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javascript indexof regex A-Za-z0-9 always returns false

I have created a JS fiddle https://jsfiddle.net/95r110s9/#&togetherjs=Emdw6ORNpc
HTML
<input id="landlordstreetaddress2" class="landlordinputs" onfocusout="validateinputentries()" />
JS
validateinputentries(){
landlordstreetaddress2 = document.getElementById('landlordstreetaddress2').value;
goodcharacters = "/^[a-zA-Z0-9#.,;:'\s]+$/gi";
for (var i = 0; i < landlordstreetaddress2.length; i++){
if (goodcharacters.indexOf(landlordstreetaddress2.charAt(i)) != -1){
console.log('Character is valid');
}
}
}
Its pulling the value from an input and running an indexOf regex expression with A-Z a-z and 0-9 with a few additional characters as well.
The problem is that it works with the entry of BCDEFG...etc and 12345...etc, but when I type "A" or "Z" or "0" or "1", it returns incorrectly.
I need it to return the same with 0123456789, ABCDEF...XYZ and abcdef...xyz
I should point out that the below does work as intended:
var badcharacters = "*|,\":<>[]`\';#?=+/\\";
badcharacter = false;
//firstname
for (var i = 0; i < landlordfirstname.value.length; i++){
if (badcharacters.indexOf(landlordfirstname.value.charAt(i)) != -1){
badcharacter = true;
break;
}
if(landlordfirstname.value.charAt(0) == " "){
badcharacter = true;
break;
}
}

String.prototype.indexOf()
The indexOf() method returns the index within the calling String object of the first occurrence of the specified value, starting the search at fromIndex. Returns -1 if the value is not found.
So, you're trying to search this value "/^[a-zA-Z0-9#.,;:'\s]+$/gi" which "never" will be found in the entered string.
You actually want to test that regexp against the entered value.
/^[a-zA-Z0-9#.,;:'\s]+$/gi.test(landlordstreetaddress2)
function validateinputentries() {
var landlordstreetaddress2 = document.getElementById('landlordstreetaddress2').value;
if (/^[a-zA-Z0-9#.,;:'\s]+$/gi.test(landlordstreetaddress2)) {
console.log('Characters are valid');
} else {
console.log('Characters are invalid');
}
}
<input id="landlordstreetaddress2" class="landlordinputs" onfocusout="validateinputentries()" />

You're trying to combine two different methods of testing a string -- one way is with a regex; the other way is by checking each character against a list of allowed characters. What you've wound up with is checking each character against a list of what would have been a regex, if you hadn't declared it as a string.
Those methods conflict with each other; you need to pick one or the other.
Check each character:
This is closest to what you were attempting. You can't use character ranges here (like a-zA-Z) as you would in a regex; you have to spell out each allowed character individually:
var validateinputentries = function() {
var address = document.getElementById('landlordstreetaddress2').value;
var goodcharacters = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789#.,;:' ";
var badcharactersfound = false;
for (var i = 0; i < address.length; i++) {
if (goodcharacters.indexOf(address.charAt(i)) == -1) {
badcharactersfound = true;
console.log("not allowed: ", address.charAt(i));
}
}
if (badcharactersfound) {
// Show validation error here
}
}
<input id="landlordstreetaddress2" class="landlordinputs" onfocusout="validateinputentries()" />
Regular Expressions
The regex version is much simpler, because the regular expression is doing most of the work. You don't need to step through the string, just test the whole string against the regex and see what comes out. In this case you're looking to see if the input contains any characters that aren't allowed, so you want to use the character exception rule: [^abc] will match any character that is not a, b, or c. You don't want to anchor the match to the beginning or the end of the string, as you were doing with the initial ^ and the trailing $; and you can leave out the + because you don't care if there are sequential bad characters, you just care if they exist at all.
var validateinputentries = function() {
var address = document.getElementById('landlordstreetaddress2').value;
var regex = new RegExp("[^a-zA-Z0-9#.,;:'\\s]","g")
var badcharactersfound = address.match(regex);
// or the above two lines could also have been written like this:
// var bad = address.match(/[^a-zA-Z0-9#.,;:'\s]/g)
// In either case the "g" operator could be omitted; then it would only return the first bad character.
if (badcharactersfound) {
console.log("Not allowed: ", badcharactersfound);
}
}
<input id="landlordstreetaddress2" class="landlordinputs" onfocusout="validateinputentries()" />

Number and Alphabet Separation in a String

I have String variables in Javascript like :
var houseNo = "62A"; var cabinNo = "5BC";
I need to fetch out the Integers and the Alphabets separate from the string where number of occurrences of each can be any number of times.
Need help to do it in the best possible way, be it through lodash or any other prototype method.
Referred to this but left in vain as don't want it through RegEx.

something like :
function decompose(string){
for(var i=0;i<string.length;i++){
if(parseInt(string[i])){ // if the char is a number?
// do whatever you want
}else{
// it's a character
}
}
}
The parseInt() function return the number of a giver char. If it is not a number, it returns NaN (not a number). if(parsInt(char)) return false if it's a char, true if it's a number

Try this:
var houseNo = "62A";
foreach(char a in houseNo)
{
if(a > 48 && a < 57)
{
/*it's a number*/
}
else
{
/*it's a letter*/
}
}
You can apply it on every string and determine what you want to do with each number or letter.

var test = "a3434dasds3432s2"
var myString = test.split("").filter(function(v) {return isNaN(v)}).join("")
var myNumber = parseInt(test.split("").filter(function(v) {return !isNaN(v)}).join(""))
best to use regex really though.

Javascript IndexOf with integers in string not working

Can anyone tell me why does this not work for integers but works for characters? I really hate reg expressions since they are cryptic but will if I have too. Also I want to include the "-()" as well in the valid characters.
String.prototype.Contains = function (str) {
return this.indexOf(str) != -1;
};
var validChars = '0123456789';
var str = $("#textbox1").val().toString();
if (str.Contains(validChars)) {
alert("found");
} else {
alert("not found");
}

Review
String.prototype.Contains = function (str) {
return this.indexOf(str) != -1;
};
This String "method" returns true if str is contained within itself, e.g. 'hello world'.indexOf('world') != -1would returntrue`.
var validChars = '0123456789';
var str = $("#textbox1").val().toString();
The value of $('#textbox1').val() is already a string, so the .toString() isn't necessary here.
if (str.Contains(validChars)) {
alert("found");
} else {
alert("not found");
}
This is where it goes wrong; effectively, this executes '1234'.indexOf('0123456789') != -1; it will almost always return false unless you have a huge number like 10123456789.
What you could have done is test each character in str whether they're contained inside '0123456789', e.g. '0123456789'.indexOf(c) != -1 where c is a character in str. It can be done a lot easier though.
Solution
I know you don't like regular expressions, but they're pretty useful in these cases:
if ($("#textbox1").val().match(/^[0-9()]+$/)) {
alert("valid");
} else {
alert("not valid");
}
Explanation
[0-9()] is a character class, comprising the range 0-9 which is short for 0123456789 and the parentheses ().
[0-9()]+ matches at least one character that matches the above character class.
^[0-9()]+$ matches strings for which ALL characters match the character class; ^ and $ match the beginning and end of the string, respectively.
In the end, the whole expression is padded on both sides with /, which is the regular expression delimiter. It's short for new RegExp('^[0-9()]+$').

Assuming you are looking for a function to validate your input, considering a validChars parameter:
String.prototype.validate = function (validChars) {
var mychar;
for(var i=0; i < this.length; i++) {
if(validChars.indexOf(this[i]) == -1) { // Loop through all characters of your string.
return false; // Return false if the current character is not found in 'validChars' string.
}
}
return true;
};
var validChars = '0123456789';
var str = $("#textbox1").val().toString();
if (str.validate(validChars)) {
alert("Only valid characters were found! String validates!");
} else {
alert("Invalid Char found! String doesn't validate.");
}
However, This is quite a load of code for a string validation. I'd recommend looking into regexes, instead. (Jack's got a nice answer up here)

You are passing the entire list of validChars to indexOf(). You need to loop through the characters and check them one-by-one.
Demo
String.prototype.Contains = function (str) {
var mychar;
for(var i=0; i<str.length; i++)
{
mychar = this.substr(i, 1);
if(str.indexOf(mychar) == -1)
{
return false;
}
}
return this.length > 0;
};
To use this on integers, you can convert the integer to a string with String(), like this:
var myint = 33; // define integer
var strTest = String(myint); // convert to string
console.log(strTest.Contains("0123456789")); // validate against chars

I'm only guessing, but it looks like you are trying to check a phone number. One of the simple ways to change your function is to check string value with RegExp.
String.prototype.Contains = function(str) {
var reg = new RegExp("^[" + str +"]+$");
return reg.test(this);
};
But it does not check the sequence of symbols in string.
Checking phone number is more complicated, so RegExp is a good way to do this (even if you do not like it). It can look like:
String.prototype.ContainsPhone = function() {
var reg = new RegExp("^\\([0-9]{3}\\)[0-9]{3}-[0-9]{2}-[0-9]{2}$");
return reg.test(this);
};
This variant will check phones like "(123)456-78-90". It not only checks for a list of characters, but also checks their sequence in string.

Thank you all for your answers! Looks like I'll use regular expressions. I've tried all those solutions but really wanted to be able to pass in a string of validChars but instead I'll pass in a regex..
This works for words, letters, but not integers. I wanted to know why it doesn't work for integers. I wanted to be able to mimic the FilteredTextBoxExtender from the ajax control toolkit in MVC by using a custom Attribute on a textBox

Remove all special characters except space from a string using JavaScript

I want to remove all special characters except space from a string using JavaScript.
For example,
abc's test#s
should output as
abcs tests.

You should use the string replace function, with a single regex.
Assuming by special characters, you mean anything that's not letter, here is a solution:
const str = "abc's test#s";
console.log(str.replace(/[^a-zA-Z ]/g, ""));

You can do it specifying the characters you want to remove:
string = string.replace(/[&\/\\#,+()$~%.'":*?<>{}]/g, '');
Alternatively, to change all characters except numbers and letters, try:
string = string.replace(/[^a-zA-Z0-9]/g, '');

The first solution does not work for any UTF-8 alphabet. (It will cut text such as Привіт). I have managed to create a function which does not use RegExp and use good UTF-8 support in the JavaScript engine. The idea is simple if a symbol is equal in uppercase and lowercase it is a special character. The only exception is made for whitespace.
function removeSpecials(str) {
var lower = str.toLowerCase();
var upper = str.toUpperCase();
var res = "";
for(var i=0; i<lower.length; ++i) {
if(lower[i] != upper[i] || lower[i].trim() === '')
res += str[i];
}
return res;
}
Update: Please note, that this solution works only for languages where there are small and capital letters. In languages like Chinese, this won't work.
Update 2: I came to the original solution when I was working on a fuzzy search. If you also trying to remove special characters to implement search functionality, there is a better approach. Use any transliteration library which will produce you string only from Latin characters and then the simple Regexp will do all magic of removing special characters. (This will work for Chinese also and you also will receive side benefits by making Tromsø == Tromso).

search all not (word characters || space):
str.replace(/[^\w ]/, '')

I don't know JavaScript, but isn't it possible using regex?
Something like [^\w\d\s] will match anything but digits, characters and whitespaces. It would be just a question to find the syntax in JavaScript.

I tried Seagul's very creative solution, but found it treated numbers also as special characters, which did not suit my needs. So here is my (failsafe) tweak of Seagul's solution...
//return true if char is a number
function isNumber (text) {
if(text) {
var reg = new RegExp('[0-9]+$');
return reg.test(text);
}
return false;
}
function removeSpecial (text) {
if(text) {
var lower = text.toLowerCase();
var upper = text.toUpperCase();
var result = "";
for(var i=0; i<lower.length; ++i) {
if(isNumber(text[i]) || (lower[i] != upper[i]) || (lower[i].trim() === '')) {
result += text[i];
}
}
return result;
}
return '';
}

const str = "abc's#thy#^g&test#s";
console.log(str.replace(/[^a-zA-Z ]/g, ""));

Try to use this one
var result= stringToReplace.replace(/[^\w\s]/g, '')
[^] is for negation, \w for [a-zA-Z0-9_] word characters and \s for space,
/[]/g for global

With regular expression
let string = "!#This tool removes $special *characters* /other/ than! digits, characters and spaces!!!$";
var NewString= string.replace(/[^\w\s]/gi, '');
console.log(NewString);
Result //This tool removes special characters other than digits characters and spaces
Live Example : https://helpseotools.com/text-tools/remove-special-characters

dot (.) may not be considered special. I have added an OR condition to Mozfet's & Seagull's answer:
function isNumber (text) {
reg = new RegExp('[0-9]+$');
if(text) {
return reg.test(text);
}
return false;
}
function removeSpecial (text) {
if(text) {
var lower = text.toLowerCase();
var upper = text.toUpperCase();
var result = "";
for(var i=0; i<lower.length; ++i) {
if(isNumber(text[i]) || (lower[i] != upper[i]) || (lower[i].trim() === '') || (lower[i].trim() === '.')) {
result += text[i];
}
}
return result;
}
return '';
}

Try this:
const strippedString = htmlString.replace(/(<([^>]+)>)/gi, "");
console.log(strippedString);

const input = `#if_1 $(PR_CONTRACT_END_DATE) == '23-09-2019' #
Test27919<alerts#imimobile.com> #elseif_1 $(PR_CONTRACT_START_DATE) == '20-09-2019' #
Sender539<rama.sns#gmail.com> #elseif_1 $(PR_ACCOUNT_ID) == '1234' #
AdestraSID<hello#imimobile.co> #else_1#Test27919<alerts#imimobile.com>#endif_1#`;
const replaceString = input.split('$(').join('->').split(')').join('<-');
console.log(replaceString.match(/(?<=->).*?(?=<-)/g));

Whose special characters you want to remove from a string, prepare a list of them and then user javascript replace function to remove all special characters.
var str = 'abc'de#;:sfjkewr47239847duifyh';
alert(str.replace("'","").replace("#","").replace(";","").replace(":",""));
or you can run loop for a whole string and compare single single character with the ASCII code and regenerate a new string.

Count number of words in string using JavaScript

I am trying to count the number of words in a given string using the following code:
var t = document.getElementById('MSO_ContentTable').textContent;
if (t == undefined) {
var total = document.getElementById('MSO_ContentTable').innerText;
} else {
var total = document.getElementById('MSO_ContentTable').textContent;
}
countTotal = cword(total);
function cword(w) {
var count = 0;
var words = w.split(" ");
for (i = 0; i < words.length; i++) {
// inner loop -- do the count
if (words[i] != "") {
count += 1;
}
}
return (count);
}
In that code I am getting data from a div tag and sending it to the cword() function for counting. Though the return value is different in IE and Firefox. Is there any change required in the regular expression? One thing that I show that both browser send same string there is a problem inside the cword() function.

[edit 2022, based on comment] Nowadays, one would not extend the native prototype this way. A way to extend the native protype without the danger of naming conflicts is to use the es20xx symbol. Here is an example of a wordcounter using that.
Old answer: you can use split and add a wordcounter to the String prototype:
if (!String.prototype.countWords) {
String.prototype.countWords = function() {
return this.length && this.split(/\s+\b/).length || 0;
};
}
console.log(`'this string has five words'.countWords() => ${
'this string has five words'.countWords()}`);
console.log(`'this string has five words ... and counting'.countWords() => ${
'this string has five words ... and counting'.countWords()}`);
console.log(`''.countWords() => ${''.countWords()}`);

I would prefer a RegEx only solution:
var str = "your long string with many words.";
var wordCount = str.match(/(\w+)/g).length;
alert(wordCount); //6
The regex is
\w+ between one and unlimited word characters
/g greedy - don't stop after the first match
The brackets create a group around every match. So the length of all matched groups should match the word count.

This is the best solution I've found:
function wordCount(str) {
var m = str.match(/[^\s]+/g)
return m ? m.length : 0;
}
This inverts whitespace selection, which is better than \w+ because it only matches the latin alphabet and _ (see http://www.ecma-international.org/ecma-262/5.1/#sec-15.10.2.6)
If you're not careful with whitespace matching you'll count empty strings, strings with leading and trailing whitespace, and all whitespace strings as matches while this solution handles strings like ' ', ' a\t\t!\r\n#$%() d ' correctly (if you define 'correct' as 0 and 4).

You can make a clever use of the replace() method although you are not replacing anything.
var str = "the very long text you have...";
var counter = 0;
// lets loop through the string and count the words
str.replace(/(\b+)/g,function (a) {
// for each word found increase the counter value by 1
counter++;
})
alert(counter);
the regex can be improved to exclude html tags for example

//Count words in a string or what appears as words :-)
function countWordsString(string){
var counter = 1;
// Change multiple spaces for one space
string=string.replace(/[\s]+/gim, ' ');
// Lets loop through the string and count the words
string.replace(/(\s+)/g, function (a) {
// For each word found increase the counter value by 1
counter++;
});
return counter;
}
var numberWords = countWordsString(string);