Develop Reference

Simple Regexp Pattern matching with escape characters

Hopefully a simple one!
I've been trying to get this to work for several hours now but am having no luck, as I'm fairly new to regexp I may be missing something very obvious here and was hoping someone could point me in the right direction. The pattern I want to match is as follows: -
At least 1 or more numbers + "##" + at least 1 or more numbers + "##" + at least 1 or more numbers
so a few examples of valid combinations would be: -
1##2##3
123#123#123
0##0##0
A few invalid combinations would be
a##b##c
1## ##1
I've got the following regexp like so: -
[\d+]/#/#[\d+]/#/#[\d+]
And am using it like so (note the double slashes as its inside a string): -
var patt = new RegExp("[\\d+]/#/#[\\d+]/#/#[\\d+]");
if(newFieldValue!=patt){newFieldValue=="no match"}
I also tried these but still nothing: -
if(!patt.text(newFieldValue)){newFieldValue==""}
if(patt.text(newFieldValue)){}else{newFieldValue==""}
But nothing I try is matching, where am I going wrong here?
Any pointers gratefully received, cheers!

1) I can't see any reason to use the RegExp constructor over a RegExp literal for your case. (The former is used primarily where the pattern needs to by dynamic, i.e. is contributed to by variables.)
2) You don't need a character class if there's only one type of character in it (so \d+ not [\d+]
3) You are not actually checking the pattern against the input. You don't apply RegEx by creating an instance of it and using ==; you need to use test() or match() to see if a match is made (the former if you want to check only, not capture)
4) You have == where you mean to assign (=)
if (!/\d+##\d+##\d+/.test(newFieldValue)) newFieldValue = "no match";

You put + inside the brackets, so you're matching a single character that's either a digit or +, not a sequence of digits. I also don't understand why you have / before each #, your description doesn't mention anything about this character.
Use:
var patt = /\d+##\d+##\d+/;

You should use the test method of the pat regex
if (!patt.test(newFieldValue)){ newFieldValue=="no match"; }
once you have a valid regular expression.

Try this regex :
^(?:\d+##){2}\d+$
Demo: http://regex101.com/r/mE8aG7
With the following regex
[\d+]/#/#[\d+]/#/#[\d+]
You would only match things like:
+/#/#5/#/#+
+/#/#+/#/#+
0/#/#0/#/#0
because the regex engine sees it like on the schema below:

Something like:
((-\s)?\d+##)+\d+

Javascript string validation using the regex object

I am complete novice at regex and Javascript. I have the following problem: need to check into a textfield the existence of one (1) or many (n) consecutive * (asterisk) character/characters eg. * or ** or *** or infinite (n) *. Strings allowed eg. *tomato or tomato* or **tomato or tomato** or as many(n)*tomato many(n)*. So, far I had tried the following:
var str = 'a string'
var value = encodeURIComponent(str);
var reg = /([^\s]\*)|(\*[^\s])/;
if (reg.test(value) == true ) {
alert ('Watch out your asterisks!!!')
}

By your question it's hard to decipher what you're after... But let me try:
Only allow asterisks at beginning or at end
If you only allow an arbitrary number (at least one) of asterisks either at the beginning or at the end (but not on both sides) like:
*****tomato
tomato******
but not **tomato*****
Then use this regular expression:
reg = /^(?:\*+[^*]+|[^*]+\*+)$/;
Match front and back number of asterisks
If you require that the number of asterisks at the biginning matches number of asterisks at the end like
*****tomato*****
*tomato*
but not **tomato*****
then use this regular expression:
reg = /^(\*+)[^*]+\1$/;
Results?
It's unclear from your question what the results should be when each of these regular expressions match? Are strings that test positive to above regular expressions fine or wrong is on you and your requirements. As long as you have correct regular expressions you're good to go and provide the functionality you require.
I've also written my regular expressions to just exclude asterisks within the string. If you also need to reject spaces or anything else simply adjust the [^...] parts of above expressions.
Note: both regular expressions are untested but should get you started to build the one you actually need and require in your code.

If I understand correctly you're looking for a pattern like this:
var pattern = /\**[^\s*]+\**/;
this won't match strings like ***** or ** ***, but will match ***d*** *d or all of your examples that you say are valid (***tomatos etc).If I misunderstood, let me know and I'll see what I can do to help. PS: we all started out as newbies at some point, nothing to be ashamed of, let alone apologize for :)
After the edit to your question I gather the use of an asterisk is required, either at the beginning or end of the input, but the string must also contain at least 1 other character, so I propose the following solution:
var pattern = /^\*+[^\s*]+|[^\s*]+\*+$/;
'****'.match(pattern);//false
' ***tomato**'.match(pattern);//true
If, however *tomato* is not allowed, you'll have to change the regex to:
var pattern = /^\*+[^\s*]+$|^[^\s*]+\*+$/;
Here's a handy site to help you find your way in the magical world of regular expressions.

What's wrong with my JavaScript regex / regex syntax?

I need a regex to use with javascript/jquery that fits these rules...
it will include 10 digits
if there is a leading 1 or +1 it should be ignored
valid characters allowed in the field are... 0-9,(), and -
I found a regex at Snipplr (the first one), but its not working. First of all, I'm not even sure if that regex fits my rules. Secondly, its allowing inputs like &^%$$#%^adfafsd. I believe the error is in my code not the regex. For example, are there supposed to be quotes around the expression?
Here is the code that is supposed to be validating the phone field...
$('#phone').bind('blur', function() {
var pattern = new RegExp("^(?:\+?1\s*(?:[.-]\s*)?)?(?:\(\s*([2-9]1[02-9]|[2-9][02-8]1|[2-9][02-8][02-9])\s*\)|([2-9]1[02-9]|[2-9][02-8]1|[2-9][02-8][02-9]))\s*(?:[.-]\s*)?([2-9]1[02-9]|[2-9][02-9]1|[2-9][02-9]{2})\s*(?:[.-]\s*)?([0-9]{4})$");
if(pattern.test($('#phone').val())){
$("#phone").addClass("error");
return false;
}else{
$("#phone").removeClass("error");
return true;
}
return true;
})

When you're not using the literal form ( /[regex]/ ), you need to escape the regex string. Try this instead:
var regex = /^(?:\+?1\s*(?:[.-]\s*)?)?(?:\(\s*([2-9]1[02-9]|[2-9][02-8]1|[2-9][02-8][02-9])\s*\)|([2-9]1[02-9]|[2-9][02-8]1|[2-9][02-8][02-9]))\s*(?:[.-]\s*)?([2-9]1[02-9]|[2-9][02-9]1|[2-9][02-9]{2})\s*(?:[.-]\s*)?([0-9]{4})$/;
if(regex.test($('#phone').val()){ ... }

if there is a leading 1 or +1 it should be ignored
it will include 10 digits
valid characters allowed in the field are... 0-9,(), and -
That could be matched with an expression like:
/^(?:\+?1)?[()-]*(?:\d[()-]*){10}$/

Javascript regexp using val().match() method

I'm trying to validate a field named phone_number with this rules:
the first digit should be 3 then another 9 digits so in total 10 number example: 3216549874
or can be 7 numbers 1234567
here i have my code:
if (!($("#" + val["htmlId"]).val().match(/^3\d{9}|\d{7}/)))
missing = true;
Why doesnt work :( when i put that into an online regexp checker shows good.

You should be using test instead of match and here's the proper code:
.test(/^(3\d{9}|\d{7})$/)
Match will find all the occurrences, while test will only check to see if at least one is available (thus validating your number).

Don't get confused by pipe. Must end each expression
if (!($("#" + val["htmlId"]).val().match(/^3\d{9}/|/\d{7}/)))
missing = true;
http://jsfiddle.net/alfabravoteam/e6jKs/

I had similar problem and my solution was to write it like:
if (/^(3\d{9}|\d{7})$/.test($("#" + val["htmlId"]).val()) == false) {
missing = true;
}

Try this, it's a little more strict.
.match(/^(3\d{9}|\d{7})$/)

Test Password with Regex

I want to test a passwprd which must have at least 6 characters and 1 number in it. What regex string I can use with JS to get this done?
UPDATED
I forgot to write it must have at least 6 alpha characters and 1 numeric character but it should also allow special characters or any other character. Can you please modify your answers? I greatly appreciated your responses

This does smell a little like a homework question, but oh well. You can actually accomplish this concisely using a single regular expression and the "look ahead" feature.
/(?=.{6}).*\d.*/.test("helelo1")
The first bit in the brackets says "peek ahead to see if there's 6 characters". Following this we check for any number of characters, followed by a number, followed by any number of characters.
It is even possible to accomplish your goal in a single regex without having the faculty of look ahead... It's just a little hard to look at the solution and not wince:
new RegExp("[0-9].....|" +
".[0-9]....|" +
"..[0-9]...|" +
"...[0-9]..|" +
"....[0-9].|" +
".....[0-9]").test("password1")

Try this:
password.match(/(?=.*\d).{6}/);
More info here.

As far as I know this is best done with a combination of string functions and regex:
if( myPass.match(/[a-zA-Z]/).length >= 6 && myPass.match(/\d/g).length ) {
// Good passwords are good!
}
EDIT: Updated to include the new stipulations. Special characters are allowed, but not required.

if (/.{6,}/.test(password) && /\d/.test(password)) {
// success
} else {
// fail
}

/^(?=[\w\d]{6,}$)(?=.*\d)/.test(password)
requires 6 or more characters (letters, numbers or _)
requires at least one digit
won't allow any special characters

This is a js to check password,
it checks min 7 chars, contains 1 Upper case and 1 digit and 1 special character and must not contain a space, hope it will help you.
pwLength = this.value.length;
if (pwLength > 7 && pwLength < 21) {
charLengthIcon.removeClass("fail").addClass("pass");
}
else charLengthIcon.removeClass("pass").addClass("fail");
if (this.value.match(/[A-Z]/g)) {
capLetterIcon.removeClass("fail").addClass("pass");
}
else capLetterIcon.removeClass("pass").addClass("fail");
if (this.value.match(/[0-9]/g)) {
numberIcon.removeClass("fail").addClass("pass");
}
else numberIcon.removeClass("pass").addClass("fail");
if (this.value.match(/[##$%!$&~*^(){}?><.,;:"'-+=|]/g)) {
splcharIcon.removeClass("fail").addClass("pass");
}
else splcharIcon.removeClass("pass").addClass("fail");
if (this.value.match(/[\s/]/g)) {
whiteSpce.removeClass("pass").addClass("fail");
}
else whiteSpce.removeClass("fail").addClass("pass");
confirmPW();
});