I am trying to pass value as 95%
numexu = 95%
"^((>|GT|>=|GE|<|LT|<=|LE|==|EQ|!=|NE)?\\s*\\d?[%]?)$
if (!regex.IsMatch(numexu))
throw new ArgumentException("Percent expression is in an invalid format.");
it is throwing exception in code.
Regards,
Regex
You are checking only for 1 number \\d?, try instead this: \\d{0,2}, this accepts 0, 1 or 2 numbers. The ? makes it 0 or 1 times matching.
I am not sure if you need to escape the %, if so then \\%. Additionally if you have only one character you can skip the brackets [%], so % (or \\%, if needed to escape)
This Function will work for your requirement
function check() {
var txtfield; txtfield =document.getElementById('txtbox').value;
var reg=/^(\d{0,2}%?$)/;
if(reg.test(txtfield)){
alert("match");
}
else { alert("Try again"); }
}
Related
How can I get the last 4 digits in $MyBuildNumber, which could also be an asterisk? I want $NewVersion to return "2.8.1.*"; however, this code does not return anything for me:
$MyBuildNumber = "MyBuildNumberIs_2.8.1.*"
$VersionRegex = "\d+[*]?\.\d+[*]?\.\d+[*]?\.\d+[*]?"
$VersionData = [regex]::matches($MyBuildNumber,$VersionRegex)
switch($VersionData.Count)
{
0
{
Write-Error "Could not find version number data in MyBuildNumber."
exit 1
}
1 {}
}
$NewVersion = $VersionData[0]
Write-Host "Version: $NewVersion"
You can use character class with digit and *:
[\d*]+\.[\d*]+\.[\d*]+\.[\d*]+
anubhava beat me to it, but here's a more compact version of the same thing because why not:
([\d*]+\.){3}[\d*]+
I'm getting an error when I try to use an if/else statement in Xcode 6 with Swift. This is what I have
} else if countElements(sender.text) == 1, 2
It's telling me:
Type 'String!' does not conform to protocol '_CollectionType'
How can I compare two values on one line?
You can use "||" = "or"
} else if countElements(sender.text) == 1 || countElements(sender.text) == 2 {
}
The other answer correctly shows the way to make 2 comparisons in Swift, but this can be done in a single statement thanks to Swift's pattern matching operator. For example:
if 1...2 ~= countElements(sender.text) {
println("count is 1 or 2")
}
Basically, the provided statement is true if the number of characters in your string is inclusively between 1 and 2.
You can also use the contains global function, providing it a range and the element to check for inclusion:
contains(1...2, countElements(sender.text))
I have a form that asks for a password and I want to validate if the password has at least eight characters, and of these eight characters at least two must be numbers and two must be letters in any order. I'm trying with this:
function validatePassword():void
{
var passVal:String = pass.text;
if(validPass(passVal))
{
trace("Password Ok");
sendForm();
}
else
{
trace("You have entered an invalid password");
}
function validPass(passVal:String):Boolean{
var pw:RegExp = /^?=.{8,}[A-Za-z]{2,}[0-9]{2,}/;
return(pw.test(passVal));
}
}
But it doesn't work. What I'm doing wrong?
Any help would be really appreciated!
use this pattern ^(?=.{8})(?=(.*\d){2})(?=(.*[A-Za-z]){2}).*$
^ anchor
(?=.{8}) look ahead for at least 8 characters
(?=(.*\d){2}) look ahead for at least 2 digits in any order
(?=(.*[A-Za-z]){2}) look ahead for at least 2 letters in any order
.*$ catch everything to the end if passed previous conditions
The problem is that your regex is forcing the numbers to follow the letters ([A-Za-z]{2,}[0-9]{2,}). While it is possible to write such a regex, I suggest using a simple length check and two regexes:
function validPass(passVal:String):Boolean{
if (passVal.length < 8)
return False;
var letterRegex:RegExp = /^.*?[A-Za-z].*?[A-Za-z].*?$/;
var numberRegex:RegExp = /^.*?\d.*?\d.*?$/;
return letterRegex.test(passVal) && numberRegex.test(passVal);
}
I am using ColdFusion 9.0.1.
I am trying to test whether a user has provided a non alphanumeric value. If they have, I want to return false. I'm pretty sure I am close, but I keep getting an error:
Complex object types cannot be converted to simple values.
I've tried multiple ways of making this work, but I can't get it to work.
Specifically, I want to allow only a through z and 0 through 9. No spaces, or special characters.
Can you help me tweak this?
<cfscript>
LOCAL.Description = trim(left(ARGUMENTS.Description, 15));
if (len(LOCAL.Description) lte 4) {
return false;
} else if (reMatchNoCase("[^A-Za-z0-9_]", LOCAL.Description) neq "") {
return false;
} else {
return true;
</cfscript>
W
reMatchNoCase returns Array which cannot be compared to the string, use ArrayLen() on the result in order to find out if there any matches
There is actually another problem in your code. First line will produce an error if the length of the description is less than 15, which means that the first IF is obsolete since it will always be false.
reMatchNoCase("[^A-Za-z0-9_]", LOCAL.Description) neq ""
It is because ReMatchNoCase returns an array, not a simple string. Either check the array length, or better yet, use ReFindNoCase instead. It returns the position of the first match, or 0 if it was not found.
You can also try the following approach:
<cfscript>
local.description = trim(local.description);
return reFind("(?i)^[A-Z0-9_]{5,}$", local.description)?true:false;
</cfscript>
I'm late to the party but reFindNoCase is the optimal solution in 2021. Here's how I would handle the code in the original question:
// best practice not to have a local var name identical to an argument var
var myValue = trim( left( arguments.description, 15 ) );
// return false if myValue is less than 4 or has special characters
return(
!len( myValue ) lte 4 &&
!reFindNoCase( "[^a-z0-9]", myValue )
);
I have been looking for a regular expression with Google for an hour or so now and can't seem to work this one out :(
If I have a number, say:
2345
and I want to find any other number with the same digits but in a different order, like this:
2345
For example, I match
3245 or 5432 (same digits but different order)
How would I write a regular expression for this?
There is an "elegant" way to do it with a single regex:
^(?:2()|3()|4()|5()){4}\1\2\3\4$
will match the digits 2, 3, 4 and 5 in any order. All four are required.
Explanation:
(?:2()|3()|4()|5()) matches one of the numbers 2, 3, 4, or 5. The trick is now that the capturing parentheses match an empty string after matching a number (which always succeeds).
{4} requires that this happens four times.
\1\2\3\4 then requires that all four backreferences have participated in the match - which they do if and only if each number has occurred once. Since \1\2\3\4 matches an empty string, it will always match as long as the previous condition is true.
For five digits, you'd need
^(?:2()|3()|4()|5()|6()){5}\1\2\3\4\5$
etc...
This will work in nearly any regex flavor except JavaScript.
I don't think a regex is appropriate. So here is an idea that is faster than a regex for this situation:
check string lengths, if they are different, return false
make a hash from the character (digits in your case) to integers for counting
loop through the characters of your first string:
increment the counter for that character: hash[character]++
loop through the characters of the second string:
decrement the counter for that character: hash[character]--
break if any count is negative (or nonexistent)
loop through the entries, making sure each is 0:
if all are 0, return true
else return false
EDIT: Java Code (I'm using Character for this example, not exactly Unicode friendly, but it's the idea that matters now):
import java.util.*;
public class Test
{
public boolean isSimilar(String first, String second)
{
if(first.length() != second.length())
return false;
HashMap<Character, Integer> hash = new HashMap<Character, Integer>();
for(char c : first.toCharArray())
{
if(hash.get(c) != null)
{
int count = hash.get(c);
count++;
hash.put(c, count);
}
else
{
hash.put(c, 1);
}
}
for(char c : second.toCharArray())
{
if(hash.get(c) != null)
{
int count = hash.get(c);
count--;
if(count < 0)
return false;
hash.put(c, count);
}
else
{
return false;
}
}
for(Integer i : hash.values())
{
if(i.intValue()!=0)
return false;
}
return true;
}
public static void main(String ... args)
{
//tested to print false
System.out.println(new Test().isSimilar("23445", "5432"));
//tested to print true
System.out.println(new Test().isSimilar("2345", "5432"));
}
}
This will also work for comparing letters or other character sequences, like "god" and "dog".
Put the digits of each number in two arrays, sort the arrays, find out if they hold the same digits at the same indices.
RegExes are not the right tool for this task.
You could do something like this to ensure the right characters and length
[2345]{4}
Ensuring they only exist once is trickier and why this is not suited to regexes
(?=.*2.*)(?=.*3.*)(?=.*4.*)(?=.*5.*)[2345]{4}
The simplest regular expression is just all 24 permutations added up via the or operator:
/2345|3245|5432|.../;
That said, you don't want to solve this with a regex if you can get away with it. A single pass through the two numbers as strings is probably better:
1. Check the string length of both strings - if they're different you're done.
2. Build a hash of all the digits from the number you're matching against.
3. Run through the digits in the number you're checking. If you hit a match in the hash, mark it as used. Keep going until you don't get an unused match in the hash or run out of items.
I think it's very simple to achieve if you're OK with matching a number that doesn't use all of the digits. E.g. if you have a number 1234 and you accept a match with the number of 1111 to return TRUE;
Let me use PHP for an example as you haven't specified what language you use.
$my_num = 1245;
$my_pattern = '/[' . $my_num . ']{4}/'; // this resolves to pattern: /[1245]{4}/
$my_pattern2 = '/[' . $my_num . ']+/'; // as above but numbers can by of any length
$number1 = 4521;
$match = preg_match($my_pattern, $number1); // will return TRUE
$number2 = 2222444111;
$match2 = preg_match($my_pattern2, $number2); // will return TRUE
$number3 = 888;
$match3 = preg_match($my_pattern, $number3); // will return FALSE
$match4 = preg_match($my_pattern2, $number3); // will return FALSE
Something similar will work in Perl as well.
Regular expressions are not appropriate for this purpose. Here is a Perl script:
#/usr/bin/perl
use strict;
use warnings;
my $src = '2345';
my #test = qw( 3245 5432 5542 1234 12345 );
my $canonical = canonicalize( $src );
for my $candidate ( #test ) {
next unless $canonical eq canonicalize( $candidate );
print "$src and $candidate consist of the same digits\n";
}
sub canonicalize { join '', sort split //, $_[0] }
Output:
C:\Temp> ks
2345 and 3245 consist of the same digits
2345 and 5432 consist of the same digits