I am working with a small issue, but I don't know how to solve it clearly. I have to validate a generated password, with some constraints:
password length: [8, 24]
password contains
at least 1 lower case character
at least 1 upper case character
at least 1 digit
at least 1 special character (printable based on ASCII code)
I've used Regex pattern, but it didn't work correctly with both cases: valid and invalid.
The first RegEx pattern:
def pattern = /(=?.{8,24})((:?[a-z]+)(:?[0-9]+)(:?[A-Z]+)(:?\W+))/
can check all invalid passwords but not for the valid one.
The second RegEx pattern:
def pattern = /(=?.{8,24})((:?[a-z]*)(:?[0-9]*)(:?[A-Z]*)(:?\W*))/
can check all valid passwords but not for the invalid one.
I am new to Groovy, so I don't know how to create the correct RegEx pattern to solve this.
Could you please help me?
Regex is not a solution to everything, and trying to come up with a single regex for a given problem is often wasting brain cycles. Just separate it out into multiple tests, for example (this Perl-like pseudo code, but you should be able to transform that to the language you are using):
sub valid_pw
{
return false if (length($_) < 8 || length($_) > 24);
# don't use [a-z], it makes for nasty surprises in e.g. fi_FI
return false if (!/[[:lower:]]/);
return false if (!/[[:upper:]]/);
return false if (!/[[:digit:]]/);
return false if (!/[[:print:]]/);
return true;
}
Why do you need a single regex for this? (also why are you placing a maximum length on the password, which is another discussion)
/^.{8,24}$/
/[a-z]/
/[A-Z]/
/\d/
/[^\d\w]/
I guess you could combine them using lookaheads (say /(?=.*[a-z])(?=.*[A-Z])...), but if you do it's probably a really good idea to comment it heavily.
Related
We can check to see if a digit is in a password, for example, by doing something like:
(?=.*\d)
Or if there's a digit and lowercase with:
(?=.*\d)(?=.*[a-z])
This will basically go on "until the end" to check whether there's a letter in the string.
However, I was wondering if it's possible in some sort of generic way to limit the scope of a lookahead. Here's a basic example which I'm hoping will demonstrate the point:
start_of_string;
middle_of_string;
end_of_string;
I want to use a single regular expression to match against start_of_string + middle_of_string + end_of_string.
Is it possible to use a lookahead/lookbehind in the middle_of_string section WITHOUT KNOWING WHAT COMES BEFORE OR AFTER IT? That is, not knowing the size or contents of the preceding/succeeding string component. And limit the scope of the lookahead to only what is contained in that portion of the string?
Let's take one example:
start_of_string = 'start'
middle_of_string = '123'
end_of_string = 'ABC'
Would it be possible to check the contents of each part but limit it's scope like this?
string = 'start123ABC'
# Check to make sure the first part has a letter, the second part has a number and the third part has a capital
((?=.*[a-z]).*) # limit scope to the first part only!!
((?=.*[0-9]).*) # limit scope to only the second part.
((?=.*[A-Z]).*) # limit scope to only the last part.
In other words, can lookaheads/lookbehinds be "chained" with other components of a regex without it screwing up the entire regex?
UPDATE:
Here would be an example, hopefully this is more helpful to the question:
START_OF_STRING = 'abc'
Does 'x' exist in it? (?=.*x) ==> False
END_OF_STRING = 'cdxoy'
Does 'y' exist in it? (?=.*y) ==> True
FULL_STRING = START_OF_STRING + END_OF_STRING
'abcdxoy'
Is it possible to chain the two regexes together in any sort of way to only wok on its 'substring' component?
For example, now (?=.*x) in the first part of the string would return True, but it should not.
`((?=.*x)(?=.*y)).*`
I think the short answer to this is "No, it's not possible.", but am looking to hear from someone who understands this to tell why it is or isn't.
In .NET and javascript you could use a positive lookahead at the start of your string component and a negative lookbehind at the end of it to "constrain" the match. Example:
.*(?=.*arrow)(?<middle>.*)(?<=.*arrow).*
helloarrowxyz
{'middle': 'arrow'}
If in pcre, python, or other you would need to either have a fixed width lookahead to constraint it from going too far forward, such as what Wiktor Stribiżew says above:
.*(?=.{0,5}arrow)(?<middle>.{0,5}).*
Otherwise, it wouldn't be possible to do without either a fixed-width lookahead or a variable width look-behind.
I'm attempting to write a regex to prevent certain user input in mathematical expressions. (e.g. '1+1' would be valid whereas'1++1' should be invalidated)
Acceptable characters include *digits 0-9* (\d works in lieu of 0-9), + - # / ( ) and white-spaces.
I've attempted to put together a regex but I cant find anything in python regular expression syntax that would validate (or consequently invalidate certain characters when typed together.
(( is ok
++, --, +-, */, are not
I hope there is a simple way to do this, but I anticipate if there isn't, I will have to write regex's for every possible combination of characters I don't want to allow together.
I've tried:
re.compile(r"[\d\s*/()+-]")
re.compile(r"[\d]\[\s]\[*]\[/]\[(]\[)]\[+]\[-]")
I expect to be able to invalidate the expression if someone were to type "1++1"
Edit: Someone suggested the below link is similar to my question...it is not :)
Validate mathematical expressions using regular expression?
Probably the way to go is by inverting your logic:
abort if the regex detects any invalid combination - those are much less compared to the amount of valid combinations.
So e.g.:
re.compile(r"++")
Also, is it possible at all to enumerate all valid terms? If the length of the term is not limit, it is impossible to enumerate all vaild terms
Perhaps one option might be to check the string for the unwanted combinations:
[0-9]\s*(?:[+-][+-]|\*/)\s*[0-9]
Regex demo | Python demo
For example
pattern = r"[0-9]\s*(?:[+-][+-]|\*/)\s*[0-9]"
strings = [
'This is test 1 -- 1',
'This is test 2',
'This is test 3+1',
'This is test 4 */4'
]
for s in strings:
res = re.search(pattern, s)
if not res:
print("Valid: " + s)
Result
Valid: This is test 2
Valid: This is test 3+1
Below is a snippet from my code. This is hardly the solution I was originally looking for but it does accomplish what I was trying to do. When a user curls to an api endpoint 1++1, it will return "Forbidden" based on the below regex for "math2 =...." Alternatively, it will return "OK" if a user curls 1+1. I hope I am understanding how Stack Overflow works and have formatted this properly...
# returns True if a valid expression, False if not.
def validate_expression(calc):
math = re.compile(r"^[\d\s*/()+-]+$")
math2 = re.compile(r"^[\d++\d]+$")
print('2' " " 'validate_expression')
if math.search(calc) is not None and math2.search(calc) is None:
return True
else:
return False
I need to check an input field for a German IBAN. The user should be allowed to leave in white spaces and input should be validated to have a starting DE and then exact 20 characters numbers and letters.
Without the white space allowance, I tried
^[DE]{2}([0-9a-zA-Z]{20})$
but I cannot find where and how I can add "white spaces anywhere allowed.
This should be simple, but I simply cannot find a solution.
Thanks for help!
Because you should use the right tool for the right task: you should not rely on regexps to validate IBAN numbers, but instead use the IBAN checksum algorithm to check the whole code is actually correct, making any regexp superfluous and redundant. i.e.: remove all spaces, rearrange the code, convert to integers, and compute remainder, here it's best explained.
Though, there am I trying to answer your question, for the fun of it:
what about:
^DE([0-9a-zA-Z]\s?){20}$
which only difference is allowing a whitespace (or not) after each occurence of a alphanumeric character.
here is the visualization:
edit: for the OP's information, the only difference is that this regexp, from #ulugbex-umirov: (?:\s*[0-9a-zA-Z]\s*) does a lookahead check to see if there's a space between the iso country code and the checksum (which only made of numerical digits), which I do not support on purpose.
And actually to support a correct IBAN syntax, which is formed of groups of 4 characters, as the wikipedia page says:
^DE\d{2}\s?([0-9a-zA-Z]{4}\s?){4}[0-9a-zA-Z]{2}$
example
If your UI is in Javascript, you can use that library for doing IBAN validation:
<script src="iban.js"></script>
<script>
// the API is now accessible from the window.IBAN global object
IBAN.isValid('hello world'); // false
IBAN.isValid('BE68539007547034'); // true
</script>
so you know this is a valid IBAN, and can validate it before the data is ever even sent to the backend. Simpler, lighter and more elegant… Why do something else?
Here is a list of IBANs from 70 Countries. I generated it with a python script i wrote based on this https://en.wikipedia.org/wiki/International_Bank_Account_Number
AL[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){2}([a-zA-Z0-9]{4}\s?){4}\s?
AD[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){2}([a-zA-Z0-9]{4}\s?){3}\s?
AT[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){4}\s?
AZ[a-zA-Z0-9]{2}\s?([a-zA-Z0-9]{4}\s?){1}([0-9]{4}\s?){5}\s?
BH[a-zA-Z0-9]{2}\s?([a-zA-Z]{4}\s?){1}([a-zA-Z0-9]{4}\s?){3}([a-zA-Z0-9]{2})\s?
BY[a-zA-Z0-9]{2}\s?([a-zA-Z0-9]{4}\s?){1}([0-9]{4}\s?){5}\s?
BE[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){3}\s?
BA[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){4}\s?
BR[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){5}([0-9]{3})([a-zA-Z]{1}\s?)([a-zA-Z0-9]{1})\s?
BG[a-zA-Z0-9]{2}\s?([a-zA-Z]{4}\s?){1}([0-9]{4}\s?){1}([0-9]{2})([a-zA-Z0-9]{2}\s?)([a-zA-Z0-9]{4}\s?){1}([a-zA-Z0-9]{2})\s?
CR[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){4}([0-9]{2})\s?
HR[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){4}([0-9]{1})\s?
CY[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){2}([a-zA-Z0-9]{4}\s?){4}\s?
CZ[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){5}\s?
DK[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){3}([0-9]{2})\s?
DO[a-zA-Z0-9]{2}\s?([a-zA-Z]{4}\s?){1}([0-9]{4}\s?){5}\s?
TL[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){4}([0-9]{3})\s?
EE[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){4}\s?
FO[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){3}([0-9]{2})\s?
FI[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){3}([0-9]{2})\s?
FR[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){2}([0-9]{2})([a-zA-Z0-9]{2}\s?)([a-zA-Z0-9]{4}\s?){2}([a-zA-Z0-9]{1})([0-9]{2})\s?
GE[a-zA-Z0-9]{2}\s?([a-zA-Z0-9]{2})([0-9]{2}\s?)([0-9]{4}\s?){3}([0-9]{2})\s?
DE[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){4}([0-9]{2})\s?
GI[a-zA-Z0-9]{2}\s?([a-zA-Z]{4}\s?){1}([a-zA-Z0-9]{4}\s?){3}([a-zA-Z0-9]{3})\s?
GR[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){1}([0-9]{3})([a-zA-Z0-9]{1}\s?)([a-zA-Z0-9]{4}\s?){3}([a-zA-Z0-9]{3})\s?
GL[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){3}([0-9]{2})\s?
GT[a-zA-Z0-9]{2}\s?([a-zA-Z0-9]{4}\s?){1}([a-zA-Z0-9]{4}\s?){5}\s?
HU[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){6}\s?
IS[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){5}([0-9]{2})\s?
IE[a-zA-Z0-9]{2}\s?([a-zA-Z0-9]{4}\s?){1}([0-9]{4}\s?){3}([0-9]{2})\s?
IL[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){4}([0-9]{3})\s?
IT[a-zA-Z0-9]{2}\s?([a-zA-Z]{1})([0-9]{3}\s?)([0-9]{4}\s?){1}([0-9]{3})([a-zA-Z0-9]{1}\s?)([a-zA-Z0-9]{4}\s?){2}([a-zA-Z0-9]{3})\s?
JO[a-zA-Z0-9]{2}\s?([a-zA-Z]{4}\s?){1}([0-9]{4}\s?){5}([0-9]{2})\s?
KZ[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){3}([0-9]{1})([a-zA-Z0-9]{3}\s?)([a-zA-Z0-9]{4}\s?){2}([a-zA-Z0-9]{2})\s?
XK[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){1}([0-9]{4}\s?){2}([0-9]{2})([0-9]{2}\s?)\s?
KW[a-zA-Z0-9]{2}\s?([a-zA-Z]{4}\s?){1}([a-zA-Z0-9]{4}\s?){5}([a-zA-Z0-9]{2})\s?
LV[a-zA-Z0-9]{2}\s?([a-zA-Z]{4}\s?){1}([a-zA-Z0-9]{4}\s?){3}([a-zA-Z0-9]{1})\s?
LB[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){1}([a-zA-Z0-9]{4}\s?){5}\s?
LI[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){1}([0-9]{1})([a-zA-Z0-9]{3}\s?)([a-zA-Z0-9]{4}\s?){2}([a-zA-Z0-9]{1})\s?
LT[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){4}\s?
LU[a-zA-Z0-9]{2}\s?([0-9]{3})([a-zA-Z0-9]{1}\s?)([a-zA-Z0-9]{4}\s?){3}\s?
MK[a-zA-Z0-9]{2}\s?([0-9]{3})([a-zA-Z0-9]{1}\s?)([a-zA-Z0-9]{4}\s?){2}([a-zA-Z0-9]{1})([0-9]{2})\s?
MT[a-zA-Z0-9]{2}\s?([a-zA-Z]{4}\s?){1}([0-9]{4}\s?){1}([0-9]{1})([a-zA-Z0-9]{3}\s?)([a-zA-Z0-9]{4}\s?){3}([a-zA-Z0-9]{3})\s?
MR[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){5}([0-9]{3})\s?
MU[a-zA-Z0-9]{2}\s?([a-zA-Z]{4}\s?){1}([0-9]{4}\s?){4}([0-9]{3})([a-zA-Z]{1}\s?)([a-zA-Z]{2})\s?
MC[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){2}([0-9]{2})([a-zA-Z0-9]{2}\s?)([a-zA-Z0-9]{4}\s?){2}([a-zA-Z0-9]{1})([0-9]{2})\s?
MD[a-zA-Z0-9]{2}\s?([a-zA-Z0-9]{2})([a-zA-Z0-9]{2}\s?)([a-zA-Z0-9]{4}\s?){4}\s?
ME[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){4}([0-9]{2})\s?
NL[a-zA-Z0-9]{2}\s?([a-zA-Z]{4}\s?){1}([0-9]{4}\s?){2}([0-9]{2})\s?
NO[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){2}([0-9]{3})\s?
PK[a-zA-Z0-9]{2}\s?([a-zA-Z0-9]{4}\s?){1}([0-9]{4}\s?){4}\s?
PS[a-zA-Z0-9]{2}\s?([a-zA-Z0-9]{4}\s?){1}([0-9]{4}\s?){5}([0-9]{1})\s?
PL[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){6}\s?
PT[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){5}([0-9]{1})\s?
QA[a-zA-Z0-9]{2}\s?([a-zA-Z]{4}\s?){1}([a-zA-Z0-9]{4}\s?){5}([a-zA-Z0-9]{1})\s?
RO[a-zA-Z0-9]{2}\s?([a-zA-Z]{4}\s?){1}([a-zA-Z0-9]{4}\s?){4}\s?
SM[a-zA-Z0-9]{2}\s?([a-zA-Z]{1})([0-9]{3}\s?)([0-9]{4}\s?){1}([0-9]{3})([a-zA-Z0-9]{1}\s?)([a-zA-Z0-9]{4}\s?){2}([a-zA-Z0-9]{3})\s?
SA[a-zA-Z0-9]{2}\s?([0-9]{2})([a-zA-Z0-9]{2}\s?)([a-zA-Z0-9]{4}\s?){4}\s?
RS[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){4}([0-9]{2})\s?
SK[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){5}\s?
SI[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){3}([0-9]{3})\s?
ES[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){5}\s?
SE[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){5}\s?
CH[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){1}([0-9]{1})([a-zA-Z0-9]{3}\s?)([a-zA-Z0-9]{4}\s?){2}([a-zA-Z0-9]{1})\s?
TN[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){5}\s?
TR[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){1}([0-9]{1})([a-zA-Z0-9]{3}\s?)([a-zA-Z0-9]{4}\s?){3}([a-zA-Z0-9]{2})\s?
AE[a-zA-Z0-9]{2}\s?([0-9]{3})([0-9]{1}\s?)([0-9]{4}\s?){3}([0-9]{3})\s?
GB[a-zA-Z0-9]{2}\s?([a-zA-Z]{4}\s?){1}([0-9]{4}\s?){3}([0-9]{2})\s?
VA[a-zA-Z0-9]{2}\s?([0-9]{3})([0-9]{1}\s?)([0-9]{4}\s?){3}([0-9]{2})\s?
VG[a-zA-Z0-9]{2}\s?([a-zA-Z0-9]{4}\s?){1}([0-9]{4}\s?){4}\s?
Original:
^[DE]{2}([0-9a-zA-Z]{20})$
Debuggex Demo
Modified:
^DE(?:\s*[0-9a-zA-Z]\s*){20}$
Debuggex Demo
This is the correct regex to match DE IBAN account numbers:
DE\d{2}[ ]\d{4}[ ]\d{4}[ ]\d{4}[ ]\d{4}[ ]\d{2}|DE\d{20}
Pass: DE89 3704 0044 0532 0130 00|||DE89370400440532013000
Fail: DE89-3704-0044-0532-0130-00
Most simple solution I can think of:
^DE(\s*[[:alnum:]]){20}\s*$
In particular, your initial [DE]{2} is wrong, as it allows 'DD', 'EE', 'ED' as well as the intended 'DE'.
To allow any amount of spaces anywhere:
^ *D *E( *[A-Za-z0-9]){20} *$
As you want to allow lower letters, also DE might be lower?
^ *[Dd] *[Ee]( *[A-Za-z0-9]){20} *$
^ matches the start of the string
$ end anchor
in between each characters there are optional spaces *
[character class] defines a set/range of characters
To allow at most one space in between each characters, replace the quantifier * (any amount of) with ? (0 or 1). If supported, \s shorthand can be used to match [ \t\r\n\f] instead of space only.
Test on regex101.com, also see the SO regex FAQ
Using Google Apps Script, I pasted Laurent's code from github into a script and added the following code to test.
// Use the Apps Script IDE's "Run" menu to execute this code.
// Then look at the View > Logs menu to see execution results.
function myFunction() {
//https://github.com/arhs/iban.js/blob/master/README.md
// var IBAN = require('iban');
var t1 = IBAN.isValid('hello world'); // false
var t2 = IBAN.isValid('BE68539007547034'); // true
var t3 = IBAN.isValid('BE68 5390 0754 7034'); // true
Logger.log("Test 1 = %s", t1);
Logger.log("Test 2 = %s", t2);
Logger.log("Test 3 = %s", t3);
}
The only thing needed to run the example code was commenting out the require('iban') line:
// var IBAN = require('iban');
Finally, instead of using client handlers to attempt a RegEx validation of IBAN input, I use a a server handler to do the validation.
Is it possible to write a regular expression that matches all strings that does not only contain numbers? If we have these strings:
abc
a4c
4bc
ab4
123
It should match the four first, but not the last one. I have tried fiddling around in RegexBuddy with lookaheads and stuff, but I can't seem to figure it out.
(?!^\d+$)^.+$
This says lookahead for lines that do not contain all digits and match the entire line.
Unless I am missing something, I think the most concise regex is...
/\D/
...or in other words, is there a not-digit in the string?
jjnguy had it correct (if slightly redundant) in an earlier revision.
.*?[^0-9].*
#Chad, your regex,
\b.*[a-zA-Z]+.*\b
should probably allow for non letters (eg, punctuation) even though Svish's examples didn't include one. Svish's primary requirement was: not all be digits.
\b.*[^0-9]+.*\b
Then, you don't need the + in there since all you need is to guarantee 1 non-digit is in there (more might be in there as covered by the .* on the ends).
\b.*[^0-9].*\b
Next, you can do away with the \b on either end since these are unnecessary constraints (invoking reference to alphanum and _).
.*[^0-9].*
Finally, note that this last regex shows that the problem can be solved with just the basics, those basics which have existed for decades (eg, no need for the look-ahead feature). In English, the question was logically equivalent to simply asking that 1 counter-example character be found within a string.
We can test this regex in a browser by copying the following into the location bar, replacing the string "6576576i7567" with whatever you want to test.
javascript:alert(new String("6576576i7567").match(".*[^0-9].*"));
/^\d*[a-z][a-z\d]*$/
Or, case insensitive version:
/^\d*[a-z][a-z\d]*$/i
May be a digit at the beginning, then at least one letter, then letters or digits
Try this:
/^.*\D+.*$/
It returns true if there is any simbol, that is not a number. Works fine with all languages.
Since you said "match", not just validate, the following regex will match correctly
\b.*[a-zA-Z]+.*\b
Passing Tests:
abc
a4c
4bc
ab4
1b1
11b
b11
Failing Tests:
123
if you are trying to match worlds that have at least one letter but they are formed by numbers and letters (or just letters), this is what I have used:
(\d*[a-zA-Z]+\d*)+
If we want to restrict valid characters so that string can be made from a limited set of characters, try this:
(?!^\d+$)^[a-zA-Z0-9_-]{3,}$
or
(?!^\d+$)^[\w-]{3,}$
/\w+/:
Matches any letter, number or underscore. any word character
.*[^0-9]{1,}.*
Works fine for us.
We want to use the used answer, but it's not working within YANG model.
And the one I provided here is easy to understand and it's clear:
start and end could be any chars, but, but there must be at least one NON NUMERICAL characters, which is greatest.
I am using /^[0-9]*$/gm in my JavaScript code to see if string is only numbers. If yes then it should fail otherwise it will return the string.
Below is working code snippet with test cases:
function isValidURL(string) {
var res = string.match(/^[0-9]*$/gm);
if (res == null)
return string;
else
return "fail";
};
var testCase1 = "abc";
console.log(isValidURL(testCase1)); // abc
var testCase2 = "a4c";
console.log(isValidURL(testCase2)); // a4c
var testCase3 = "4bc";
console.log(isValidURL(testCase3)); // 4bc
var testCase4 = "ab4";
console.log(isValidURL(testCase4)); // ab4
var testCase5 = "123"; // fail here
console.log(isValidURL(testCase5));
I had to do something similar in MySQL and the following whilst over simplified seems to have worked for me:
where fieldname regexp ^[a-zA-Z0-9]+$
and fieldname NOT REGEXP ^[0-9]+$
This shows all fields that are alphabetical and alphanumeric but any fields that are just numeric are hidden. This seems to work.
example:
name1 - Displayed
name - Displayed
name2 - Displayed
name3 - Displayed
name4 - Displayed
n4ame - Displayed
324234234 - Not Displayed
I am wanting to create a regular expression for the following scenario:
If a string contains the percentage character (%) then it can only contain the following: %20, and cannot be preceded by another '%'.
So if there was for instance, %25 it would be rejected. For instance, the following string would be valid:
http://www.test.com/?&Name=My%20Name%20Is%20Vader
But these would fail:
http://www.test.com/?&Name=My%20Name%20Is%20VadersAccountant%25
%%%25
Any help would be greatly appreciated,
Kyle
EDIT:
The scenario in a nutshell is that a link is written to an encoded state and then launched via JavaScript. No decoding works. I tried .net decoding and JS decoding, each having the same result - The results stay encoded when executed.
Doesn't require a %:
/^[^%]*(%20[^%]*)*$/
Which language are you using?
Most languages have a Uri Encoder / Decoder function or class.
I would suggest you decode the string first and than check for valid (or invalid) characters.
i.e. something like /[\w ]/ (empty is a space)
With a regex in the first place you need to respect that www.example.com/index.html?user=admin&pass=%%250 means that the pass really is "%250".
Another solution if look-arounds are not available:
^([^%]|%([013-9a-fA-F][0-9a-fA-F]|2[1-9a-fA-F]))*$
Reject the string if it matches %[^2][^0]
I think that would find what you need
/^([^%]|%%|%20)+$/
Edit: Added case where %% is valid string inside URI
Edit2: And fixed it for case where it should fail :-)
Edit3:
In case you need to use it in editor (which would explain why you can't use more programmatic way), then you have to correctly escape all special characters, for example in Vim that regex should lool:
/^\([^%]\|%%\|%20\)\+$/
Maybe a better approach is to deal with that validation after you decode that string:
string name = HttpUtility.UrlDecode(Request.QueryString["Name"]);
/^([^%]|%20)*$/
This requires a test against the "bad" patterns. If we're allowing %20 - we don't need to make sure it exists.
As others have said before, %% is valid too... and %%25would be %25
The below regex matches anything that doesn't fit into the above rules
/(?<![^%]%)%(?!(20|%))/
The first brackets check whether there is a % before the character (meaning that it's %%) and also checks that it's not %%%. it then checks for a %, and checks whether the item after doesn't match 20
This means that if anything is identified by the regex, then you should probably reject it.
I agree with dominic's comment on the question. Don't use Regex.
If you want to avoid scanning the string twice, you can just iteratively search for % and then check that it is being followed by 20 and nothing else. (Update: allow a % after to be interpreted as a literal %nnn sequence)
// pseudo code
pos = 0
while (pos = mystring.find(pos, '%'))
{
if mystring[pos+1] = "%" then
pos = pos + 2 // ok, this is a literal, skip ahead
else if mystring.substring(pos,2) != "20"
return false; // string is invalid
end if
}
return true;