I am trying to match a pattern in javascript.
Following is the example:
var pattern = "/^[a-z0-9]+$/i"; // This is should accept on alpha numeric characters.
var reg = new RegExp(pattern);
console.log("Should return false : "+reg.test("This $hould return false"));
console.log("Should return true : "+reg.test("Thisshouldreturntrue"));
When i run this I am getting both the results as false.
I do think that I am missing something simple. But little bit confused.
Thanks in advance.
You need not use slashes if you are using RegExp constructor. You either use enclosing slashes without double quotes to denote a regular expression or you pass a string (usually enclosed in quotes) to RegExp constructor:
var pattern = "^[a-z0-9]+$"; // This is should accept on alpha numeric characters.
var reg = new RegExp(pattern, "i");
console.log("Should return false : "+reg.test("This $hould return false"));
console.log("Should return true : "+reg.test("Thisshouldreturntrue"));
Your pattern is wrong. You don't need to use RegExp constructor here. And you need either ingnore case flag or add uppercase letters to range.
var reg = /^[a-zA-Z0-9]+$/;
console.log("Should return false : "+reg.test("This $hould return false"));
console.log("Should return true : "+reg.test("Thisshouldreturntrue"));
Related
I'm attempting one of the beginner coderByte challenges, Simple Symbols. Challenge summary below.
"Using the JavaScript language, have the function SimpleSymbols(str) take the str parameter being passed and determine if it is an acceptable sequence by either returning the string true or false. The str parameter will be composed of + and = symbols with several letters between them (ie. ++d+===+c++==a) and for the string to be true each letter must be surrounded by a + symbol. So the string to the left would be false. The string will not be empty and will have at least one letter."
function SimpleSymbols(str){
var RegExp = /\+\w\+/gi;
var regexp1 = /^\w/gi;
var regexp2 = /\w$/g;
if(regexp1.test(str) == true){
return false
} else if(regexp2.test(str) == true){
return false
} else if(RegExp == true){
return true
}
}
console.log(SimpleSymbols('+d+=3=+s+'));
console.log(SimpleSymbols('f++d+'));
The first regular expression I'm testing, /^\w/gi, comes back undefined, and I can't figure out why?
https://regex101.com/ is a great tool I've used before, and my expression does identify f as the first character in the string, but when I test it in codepen, it comes back undefined in the console.
Code
See regex in use here
^[+=\d]*\+(?:[a-z]\+[+=\d]*)+$
Alternatively, using the opposite logic (catching invalid strings instead of valid ones), you can use (?:^|[^+])[a-z]|[a-z](?:[^+]|$)
Usage
Please note the valid/invalid strings below have been created according to the OP's explanation of valid and invalid strings: That each letter must be surrounded by a + symbol. and that the plus sign + may be shared between characters such that +a+a+ is valid (specified in comments below the question).
var a = [
// valid
"++d+===+c++==+a++",
"+a+a+a+",
"+a++a+",
"+a+",
// invalid
"++d+===+c++==a",
"+=d+",
"+dd+",
"+d=+",
"+d+d",
"d+d+"
];
var r = /^[+=\d]*\+(?:[a-z]\+[+=\d]*)+$/mi;
a.forEach(function(s){
console.log(r.test(s));
});
Explanation
^ Assert position at the start of the line
[+=\d]* Match any number of characters in the set (+, =, or digit)
\+ Match a literal plus sign +
(?:[a-z]\+[+=\d]*)+ Match one or more of the following
[a-z] Match a lowercase ASCII letter
\+ Match a literal plus sign +
[+=\d]* Match any number of characters in the set (+, =, or digit)
$ Assert position at the end of the line
It's returning undefined because your expression does not meet any of the criteria. Since you have no else {} defined, than nothing gets returned. Thus you get undefined. Try this:
function SimpleSymbols(str){
var RegExp = /\+\w\+/gi;
var regexp1 = /^\w/gi;
var regexp2 = /\w$/g;
if(regexp1.test(str) == true){
return false
} else if(regexp2.test(str) == true){
return false
} else if(RegExp == true){
return true
} else {
return "catch all here";
}
}
console.log(SimpleSymbols('+d+=3=+s+'));
console.log(SimpleSymbols('f++d+'));
You can use a single regex and the string.test() method (which returns just true/false).
Below are 2 different ways (regex) to do it .
First requires a separate + between word chars. Example +a++b+ (true)
^
(?: [+=]* \+ \w \+ [+=]* )+
$
Second can take a common + between word chars. Example +a+b+ (true)
^
(?:
[+=]* \+ \w
(?= \+ )
)+
[+=]*
$
var patt1 = new RegExp("^(?:[+=]*\\+\\w\\+[+=]*)+$");
function SimpleSymbols_1(str){
return patt1.test(str);
}
var patt2 = new RegExp("^(?:[+=]*\\+\\w(?=\\+))+[+=]*$");
function SimpleSymbols_2(str){
return patt2.test(str);
}
console.log(SimpleSymbols_1('+d+=3=+s+'));
console.log(SimpleSymbols_1('f++d+'));
console.log(SimpleSymbols_1('+a+b+c+'));
console.log(SimpleSymbols_2('+a+b+c+'));
console.log(SimpleSymbols_2('+a+=+c+'));
console.log(SimpleSymbols_2('+a++c+'));
Thank you all for throwing some support/comments my way. Again, I am new to JavaScript and Regular Expressions are fairly foreign to me, though I am gaining some traction in understanding them. Here is the updated solution I posted. It's quite convoluted and perhaps a more inelegant and non-simple way to come to the right answer, but it worked.
function SimpleSymbols(str){
var RegExp = /\+[a-z]\+/gi;
var regexp1 = /^[a-z]/gi;
var regexp2 = /[a-z]$/g;
var regexp3 = /[a-z]\=/gi;
var regexp4 = /\=[a-z]/gi;
if(regexp1.test(str) === true){
return false
} else if(regexp2.test(str) === true){
return false
} else if(regexp3.test(str) === true){
return false
} else if(regexp4.test(str) === true){
return false
} else {
return true
}
}
console.log(SimpleSymbols('+d+=3=+s+'));
console.log(SimpleSymbols('f++d+'));
console.log(SimpleSymbols('+d===+a+'));
console.log(SimpleSymbols('+a='));
console.log(SimpleSymbols('2+a+a+'));
console.log(SimpleSymbols('==a+'));
I was sure there had to be a way to use only one regular expression, but again, I'm still very much a novice.
Thanks again everyone.
I need to format a string for comparison purpose.
Lets say we have
Multiple Choice
I want to convert it to
multiplechoice
So white spaces removed, any special characters removed and lowercase.
I need to do this in SAPUI5 while comparing a value which I get from a model.
if (oCurrentQuestionModel.getProperty("/type") === "multiple choice")
How can I achieve this?
You can do it as:
var str = "Multiple Choice";
var strLower = str.toLowerCase();
strLower.replace(/\s/g, '');
Working demo.
The Regex
\s is the regex for "whitespace", and g is the "global" flag, meaning match all \s (whitespaces).
function cleaner(str) {
if (str) {
var strLower = str.toLowerCase();
return strLower.replace(/\W/g, '');
}
return false;
}
I have a variable which contain a string and I want to return only the letters from regular expression (“b” and “D”) or any letter that I indicate on regular expression from match().
var kk = "AaBbCcDd".match(/b|D/g);
kk.forEach(function(value,index){
console.log(value,index)
});
My problem is that regular expression I think because is returning b and D but the index is not the index from kk variable and I'm not really sure, why ... so if someone can help me a little bit because I stuck
The match method from javascript only returns an array with the given match:
https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Global_Objects/String/match
You would need to implement a new function which will loop through all characters of your string and return the given index of the matches.
This method could use the function search from String.prototype: https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Global_Objects/String/search
You have to write a new function to get the index of the matched regex like a sample below:-
var re = /bar/g,
str = "foobarfoobar";
while ((match = re.exec(str)) != null) {
alert("match found at " + match.index);
}
Hope this will help you
Actually this is the answer :
var kk = "AaBbCcDd".match(/B?d?/g);
kk.forEach(function(value,index){
console.log(value,index)
});
if someone will encounter this scenario ...
The match() regular expresion B?d? will return an array indicating the position of "B" and "d" of the initial array kk.
I am having some trouble with my regex in javascript.
I have the following code, that I think should match, but it doesn't.
var rgx = new RegExp("{\d+:(\d+)}");
if (rgx.test("{0:00000}") == true) {
alert("match");
}
else
{
alert("no match");
}
I am unsure if I should use test() here. I really want to catch the group, in my regex but exec() seems to give me the same result.
So what am I doing wrong?
The problem is that you need to escape the \ character in your regex:
var rgx = new RegExp("{\\d+:(\\d+)}");
Alternatively, you can use the literal syntax:
var rgx = /{\d+:(\d+)}/;
To capture the results, you should also use the .match function as opposed to test or exec. It will return null if it doesn't match and an array of at least one element if it does match.
There are multiple issues with the regex:
var rgx = new RegExp("{\d+:(\d+)}");
First (first noted by syazdani), you must string-escape the backslashes:
var rgx = new RegExp("{\\d+:(\\d+)}");
or better yet use a regex literal:
var rgx = /{\d+:(\d+)}/
Second, { and } have a special meaning in regex and should be escaped:
var rgx = /\{\d+:(\d+)\}/
Third, as noted by Ian, you might want to ensure the entire string is matched:
var rgx = /^\{\d+:(\d+)\}$/
RegExp#test returns a boolean true/false whether the string matches.
RegExp#exec returns an array holding the match and all captured groups if the string is matched, or null if the string is not matched:
var matches = /\{\d+:(\d+)\}/.exec("{0:000000}");
if(matches){
console.log(matches[1]); //logs "000000"
}
How can I quickly validate if a string is alphabetic only, e.g
var str = "!";
alert(isLetter(str)); // false
var str = "a";
alert(isLetter(str)); // true
Edit : I would like to add parenthesis i.e () to an exception, so
var str = "(";
or
var str = ")";
should also return true.
Regular expression to require at least one letter, or paren, and only allow letters and paren:
function isAlphaOrParen(str) {
return /^[a-zA-Z()]+$/.test(str);
}
Modify the regexp as needed:
/^[a-zA-Z()]*$/ - also returns true for an empty string
/^[a-zA-Z()]$/ - only returns true for single characters.
/^[a-zA-Z() ]+$/ - also allows spaces
Here you go:
function isLetter(s)
{
return s.match("^[a-zA-Z\(\)]+$");
}
If memory serves this should work in javascript:
function containsOnlyLettersOrParenthesis(str)
(
return str.match(/^([a-z\(\)]+)$/i);
)
You could use Regular Expressions...
functions isLetter(str) {
return str.match("^[a-zA-Z()]+$");
}
Oops... my bad... this is wrong... it should be
functions isLetter(str) {
return "^[a-zA-Z()]+$".test(str);
}
As the other answer says... sorry