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Wix Form Validation

First off, I am a complete noob. I have just recently started getting good with python and have close to zero knowledge regarding javascript and corvid for wix. I am looking through documentation but I can't figure out how to do the following: When a user fills out a field, let's say a phone number field, how do you validate whether that field has any alphabetical characters or "+" or "-" etc or not?
I am trying to do something like this:
$w.onReady(function () {
$w("#input1").onCustomValidation( (value, reject) => {
if( input1 is anything but an integer ) {
reject("Only Numbers. No '-', '.', '()', '+', or any alphabetical characters");
}
} );
});
I feel like I am close but have no idea. Any help is appreciated

To answer your question:
Includes Function
You would want to use the .includes() function native to javascript which returns a boolean of if it is or isn't included. If you need the string index of each occurrence you would need to use a for loop as shown below.
var string = "+54 2453-2534-242";
for(int i = 0; i < string.length(); i++){
if(string.substr(i,1) === "-"){
console.log(i);
}
}
Includes Function
Replace Function
Below is an example of how to use this function.
If you wish to remove these or one of these characters you can use the .replace() function or the .split() and .join() functions.
const string = "+53 573-2566-242";
const character_1 = "+";
const character_2= "-";
// Includes
console.log(string + ".includes(+):", string.includes(character_1));
console.log(string + ".includes(-):",string.includes(character_2));
// Replace Singular
var new_string = string.replace(character_1,'');
console.log(new_string);
// Replace Multiple
var new_string2 = string.split(character_2).join("");
console.log(new_string2);
Replace Funtion
If you are still stuck, please feel free to comment
Edit
To check if there are any alpha-numeric characters in a string. You can simply use the .replace() function and compare it. As seen below:
var string = "abc546"; // Your Phone Number Input
var string_converted = string.replace(/\D/g,'');
if(string !== string_converted){
console.log("Contains Characters that are not of type NUMBER!");
console.log(`${string} vs ${string_converted}`);
}
In your case, you could use the code below:
$w.onReady(function () {
$w("#input1").onCustomValidation( (value, reject) => {
// Assuing *value* is your input
var converted = value.replace(/\D/g,'');
if(value !== converted){
// Regective Statment Here..
}else{
// All good! (No Alph-numeric Characters)
}
});
});

Algorithm - Search and Replace a string

I am doing a algorithm in freeCodeCamp.(https://www.freecodecamp.org/learn/javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/search-and-replace)
The task is as below:
Perform a search and replace on the sentence using the arguments provided and return the new sentence.
First argument is the sentence to perform the search and replace on.
Second argument is the word that you will be replacing (before).
Third argument is what you will be replacing the second argument with (after).
Note:
Preserve the case of the first character in the original word when you are replacing it. For example if you mean to replace the word "Book" with the word "dog", it should be replaced as "Dog"
**
myReplace("Let us get back to more Coding", "Coding", "algorithms") should return "Let us get back to more Algorithms".
myReplace("Let us go to the store", "store", "mall") should return "Let us go to the mall".
**
//if the before is uppercase, the after should be uppercase also
// str = str.replace(before, after);
var regex = /[A-Z]+/g; //check for uppercase
var newStr = "";
console.log(regex.test(before));
if (regex.test(before)) {
//if uppercase, return true, "after" convert to uppercase
after = after.toUpperCase();
newStr = after[0];
for (var i = 1; i < after.length; i++) {
//start at index=1 letter, all convert to
newStr += after[i].toLowerCase();
}
console.log(newStr);
str = str.replace(before, newStr);
} else {
str = str.replace(before, after);
}
// console.log(newStr);
console.log(str);
return str;
}
I think there should be OK for the code, but can anyone help find why the if statement can't work.
Much thanks!

The problem is that you're calling regex.test() multiple times on the same regular expression instance.
[...]
var regex = /[A-Z]+/g; //check for uppercase
var newStr = "";
console.log(regex.test(before));
if (regex.test(before)) {
//if uppercase, return true, "after" convert to uppercase
after = after.toUpperCase();
[...]
If your string is Hello_there, the first regex.test() will return true, because Hello matched. If you call regex.test() again with the same regex instance, it will have advanced in the string, and try to match starting with _there. In this case, it will fail, because _there does not begin with a capital letter between A and Z.
There are a lot of ways to fix this issue. Perhaps the simplest is to store the result of the first call to a variable, and use it everywhere you're calling regex.test():
[...]
var regex = /[A-Z]+/g; //check for uppercase
var newStr = "";
var upper_check = regex.test(before);
console.log(upper_check);
if (upper_check) {
[...]

It seems overkill to use a regex, when you really need to only check the first character. Your regex will find uppercase letters anywhere...
If the assignment is to only change one occurrence, then a regex is not really the right tool here: it does not really help to improve the code nor the efficiency. Just do:
function myReplace(str, before, after) {
if (before[0] === before[0].toUpperCase()) {
after = after[0].toUpperCase() + after.slice(1);
} else {
after = after[0].toLowerCase() + after.slice(1);
}
return str.replace(before, after);
}

function myReplace(str, before, after) {
var upperRegExp = /[A-Z]/g
var lowerRegExp = /[a-z]/g
var afterCapitalCase = after.replace(/^./, after[0].toUpperCase());
if (before[0].match(upperRegExp)) {
return str.replace(before, afterCapitalCase)
} else if (after[0].match(upperRegExp) && before[0].match(lowerRegExp)) {
return str.replace(before, after.toLowerCase());
} else {
return str.replace(before, after)
}
}

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()" />

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.