Develop Reference

Javascript regex doesn't give error message when it should

I try to input the following value "f+11111111111", and my code doesn't display error message. I tried to remove regex2 condition, and then it worked. But how come? It's an AND condition, why does it act like it's an OR operator?
function validatePhone()
{
var phone = document.getElementById("phone").value;
var regex1 =/^[\+\d][\s(\-\d][\s()\-\d]{8,}/g;
var regex2 = /\D*(\d\D*){11}/g;
if (!phone.match(regex1) && !phone.match(regex2)) {
producePrompt("Error", "comment_phone_prompt");
return false;
}
}
function producePrompt(message, promptLocation)
{
document.getElementById(promptLocation).innerHTML = message;
}

Your second regular expression /\D*(\d\D*){11}/g matches the given string f+11111111111 and therefore the whole condition evaluates to false. You can visualise your regular expressions using regexper.com.
I'm not sure what you're trying to do but mind that even this string is matched by the second regex: 'f+1dsadsadasda123131231dsdadai-094i-)#)#(#)_(#_!' Is this what you want?

Second regex matches everything. Single regex is enough.
function validatePhone() {
var phone = document.getElementById("phone").value;
var regex1 = /^(\+\d)?(\(|-)?\d{3}(\)|-)?\d{3}-?\d{4}$/;
//match +1-800-123-45678, (555)123-4567, and so on
if (!regex1.test(phone)){
producePrompt("Error", "comment_phone_prompt");
return false;
}
return true;//both sides should return
}

Javascript check for 1 special character and 2 digits

How to create a javascript validation for a password field which must contain at least one special character and at least two digits ?

Exact regular expression that perfect match to your query is below, it is tested ...
^(?=.*?[0-9].*?[0-9])(?=.*[!##$%])[0-9a-zA-Z!##$%]{8,}$

function check(str){
var temp = str;
if(/^[a-zA-Z0-9- ]*$/.test(str) == false && temp.replace(/[^0-9]/g,"").length>1) return true;
return false;
}

javascript regex does not work for sentence string

I am writing a function which takes string as an argument. Then if the string begins with capital letter then return true otherwise return false. But my current function only works for one word string which I want it to work for both one word and a whole sentence. How can I improve my code to achieve this? Secondly, it should not work when numbers are passed inside sentence. How can I do this?
Here is my code
function takeString (str) {
var regex = /^[A-Za-z]+$/;
if (str.match(regex)) {
if (str.charAt(0) === str.toUpperCase().charAt(0)) {
alert('true');
return true;
} else {
alert('false');
return false;
}
} else {
alert('Only letters please.');
}
}
takeString('This is'); // shows Only letters please which is wrong. this should work
takeString('String); // returns true which right
takeString('string'); // returns false which is right
takeString('This is 12312321'); // shows only letters please which is right bcoz it has digits
takeString('12312312'); // show Only letters please which is right.
​

Spaces aren't letters. You have to add them into your character set:
> 'This is a string'.match(/^[A-Za-z]+$/);
null
> 'This is a string'.match(/^[A-Za-z\s]+$/);
["This is a string"]
\s matches all whitespace, so if you don't want to match tabs, replace \s with a space.
Here's a slightly more condensed version of your code:
function takeString(str) {
return str.match(/^[A-Z][A-Za-z ]*$/);
}

along with the regex advice given by Blender, you'll want to also do the following (in order to satisfy the need to check each word ... assuming words are space or tab separated only:
use the split function to break the string into words ( var mywords = str.split(/\s+/) )
iterate over mywords array returned by split, checking each array element against the regex
return an error if the regex doesnt match
return success if you match every word
takeString (str) {
var mywords = str.split(/\s+/);
for (i = 0; i < mywords.length; i++) {
if (str.match(/^[A-Z][A-Za-z]*$/) != true) {
return false;
}
}
return true;
}
(someone needs to check my js ... )

javascript code to check special characters

I have JavaScript code to check if special characters are in a string. The code works fine in Firefox, but not in Chrome. In Chrome, even if the string does not contain special characters, it says it contains special characters.
var iChars = "~`!#$%^&*+=-[]\\\';,/{}|\":<>?";
for (var i = 0; i < chkfile.value.length; i++)
{
if (iChars.indexOf(chkfile.value.charAt(i)) != -1)
{
alert ("File name has special characters ~`!#$%^&*+=-[]\\\';,/{}|\":<>? \nThese are not allowed\n");
return false;
}
}
Suppose I want to upload a file desktop.zip from any Linux/Windows machine.
The value of chkfile.value is desktop.zip in Firefox, but in Chrome the value of chkfile.value is c://fakepath/desktop.zip. How do I get rid of c://fakepath/ from chkfile.value?

You can test a string using this regular expression:
function isValid(str){
return !/[~`!#$%\^&*+=\-\[\]\\';,/{}|\\":<>\?]/g.test(str);
}

Try This one.
function containsSpecialCharacters(str){
var regex = /[ !##$%^&*()_+\-=\[\]{};':"\\|,.<>\/?]/g;
return regex.test(str);
}

Directly from the w3schools website:
var str = "The best things in life are free";
var patt = new RegExp("e");
var res = patt.test(str);
To combine their example with a regular expression, you could do the following:
function checkUserName() {
var username = document.getElementsByName("username").value;
var pattern = new RegExp(/[~`!#$%\^&*+=\-\[\]\\';,/{}|\\":<>\?]/); //unacceptable chars
if (pattern.test(username)) {
alert("Please only use standard alphanumerics");
return false;
}
return true; //good user input
}

Did you write return true somewhere? You should have written it, otherwise function returns nothing and program may think that it's false, too.
function isValid(str) {
var iChars = "~`!#$%^&*+=-[]\\\';,/{}|\":<>?";
for (var i = 0; i < str.length; i++) {
if (iChars.indexOf(str.charAt(i)) != -1) {
alert ("File name has special characters ~`!#$%^&*+=-[]\\\';,/{}|\":<>? \nThese are not allowed\n");
return false;
}
}
return true;
}
I tried this in my chrome console and it worked well.

You could also do it this way.
specialRegex = /[^A-Z a-z0-9]/
specialRegex.test('test!') // evaluates to true
Because if its not a capital letter, lowercase letter, number, or space, it could only be a special character

If you don't want to include any special character, then try this much simple way for checking special characters using RegExp \W Metacharacter.
var iChars = "~`!#$%^&*+=-[]\\\';,/{}|\":<>?";
if(!(iChars.match(/\W/g)) == "") {
alert ("File name has special characters ~`!#$%^&*+=-[]\\\';,/{}|\":<>? \nThese are not allowed\n");
return false;
}

Checking form for special characters in JavaScript.

I am trying to make a function that checks for special characters such as !##$%^&*~ when I input a password. I've been using regular expressions to check for everything else, but does anyone know how I can make it so the function checks the password for at least one of these special characters?
Here's what I have:
function validateEmail(email)
{
var emailPattern = /^[a-zA-Z0-9._-]+#[a-zA-Z0-9.-]+\.[a-zA-Z]{2,3}$/;
return emailPattern.test(email);
}
function validatePassword(password)
{
var passwordPattern = /(?=.*\d)(?=.*[a-z])(?=.*[A-Z])./;
return passwordPattern.test(password)
}
function validate()
{
var email = user.email.value;
if(validateEmail(user.email.value))
user.validEmail.value = "OK";
else
user.validEmail.value = "X";
if(validatePassword(user.password.value))
user.validPassword.value = "OK";
else
user.validPassword.value = "X";
}

You can match any non-(letters, digits, and underscores) characters with \W.
So to check the password if has any special character you can just simply use:
if (password.match(/\W/)) {
alert('you have at least one special character');
}
to use it in your function you can replace the whole regex with:
var passwordPattern = /^[\w\W]*\W[\w\W]*$/;
that will return true if the string has at least one special character.