i am trying to validate input string to chech whether it contains '+' symbol anywhere in the string. i used for of loop but didnt get what is exprected.
const isMobileValidWithoutPlus = funcLib.isValidMobileWithoutPlus(mobileNumber);
isValidMobileWithoutPlus(mobileNumber) {
if (!mobileNumber) {
return false;
}
const checkRegex = new RegExp('\\+?\\d+');
return checkRegex.test(mobileNumber);
}
but able to get desired out.
The regex for this would be
const rgx = new RegExp(/\+/gm);
Your regular expression checks if you have a string that can either start with + or not, and is followed by one or more numbers. But you're saying you want to just check if there's a "+" anywhere in the number. For that you can use this regex above.
Also, do you need to use a regex?
You can do this using indexOf on a string if using regex is not a must.
let number = "+001234";
function hasPlus(number) {
return number.indexOf('+') !== -1;
}
Regular expressions are generally useful when you don't have one specific string that you're looking for, or when you want to find all the apparitions of a regex in a longer string. In your case, checking if a string contains "+", it isn't necessary to use them.
Related
I'm trying to evaluate a string against a set list of parameters with RegExp in Flutter. For example, the string must contain at least:
One capital letter
One lowercase letter
One number from 0-9
One special character, such as $ or !
This is basically for a password entry field of an application. I have set things up, firstly using validateStructure as follows:
abstract class PasswordValidator {
bool validateStructure(String value);
}
Then, I have used the RegExp function as follows:
class PasswordValidatorSpecial implements PasswordValidator {
bool validateStructure(String value) {
String pattern =
r'^(?=.*?[A-Z])(?=.*?[a-z])(?=.*?[0-9])(?=.*?[!##\$&*~£]).{8,}$';
RegExp regEx = new RegExp(pattern);
return regEx.hasMatch(value);
}
}
This does work well, in a sense that when I pass a string/password through it, it does tell me if at least one of the criteria is not met. However, what I would like to do is for the output to be more specific, telling me which of those criteria isn't met.
For example, if the password were to have everything but a number (from 0-9) I would want to be able to get the output to specifically say that a number is missing, but everything else is present.
How would I adapt my code to be able to do that? I thought perhaps by using conditional 'if' statement, although I don't know how that would work. Thanks!
That's right, you can use RegExr to check your RegExp, separate each part and use them separately to have a custom error. Also instead of return a bool value, you can return a String value, such as the following function:
String validateStructure(String value) {
String patternUpperCaseCharacters = r'^(?=.*?[A-Z])';
String patternLowerCaseCharacters = r'^(?=.*?[a-z])';
String patternNumbers = r'^(?=.*?[0-9])';
String patternSpecialCharacters = r'^(?=.*?[!##\$&*~£])';
RegExp regEx = new RegExp(patternUpperCaseCharacters);
if (regEx.hasMatch(value)) {
regEx = new RegExp(patternLowerCaseCharacters);
if (regEx.hasMatch(value)) {
return "More errors";
} else {
return "You need at least one lowercase letter";
}
} else {
return "You need at least one capital letter";
}
}
in java, I am trying to find if a given string has one of the many sub strings using multiple ORs in a single If condition and if any of the sub string exists, remove it. I am not sure how to do it. Also, this string search needs to be case insensitive.
Here is the sample code
if (inputString contains any of the subStrings i.e. "_LOCATION" OR "_MANAGEMENT" Or "_ZIPCODE")
{
remove the subString from inPutString
}
Ex: Given the string - "STATE_CAPITAL_LOCATION_MANAGEMENT_PHONE_EMAIL_zipcode"
Resulting string should be - "STATE_CAPITAL_PHONE_EMAIL"
What is the best way to do it.
Thanks
Using separate If statements makes more easier.
Try this code:
String a="STATE_CAPITAL_LOCATION_MANAGEMENT_PHONE_EMAIL_zipcode";
if(a.contains("_LOCATION"))// relace with your string
{
a=a.replace("_LOCATION","");
System.out.println(a);
}
if(a.contains("_MANAGEMENT"))// relace with your string
{
a=a.replace("_MANAGEMENT","");
System.out.println(a);
}
// .....
I need to filter strings based on two requirements
1) they must start with "city_date"
2) they should not have "metro" anywhere in the string.
This need to be done in just one check.
To start I know it should be like this but dont know hoe to eliminate strings with "metro"
string pattern = "city_date_"
Added: I need to use the regex for a SQL LIKE statement. hence i need it in a string.
Use a negative lookahead assertion (I don't know if this is supported in your regex lib)
string pattern = "^city_date(?!.*metro)"
I also added an anchor ^ at the start, that will match the start of the string.
The negative lookahead assertion (?!.*metro) will fail, if there is the string "metro" somewhere ahead.
Regular expressions are usually far more expensive than direct comparisons. If direct comparisons can easily express the requirements, use them. This problem doesn't need the overhead of a regular expression. Just write the code:
std::string str = /* whatever */
const std::string head = "city_date";
const std::string exclude = "metro";
if (str.compare(head, 0, head.size) == 0 && str.find(exclude) == std::string::npos) {
// process valid string
}
by using javascript
input="contains the string your matching"
var pattern=/^city_date/g;
if(pattern.test(input)) // to match city_data at the begining
{
var patt=/metro/g;
if(patt.test(input)) return "false";
else return input; //matched string without metro
}
else
return "false"; //unable to match city_data
I need to identify (potentially nested) capture groups within regular expressions and create a tree. The particular target is Java-1.6 and I'd ideally like Java code. A simple example is:
"(a(b|c)d(e(f*g))h)"
which would be parsed to
"a(b|c)d(e(f*g))h"
... "b|c"
... "e(f*g)"
... "f*g"
The solution should ideally account for count expressions, quantifiers, etc and levels of escaping. However if this is not easy to find a simpler approach might suffice as we can limit the syntax used.
EDIT. To clarify. I want to parse the regular expression string itself. To do so I need to know the BNF or equivalent for Java 1.6 regexes. I am hoping someone has already done this.
A byproduct of a result would be that the process would test for validity of the regex.
Consider stepping up to an actual parser/lexer:
http://www.antlr.org/wiki/display/ANTLR3/FAQ+-+Getting+Started
It looks complicated, but if your language is fairly simple, it's fairly straightforward. And if it's not, doing it in regexes will probably make your life hell :)
I came up with a partial solution using an XML tool (XOM, http://www.xom.nu) to hold the tree. First the code, then an example parse. First the escaped characters (\ , ( and ) ) are de-escaped (here I use BS, LB and RB), then remaining brackets are translated to XML tags, then the XML is parsed and the characters re-escaped. What is needed further is a BNF for Java 1.6 regexes doe quantifiers such as ?:, {d,d} and so on.
public static Element parseRegex(String regex) throws Exception {
regex = regex.replaceAll("\\\\", "BS");
regex.replaceAll("BS\\(", "LB");
regex.replaceAll("BS\\)", "RB");
regex = regex.replaceAll("\\(", "<bracket>");
regex.replaceAll("\\)", "</bracket>");
Element regexX = new Builder().build(new StringReader(
"<regex>"+regex+"</regex>")).getRootElement();
extractCaptureGroupContent(regexX);
return regexX;
}
private static String extractCaptureGroupContent(Element regexX) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < regexX.getChildCount(); i++) {
Node childNode = regexX.getChild(i);
if (childNode instanceof Text) {
Text t = (Text)childNode;
String s = t.getValue();
s = s.replaceAll("BS", "\\\\").replaceAll("LB",
"\\(").replaceAll("RB", "\\)");
t.setValue(s);
sb.append(s);
} else {
sb.append("("+extractCaptureGroupContent((Element)childNode)+")");
}
}
String capture = sb.toString();
regexX.addAttribute(new Attribute("capture", capture));
return capture;
}
example:
#Test
public void testParseRegex2() throws Exception {
String regex = "(.*(\\(b\\))c(d(e)))";
Element regexElement = ParserUtil.parseRegex(regex);
CMLUtil.debug(regexElement, "x");
}
gives:
<regex capture="(.*((b))c(d(e)))">
<bracket capture=".*((b))c(d(e))">.*
<bracket capture="(b)">(b)</bracket>c
<bracket capture="d(e)">d
<bracket capture="e">e</bracket>
</bracket>
</bracket>
</regex>
What's the easiest way to do an "instring" type function with a regex? For example, how could I reject a whole string because of the presence of a single character such as :? For example:
this - okay
there:is - not okay because of :
More practically, how can I match the following string:
//foo/bar/baz[1]/ns:foo2/#attr/text()
For any node test on the xpath that doesn't include a namespace?
(/)?(/)([^:/]+)
Will match the node tests but includes the namespace prefix which makes it faulty.
I'm still not sure whether you just wanted to detect if the Xpath contains a namespace, or whether you want to remove the references to the namespace. So here's some sample code (in C#) that does both.
class Program
{
static void Main(string[] args)
{
string withNamespace = #"//foo/ns2:bar/baz[1]/ns:foo2/#attr/text()";
string withoutNamespace = #"//foo/bar/baz[1]/foo2/#attr/text()";
ShowStuff(withNamespace);
ShowStuff(withoutNamespace);
}
static void ShowStuff(string input)
{
Console.WriteLine("'{0}' does {1}contain namespaces", input, ContainsNamespace(input) ? "" : "not ");
Console.WriteLine("'{0}' without namespaces is '{1}'", input, StripNamespaces(input));
}
static bool ContainsNamespace(string input)
{
// a namspace must start with a character, but can have characters and numbers
// from that point on.
return Regex.IsMatch(input, #"/?\w[\w\d]+:\w[\w\d]+/?");
}
static string StripNamespaces(string input)
{
return Regex.Replace(input, #"(/?)\w[\w\d]+:(\w[\w\d]+)(/?)", "$1$2$3");
}
}
Hope that helps! Good luck.
Match on :? I think the question isn't clear enough, because the answer is so obvious:
if(Regex.Match(":", input)) // reject
You might want \w which is a "word" character. From javadocs, it is defined as [a-zA-Z_0-9], so if you don't want underscores either, that may not work....
I dont know regex syntax very well but could you not do:
[any alpha numeric]\*:[any alphanumeric]\*
I think something like that should work no?
Yeah, my question was not very clear. Here's a solution but rather than a single pass with a regex, I use a split and perform iteration. It works as well but isn't as elegant:
string xpath = "//foo/bar/baz[1]/ns:foo2/#attr/text()";
string[] nodetests = xpath.Split( new char[] { '/' } );
for (int i = 0; i < nodetests.Length; i++)
{
if (nodetests[i].Length > 0 && Regex.IsMatch( nodetests[i], #"^(\w|\[|\])+$" ))
{
// does not have a ":", we can manipulate it.
}
}
xpath = String.Join( "/", nodetests );