I am handling user input in my program by using regular exp.
the string contains /_MyWord/ and only a-z is accepted before /_MyWord/.
the string not contain /s/123, /s/32A and atr/will in the beginning.
My try:
^(?!.*/s/123)(?!.*/s/32A )(?!.*atr/will)([/a-z]+)/_MyWord/(.*)$
Example:
/s/123/QWERERTYU/_MyWord/45454545 -> fail
/DFGH/FGHJK/GHJK/_MyWord/DFGHJ452 -> OK
HiCanYouHelpMe/_MyWord/fgh -> OK
/_MyWord/HiCanYouHelpMefgh -> OK
Can anyone help me to finish the Regular Exp string
If I got your question correctly, try this regex:
^(?!.*\/s\/123)(?!.*\/s\/32A)(?!.*atr\/will)([\/a-zA-Z]*)\/_MyWord\/(.*)$
Unescaped: ^(?!.*/s/123)(?!.*/s/32A)(?!.*atr/will)([/a-zA-Z]*)/_MyWord/(.*)$
Changed ([\/a-z]+) to ([\/a-zA-Z]*) to include lower and upper case as well as support none (e.g /_MyWord/Test)
Regex101 Demo
Works for
/DFGH/FGHJK/GHJK/_MyWord/DFGHJ452
HiCanYouHelpMe/_MyWord/fgh
/_MyWord/HiCanYouHelpMefgh
Doesn't match:
/s/123/QWERERTYU/_MyWord/45454545
atr/will/DFGH/FGHJK/GHJK/_MyWord/DFGHJ452
Also, you really don't need lookaheads for /s/123 and /s/32A since they contain numbers so they will automatically be rejected because your condition includes [a-zA-Z]. So you might want to remove (?!.*\/s\/123)(?!.*\/s\/32A) from the beginning.
Related
I need split sentence to words removing redundant characters.
I prepared regexp for that:
val wordCharacters = """[^A-z'\d]""".r
right now I have rule which can be used to handle task in next way:
wordCharacters.split(words)
.filterNot(_.isEmpty)
where words any sentence I need to parse.
But issue is that in case I try to handle "car: carpet, as,,, java: javascript!!&#$%^&" I get one more word ^. Trying to change my regex and without ^ I'm getting much more issues for different cases...
Is any ideas how to solve it?
P.S.
If somebody want to play with it try link or code below please:
val wordCharacters = """[^A-z'\d]""".r
val stringToInt =
wordCharacters.split("car: carpet, as,,, java: javascript!!&#$%^&")
.filterNot(_.isEmpty)
.toList
println(stringToInt)
Expected result is:
List(car, carpet, as, java, javascript)
The part A-z is not exactly what you want. Probably you assume that lower a comes immediately after upper Z, but there are some other characters in between, and one of them is ^.
So, correcting the regex as
"""[^A-Za-z'\d]""".r
would fix the issue.
Have a look at the order of characters:
https://en.wikipedia.org/wiki/List_of_Unicode_characters
I'd be tempted to start with \W and expand from there.
"\\W+".r.split("car: carpet, as,,, java: javascript!!&#$%^&")
//res0: Array[String] = Array(car, carpet, as, java, javascript)
In have following code:
set[str] noNnoE = { v | str v <- eu, (/\b[^eEnN]*\b/ := v) };
The goal is to filter out of a set of strings (called 'eu'), those strings that have no 'e' or 'n' in them (both upper- and lowercase). The regular expression I've provided:
/\b[^eEnN]?\b/
seems to work like it should, when I try it out in an online regex-tester.
When trying it out in the Rascel terminal it doesn't seem to work:
rascal>/\b[^eEnN]*\b/ := "Slander";
bool: true
I expected no match. What am I missing here? I'm using the latest (stable) Rascal release in Eclipse Oxygen1a.
Actually, the online regex-tester is giving the same match that we are giving. You can look at the match as follows:
if (/<w1:\b[^eEnN]?\b>/ := "Slander")
println("The match is: |<w1>|");
This is assigning the matched string to w1 and then printing it between the vertical bars, assuming the match succeeds (if it doesn't, it returns false, so the body of the if will not execute). If you do this, you will get back a match to the empty string:
The match is: ||
The online regex tester says the same thing:
Match 1
Full match 0-0 ''
If you want to prevent this, you can force at least one occurrence of the characters you are looking for by using a +, versus a ?:
rascal>/\b[^eEnN]+\b/ := "Slander";
bool: false
Note that you can also make the regex match case insensitive by following it with an i, like so:
/\b[^en]+\b/i
This may make it easier to write if you need to add more characters into the character class.
This solution (/\b[^en]+\b/i) doesn't work for strings consisting of two words, such as the Czech Republic.
Try /\b[^en]+\b$/i. That seems to work for me.
I am trying to have my regex match the following:
169.254.0.0-169.254.254.255
Could anyone please help how can I achieve this.
so far I have this:
169\.254\.([1-9]{1,2}|[1-9]{1,2}[1-4])
but it would also pick up 169.254.255.1 which should not be one of the matches.
Please help!
thanks
This is the regex I use for general IP validation:
(([0-9](?!\d)|[1-9][0-9](?!\d)|1[0-9]{2}(?!\d)|2[0-4][0-9](?!\d)|25[0-5](?!\d))[.]?){4}
Breakdown:
1.`[0-9](?!\d)` -> Any Number 0 through 9 (The `(?!\d)` makes sure it only grabs stand alone digits)
2.`|[1-9][0-9](?!\d)` -> Or any number 10-99 (The `(?!\d)` makes sure it only grabs double digit entries)
3.`|1[0-9]{2}` -> Or any number 100-199
4.`|2[0-4][0-9]` -> Or any number 200-249
5.`|25[0-5]` -> Or any number 250-255
6.`[.]?` -> With or without a `.`
7.`{4}` -> Lines 1-6 exactly 4 times
This hasn't failed my yet for IP address validation.
For your specific case, this should do it:
(169\.254\.)((([0-9](?!\d)|[1-9][0-9](?!\d)|1[0-9]{2}|2[0-4][0-9]|25[0-4])[.])(([0-9](?!\d)|[1-9][0-9](?!\d)|1[0-9]{2}|2[0-4][0-9]|25[0-5])))
This is very long because I couldn't figure out how to get 169.254.(0-254).255 to check without getting 169.254.255.1 to fail
Edit: Fixed due to comments
the regex ([0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-4]) matches 0-254.
see this page for more discussion
I've written an article that provides regular expressions for all the components of a generic URI (as defined in RFC3986: Uniform Resource Identifier (URI): Generic Syntax)
See: Regular Expression URI Validation
One of the components of a generic URI is an IPv4 address. Here is the free-spacing mode Python version from that article:
re_python_rfc3986_IPv4address = re.compile(r""" ^
# RFC-3986 URI component: IPv4address
(?: (?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?) \.){3} # (dec-octet "."){3}
(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?) # dec-octet "."
$ """, re.VERBOSE)
And the un-commented JavaScript version:
var re_js_rfc3986_IPv4address = /^(?:(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.){3}(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)$/;
I need to check an input field for a German IBAN. The user should be allowed to leave in white spaces and input should be validated to have a starting DE and then exact 20 characters numbers and letters.
Without the white space allowance, I tried
^[DE]{2}([0-9a-zA-Z]{20})$
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