuation('hell,o'));
console.log(find_punctuation('hello'));
console.log(find_punctuation('cat?erpillar'));
console.log(find_punctuation('grey cat'));

Use a regular expression:
function find_punctuation(str) {
let result = str.match(/^(.*?)[.?: ]/);
if (result) {
return result[1];
} else {
return false;
}
}
console.log(find_punctuation('he.y'));
console.log(find_punctuation('hey'));

Try this:
function find_punctuation(str){
let result = "";
for (let i = 0; i < str.length; i ++ )
if (str[i] === "." || str[i] === "?" || str[i] === " ") break
else result += str[i]
return result
}
console.log(find_punctuation('he.y'));

how to fix: Array index is out of range

i have this problem
Have the function WildcardCharacters(str) read str which will contain two strings separated by a space. The first string will consist of the following sets of characters: +, , and {N} which is optional. The plus (+) character represents a single alphabetic character, the asterisk () represents a sequence of the same character of length 3 unless it is followed by {N} which represents how many characters should appear in the sequence where N will be at least 1. Your goal is to determine if the second string exactly matches the pattern of the first string in the input.
i have the solution with js, but now i am trying to solve with swift, i need your help.
this work with js, and i'm using swift v4.2
code with js
function WildcardCharacters(str) {
// code goes here
let strArr= str.split(' ')
let specChar = strArr[0]
let charStr = strArr[1].split('')
let arr = specChar.split('')
let letters = /^[A-Za-z]+$/
let i = 0
while(i< arr.length){
if(arr[i] == '+'){
if(!charStr[0].match(letters)) return "false"
charStr = charStr.slice(1,charStr.length)
}
else if(arr[i] == '*'){
let curr = charStr[0]
let j = 1, k = 0
if(arr[i+1] != undefined && arr[i+1] == '{'){
k = arr[i+2]
i = i+4
}else{
k = 3
i++
}
while(j < k){
charStr = charStr.slice(1,charStr.length)
if(charStr[0] != curr) return "false"
j++
}
charStr = charStr.slice(1,charStr.length)
continue
}
i++
}
if(charStr.length != 0) return 'false'
return "true"
}
// keep this function call here
WildcardCharacters("+++++* abcdemmmmmm");
code by Smakar20 (user github)
code with swift
extension String {
func isAlphanumeric() -> Bool {
return self.rangeOfCharacter(from: CharacterSet.alphanumerics.inverted) == nil && self != ""
}
func isAlphanumeric(ignoreDiacritics: Bool = false) -> Bool {
if ignoreDiacritics {
return self.range(of: "[^a-zA-Z0-9]", options: .regularExpression) == nil && self != ""
}
else {
return self.isAlphanumeric()
}
}
}
func wild (str: String) -> String
{
let strArr = str.split(separator: " ")
let specChar = strArr[0]
var charStr = Array(strArr[1])
let arr = Array(specChar)
var i = 0
while (i<arr.count)
{
if arr[i] == "+"
{
if !String(charStr[0]).isAlphanumeric()
{
return "false"
}
charStr = Array(charStr[0...charStr.count])
}
else if arr[i] == "*"
{
let curr = charStr[0]
var j = 0
var k = 0
if String(arr[i+1]).isAlphanumeric() != true && arr[i+1] == "{"
{
k = Int(String(arr[i+2]))!
i = i+4
}
else
{
k = 3
i += 1
}
while (j<k)
{
charStr = Array(charStr[1...charStr.count])
if charStr[0] != curr
{
return "false"
}
j += 1
}
charStr = Array(charStr[1...charStr.count])
continue
}
i += 1
}
if charStr.count != 0
{
return "false"
}
return "true"
}
wild(str: "+++++* abcdemmmmmm")
if str is "++*{5} gheeeee" then the second string in this case does match the pattern, so your program should return the string true. If the second string does not match the pattern your program should return the string false.
heeelp

Convert camel case to human readable string?

Is there a reg exp or function that will convert camel case, css and underscore to human readable format? It does not need to support non-humans at this time. Sorry aliens. :(
Examples:
helloWorld -> "Hello World"
hello-world -> "Hello World"
hello_world -> "Hello World"

Split by non-words; capitalize; join:
function toCapitalizedWords(name) {
var words = name.match(/[A-Za-z][a-z]*/g) || [];
return words.map(capitalize).join(" ");
}
function capitalize(word) {
return word.charAt(0).toUpperCase() + word.substring(1);
}

Extract all words with a regular expression. Capitalize them. Then, join them with spaces.
Example regexp:
/^[a-z]+|[A-Z][a-z]*/g
/ ^[a-z]+ // 1 or more lowercase letters at the beginning of the string
| // OR
[A-Z][a-z]* // a capital letter followed by zero or more lowercase letters
/g // global, match all instances
Example function:
var camelCaseToWords = function(str){
return str.match(/^[a-z]+|[A-Z][a-z]*/g).map(function(x){
return x[0].toUpperCase() + x.substr(1).toLowerCase();
}).join(' ');
};
camelCaseToWords('camelCaseString');
// Camel Case String
camelCaseToWords('thisIsATest');
// This Is A Test

Here is the ActionScript version based on the idea from Ricks C example code. For JavaScript version remove the strong typing. For example, change var value:String to var value. Basically remove any declaration that starts with a semicolon, :String, :int, etc.
/**
* Changes camel case to a human readable format. So helloWorld, hello-world and hello_world becomes "Hello World".
* */
public static function prettifyCamelCase(value:String=""):String {
var output:String = "";
var len:int = value.length;
var char:String;
for (var i:int;i<len;i++) {
char = value.charAt(i);
if (i==0) {
output += char.toUpperCase();
}
else if (char !== char.toLowerCase() && char === char.toUpperCase()) {
output += " " + char;
}
else if (char == "-" || char == "_") {
output += " ";
}
else {
output += char;
}
}
return output;
}
JavaScript version:
/**
* Changes camel case to a human readable format. So helloWorld, hello-world and hello_world becomes "Hello World".
* */
function prettifyCamelCase(str) {
var output = "";
var len = str.length;
var char;
for (var i=0 ; i<len ; i++) {
char = str.charAt(i);
if (i==0) {
output += char.toUpperCase();
}
else if (char !== char.toLowerCase() && char === char.toUpperCase()) {
output += " " + char;
}
else if (char == "-" || char == "_") {
output += " ";
}
else {
output += char;
}
}
return output;
}

You can use a replacement function for String.replace, e.g.
function capitalize(s) {
return s[0].toUpperCase() + s.slice(1);
}
function replacer1(match, p1, p2, p3, offset, s) {
return p1 + capitalize(p2) + ' ' + p3;
}
var s1 = "helloWorld";
var r1 = s1.replace(/(^|[^a-z])([a-z]+)([A-Z])/, replacer1);
console.log(r1);
hello-world and hello_world work similar.
See JSFiddle

I don't know if there is already a built in method to do this but you could loop through the string and every time you see a character that you want to split on do so.
In your case something like:
my_str = 'helloWorld';
returnString = '';
for(var i = 0; i < my_str.length; i++) {
if(i == 0) {
returnString += (my_str[i] + 32); // capitalize the first character
}
else if(my_str[i] > 'A' || my_str[i] < 'Z') {
returnString += ' ' + my_str[i]; // add a space
}
else if(my_str[i] == '-' || my_str[i] == '_') {
returnString += ' ';
}
else {
returnString += my_string[i];
}
}
return returnString;
Edit:
After the numerous comments I have come to realize that I put up some broken code :P
Here is a tested version of it:
my_str = 'helloWorld';
function readable(str) {
// and this was a mistake about javascript/actionscript being able to capitalize
// by adding 32
returnString = str[0].toUpperCase();
for(var i = 1; i < str.length; i++) {
// my mistakes here were that it needs to be between BOTH 'A' and 'Z' inclusive
if(str[i] >= 'A' && str[i] <= 'Z') {
returnString += ' ' + str[i];
}
else if(str[i] == '-' || str[i] == '_') {
returnString += ' ';
}
else {
returnString += str[i];
}
}
return returnString;
}

Non elegant one liner using regex replaces with functions.
replace 1 - upper case first letter and remove _-
replace 2 - add space between lower case letters and upper case letters
var titleCase = s => s
.replace(/(^|[_-])([a-z])/g, (a, b, c) => c.toUpperCase())
.replace(/([a-z])([A-Z])/g, (a, b, c) => `${b} ${c}`);
console.log(titleCase("helloWorld"));
console.log(titleCase("hello-world"));
console.log(titleCase("hello_world"));

If using a library is an option, Lodash's startCase or lowerCase might be useful:
https://lodash.com/docs/#startCase
https://lodash.com/docs/#lowerCase

const result = _.chain("hello-world")
.snakeCase()
.split("_")
.map(w => _.capitalize(w))
.join(" ")
.value()
console.log(result)
<script src="
https://cdn.jsdelivr.net/npm/lodash#4.17.21/lodash.min.js
"></script>

Constructing Regular Expressions to match numeric ranges

I'm looking for a way to construct regular expressions to match numeric inputs specified by a given integer range, ie. if I pass in a range of 1,3-4 then a regex would be returned matching just 1, 3 and 4.
I wrote the following method to try and do this:
function generateRegex(values) {
if (values == "*") {
return new RegExp("^[0-9]+$");
} else {
return new RegExp("^[" + values + "]+$");
}
}
I'm having issues however as sometimes I need to match double digits, such as "8-16", and I also need to ensure that if I am passed a single digit value, such as "1", that the generated regex matches only 1, and not say 11.
I really would like this to remain a pretty small snippet of code, but am not sure enough about regexs to know how to do this. Would be massively grateful for any help!
EDIT: I realise I wasn't clear, with my original paragraph, so have edited it. I realise the regex's that I originally generated do not work at all

Regexes don't know anything about numbers, only digits. So [8-16] is invalid because you say match between 8 and 1 (instead of 1 and 8 e.g.) plus the digit 6.
If you want to match numbers, you have to consider them lexically. For example, to match numbers between 1 and 30, you have to write something like (other regexes exist):
/^(30|[1-2]\d|[1-9])$/

I was sure it was 4-8 hours :-) In the end (and in its uselessness) it was a good exercise in composing Regexes. You are free to try it. If we exclude one use of continue and the use of the Array constructor, it's fully jsLint ok.
var BuildRegex = function(matches) {
"use strict";
var splits = matches.split(','),
res = '^(',
i, subSplit, min, max, temp, tempMin;
if (splits.length === 0) {
return new RegExp('^()$');
}
for (i = 0; i < splits.length; i += 1) {
if (splits[i] === '*') {
return new RegExp('^([0-9]+)$');
}
subSplit = splits[i].split('-');
min = BuildRegex.Trim(subSplit[0], '0');
if (min === '') {
return null;
}
if (subSplit.length === 1) {
res += min;
res += '|';
continue;
} else if (subSplit.length > 2) {
return null;
}
max = BuildRegex.Trim(subSplit[1], '0');
if (max === '') {
return null;
}
if (min.length > max.length) {
return null;
}
// For 2-998 we first produce 2-9, then 10-99
temp = BuildRegex.DifferentLength(res, min, max);
tempMin = temp.min;
if (tempMin === null) {
return null;
}
res = temp.res;
// Then here 100-998
res = BuildRegex.SameLength(res, tempMin, max);
}
res = res.substr(0, res.length - 1);
res += ')$';
return new RegExp(res);
};
BuildRegex.Repeat = function(ch, n) {
"use strict";
return new Array(n + 1).join(ch);
};
BuildRegex.Trim = function(str, ch) {
"use strict";
var i = 0;
while (i < str.length && str[i] === ch) {
i += 1;
}
return str.substr(i);
};
BuildRegex.IsOnlyDigit = function(str, start, digit) {
"use strict";
var i;
for (i = start; i < str.length; i += 1) {
if (str[i] !== digit) {
return false;
}
}
return true;
};
BuildRegex.RangeDigit = function(min, max) {
"use strict";
if (min === max) {
return min;
}
return '[' + min + '-' + max + ']';
};
BuildRegex.DifferentLength = function(res, min, max) {
"use strict";
var tempMin = min,
i, tempMax;
for (i = min.length; i < max.length; i += 1) {
tempMax = BuildRegex.Repeat('9', i);
res = BuildRegex.SameLength(res, tempMin, tempMax);
tempMin = '1' + BuildRegex.Repeat('0', i);
}
if (tempMin > tempMax) {
return null;
}
return {
min: tempMin,
res: res
};
};
BuildRegex.SameLength = function(res, min, max) {
"use strict";
var commonPart;
// 100-100
if (min === max) {
res += min;
res += '|';
return res;
}
for (commonPart = 0; commonPart < min.length; commonPart += 1) {
if (min[commonPart] !== max[commonPart]) {
break;
}
}
res = BuildRegex.RecursivelyAddRange(res, min.substr(0, commonPart), min.substr(commonPart), max.substr(commonPart));
return res;
};
BuildRegex.RecursivelyAddRange = function(res, prefix, min, max) {
"use strict";
var only0Min, only9Max, i, middleMin, middleMax;
if (min.length === 1) {
res += prefix;
res += BuildRegex.RangeDigit(min[0], max[0]);
res += '|';
return res;
}
// Check if
only0Min = BuildRegex.IsOnlyDigit(min, 1, '0');
only9Max = BuildRegex.IsOnlyDigit(max, 1, '9');
if (only0Min && only9Max) {
res += prefix;
res += BuildRegex.RangeDigit(min[0], max[0]);
for (i = 1; i < min.length; i += 1) {
res += '[0-9]';
}
res += '|';
return res;
}
middleMin = min;
if (!only0Min) {
res = BuildRegex.RecursivelyAddRange(res, prefix + min[0], min.substr(1), BuildRegex.Repeat('9', min.length - 1));
if (min[0] !== '9') {
middleMin = String.fromCharCode(min.charCodeAt(0) + 1) + BuildRegex.Repeat('0', min.length - 1);
} else {
middleMin = null;
}
}
middleMax = max;
if (!only9Max) {
if (max[0] !== '0') {
middleMax = String.fromCharCode(max.charCodeAt(0) - 1) + BuildRegex.Repeat('9', max.length - 1);
} else {
middleMax = null;
}
}
if (middleMin !== null && middleMax !== null && middleMin[0] <= middleMax[0]) {
res = BuildRegex.RecursivelyAddRange(res, prefix + BuildRegex.RangeDigit(middleMin[0], middleMax[0]), middleMin.substr(1), middleMax.substr(1));
}
if (!only9Max) {
res = BuildRegex.RecursivelyAddRange(res, prefix + max[0], BuildRegex.Repeat('0', max.length - 1), max.substr(1));
}
return res;
};
// ----------------------------------------------------------
var printRegex = function(p) {
"use strict";
document.write(p + ': ' + BuildRegex(p) + '<br>');
};
printRegex('*');
printRegex('1');
printRegex('1,*');
printRegex('1,2,3,4');
printRegex('1,11-88');
printRegex('1,11-88,90-101');
printRegex('1-11111');
printRegex('75-11119');
Test here http://jsfiddle.net/dnqYV/
The C# version is here http://ideone.com/3aEt3E

I'm not sure there is a (sane) way to test integer ranges with RegExp. I believe you're fixated on RegExp, where there are much simpler (more flexible) approaches. Take a look at IntRangeTest().
var range = new IntRangeTest('0,10-20');
console.log(
"0,10-20",
range.test("") == false,
range.test("-5") == false,
range.test("0") == true,
range.test("5") == false,
range.test("11") == true,
range.test("123.23") == false
);
If you feel like it, you can easily add this to Number.prototype. You could also quite easily make this an extension to RegExp, if that's what you're worried about.

Ok so it seems that there are 4 main cases that I need to address:
Single digits, ie 1, would simply generate the regex /^1$/
Multiple digits, ie 12, would require the regex /^12&/
Single digit ranges, ie 3-6, would generate the regex /^[3-6]$/
And finally, multiple digit ranges work in a similar method to multiple digits but with a range, ie 11-14 would become /^1[1-4]$/. These would need to be split into multiple regexes if they span over multiple start digits, Ie 23-31 would become /^2[3-9]|3[0-1]$/
Therefore, all I need to do is identify each of these cases and create a compound regex using | like xanatos suggested. Ie, to match all of the above criteria would generate a regex like:
/^( 1 | 12 | [3-6] | 1[1-4] | 2[3-9]|3[0-1] )$/
Do other agree this seems like a decent way to progress?

How to get the next letter of the alphabet in Javascript?

I am build an autocomplete that searches off of a CouchDB View.
I need to be able to take the final character of the input string, and replace the last character with the next letter of the english alphabet. (No need for i18n here)
For Example:
Input String = "b"
startkey = "b"
endkey = "c"
OR
Input String = "foo"
startkey = "foo"
endkey = "fop"
(in case you're wondering, I'm making sure to include the option inclusive_end=false so that this extra character doesn't taint my resultset)
The Question
Is there a function natively in Javascript that can just get the next letter of the alphabet?
Or will I just need to suck it up and do my own fancy function with a base string like "abc...xyz" and indexOf()?

my_string.substring(0, my_string.length - 1)
+ String.fromCharCode(my_string.charCodeAt(my_string.length - 1) + 1)

// This will return A for Z and a for z.
function nextLetter(s){
return s.replace(/([a-zA-Z])[^a-zA-Z]*$/, function(a){
var c= a.charCodeAt(0);
switch(c){
case 90: return 'A';
case 122: return 'a';
default: return String.fromCharCode(++c);
}
});
}

A more comprehensive solution, which gets the next letter according to how MS Excel numbers it's columns... A B C ... Y Z AA AB ... AZ BA ... ZZ AAA
This works with small letters, but you can easily extend it for caps too.
getNextKey = function(key) {
if (key === 'Z' || key === 'z') {
return String.fromCharCode(key.charCodeAt() - 25) + String.fromCharCode(key.charCodeAt() - 25); // AA or aa
} else {
var lastChar = key.slice(-1);
var sub = key.slice(0, -1);
if (lastChar === 'Z' || lastChar === 'z') {
// If a string of length > 1 ends in Z/z,
// increment the string (excluding the last Z/z) recursively,
// and append A/a (depending on casing) to it
return getNextKey(sub) + String.fromCharCode(lastChar.charCodeAt() - 25);
} else {
// (take till last char) append with (increment last char)
return sub + String.fromCharCode(lastChar.charCodeAt() + 1);
}
}
return key;
};

Here is a function that does the same thing (except for upper case only, but that's easy to change) but uses slice only once and is iterative rather than recursive. In a quick benchmark, it's about 4 times faster (which is only relevant if you make really heavy use of it!).
function nextString(str) {
if (! str)
return 'A' // return 'A' if str is empty or null
let tail = ''
let i = str.length -1
let char = str[i]
// find the index of the first character from the right that is not a 'Z'
while (char === 'Z' && i > 0) {
i--
char = str[i]
tail = 'A' + tail // tail contains a string of 'A'
}
if (char === 'Z') // the string was made only of 'Z'
return 'AA' + tail
// increment the character that was not a 'Z'
return str.slice(0, i) + String.fromCharCode(char.charCodeAt(0) + 1) + tail
}

Just to explain the main part of the code that Bipul Yadav wrote (can't comment yet due to lack of reps). Without considering the loop, and just taking the char "a" as an example:
"a".charCodeAt(0) = 97...hence "a".charCodeAt(0) + 1 = 98 and String.fromCharCode(98) = "b"...so the following function for any letter will return the next letter in the alphabet:
function nextLetterInAlphabet(letter) {
if (letter == "z") {
return "a";
} else if (letter == "Z") {
return "A";
} else {
return String.fromCharCode(letter.charCodeAt(0) + 1);
}
}

var input = "Hello";
var result = ""
for(var i=0;i<input.length;i++)
{
var curr = String.fromCharCode(input.charCodeAt(i)+1);
result = result +curr;
}
console.log(result);

I understand the original question was about moving the last letter of the string forward to the next letter. But I came to this question more interested personally in changing all the letters in the string, then being able to undo that. So I took the code written by Bipul Yadav and I added some more code. The below code takes a series of letters, increments each of them to the next letter maintaining case (and enables Zz to become Aa), then rolls them back to the previous letter (and allows Aa to go back to Zz).
var inputValue = "AaZzHello";
console.log( "starting value=[" + inputValue + "]" );
var resultFromIncrementing = ""
for( var i = 0; i < inputValue.length; i++ ) {
var curr = String.fromCharCode( inputValue.charCodeAt(i) + 1 );
if( curr == "[" ) curr = "A";
if( curr == "{" ) curr = "a";
resultFromIncrementing = resultFromIncrementing + curr;
}
console.log( "resultFromIncrementing=[" + resultFromIncrementing + "]" );
inputValue = resultFromIncrementing;
var resultFromDecrementing = "";
for( var i2 = 0; i2 < inputValue.length; i2++ ) {
var curr2 = String.fromCharCode( inputValue.charCodeAt(i2) - 1 );
if( curr2 == "#" ) curr2 = "Z";
if( curr2 == "`" ) curr2 = "z";
resultFromDecrementing = resultFromDecrementing + curr2;
}
console.log( "resultFromDecrementing=[" + resultFromDecrementing + "]" );
The output of this is:
starting value=[AaZzHello]
resultFromIncrementing=[BbAaIfmmp]
resultFromDecrementing=[AaZzHello]