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Regex expression for exactly known pattern without "cutting into" the string not working

I am currently developing a web-application where I work with java, javascript, html, jquery, etc. and at some point I need to check that whether an input matches a known pattern and only proceed if it is true.
The pattern should be [at least one but max 3 numbers between 0-9]/[exactly 4 numbers between 0-9], so the only acceptable variations should be like
1/2014 or 23/2015 or 123/2016.
and nothing else, and I CANNOT accept something like 1234/3012 or anything else, and this is my problem right here, it accepts everything in which it can find the above pattern, so like from 12345/6789 it accepts and saves 345/6789.
I am a total newbie with regex, so I checked out http://regexr.com and this is the code I have in my javascript:
$.validator.addMethod("hatarozat", function(value, element) {
return (this.optional(element) || /[0-9]{1,3}(?:\/)[0-9]{4}/i.test(value));
}, "Hibás határozat szám!");
So this is my regex: /[0-9]{1,3}(?:\/)[0-9]{4}/i
which I built up using the above website. What could be the problem, or how can I achived what I described? I tried /^[0-9]{1,3}(?:\/)[0-9]{4}$/ibut this doesn't seem to work, please anyone help me, I have everything else done and am getting pretty stressed over something looking so simple yet I cannot solve it. Thank you!

Your last regex with the anchors (^ and $) is a correct regex. What prevents your code from working is this.optional(element) ||. Since this is a static thing, and is probably true, so it does not show any error (as || is an OR condition, if the first is true, the whole returns true, the regex is not checked at all).
So, use
return /^[0-9]{1,3}\/[0-9]{4}$/.test(value);
Note you do not need the (?:...) with \/ as the grouping does not do anything important here and is just redundant. The anchors are important, since you want the whole string to match the pattern (and ^ anchors the regex at the start of the string and $ does that at the end of the string.)

You need use the the following special characters in your regex expression:
^ and $
or \b
so 2 regexp will be correct:
/\b[0-9]{1,3}(?:\/)[0-9]{4}\b/i;
or
/^[0-9]{1,3}(?:\/)[0-9]{4}$/i

Regex for one string or a pattern

i am trying to write a regent for a line in this format:
Integer;characters etc;decimal number;decimal number;decimal
number;characters etc
For example:
989898;ABCD - eft;9809.98078;9878.876;8768.654;ABCD
For this, I have writen this which is working fine :
/[0-9]*;(.)*;[0-9\.]*;[0-9\.]*;[0-9\.]*;(.)/gi
Now, this problem is, there can be N.A. in the place of decimal number for which I am unable to get the refer right.

You can use alternation for this. Also, you can improve your regex a bit - for example [^;]* is more specific than .*, you can avoid repetition, and you don't need to escape the dot inside a character class:
/[0-9]*;[^;]*;(?:N\.A\.;|[0-9.]*;){3}.*/gi

Capturing optional part of URL with RegExp

While writing an API service for my site, I realized that String.split() won't do it much longer, and decided to try my luck with regular expressions. I have almost done it but I can't find the last bit. Here is what I want to do:
The URL represents a function call:
/api/SECTION/FUNCTION/[PARAMS]
This last part, including the slash, is optional. Some functions display a JSON reply without having to receive any arguments. Example: /api/sounds/getAllSoundpacks prints a list of available sound packs. Though, /api/sounds/getPack/8Bit prints the detailed information.
Here is the expression I have tried:
req.url.match(/\/(.*)\/(.*)\/?(.*)/);
What am I missing to make the last part optional - or capture it in whole?

This will capture everything after FUNCTION/ in your URL, independent of the appearance of any further / after FUNCTION/:
FUNCTION\/(.+)$
The RegExp will not match if there is no part after FUNCTION.

This regex should work by making last slash and part after optional:
/^\/[^/]*\/[^/]*(?:\/.*)?$/
This matches all of these strings:
/api/SECTION/FUNCTION/abc
/api/SECTION
/api/SECTION/
/api/SECTION/FUNCTION

Your pattern /(.*)/(.*)/?(.*) was almost correct, it's just a bit too short - it allows 2 or 3 slashes, but you want to accept anything with 3 or 4 slashes. And if you want to capture the last (optional) slash AND any text behind it as a whole, you simply need to create a group around that section and make it optional:
/.*/.*/.*(?:/.+)?
should do the trick.
Demo. (The pattern looks different because multiline mode is enabled, but it still works. It's also a little "better" because it won't match garbage like "///".)

JScript: how to check if a string contains a specific word?

I am trying to find a way to check if a string contains a specific sequence of characters in JScript.
In my case, I am trying to see if the string is "DPObject" followed by a number. Such as "DPObject3" or "DPObject14".
Thank you!

if (/DPObject\d+/.test(string)) {....}

Javascript String has an indexOf method you can use to check if a String contains a particular substring .
If you need to test for patterns , like "DPObject" followed by an integer , probably you need to use Regexes . ( http://www.regular-expressions.info )
It's javascript , or js for short - not JScript .

Then you should use a regular expression. I think this would be something like :
var re = new RegExp("^DPObject([0-9]+)$");
re.test(someString);
This ensures there is at least only one digit after DPObject.
The "^" at the beginning is to ensure the string starts with DPObject. Check references on regexps for this kind of problems :)
edit: added "$" to mark the end of the string, the updated should be more "solid"

There are a couple of ways:
Use Javascripts indexOf method
Use Javascript Regular Expressions
Use JQuery's contains function
Regular expressions are the most powerful and elegant way of doing it. They syntax makes sense after a while (honestly). ;-)
Good luck.

Javascript Regular Expressions Lookbehind Failing

I am hoping that this will have a pretty quick and simple answer. I am using regular-expressions.info to help me get the right regular expression to turn URL-encoded, ISO-8859-1 pound sign ("%A3"), into a URL-encoded UTF-8 pound sign ("%C2%A3").
In other words I just want to swap %A3 with %C2%A3, when the %A3 is not already prefixed with %C2.
So I would have thought the following would work:
Regular Expression: (?!(\%C2))\%A3
Replace With: %C2%A3
But it doesn't and I can't figure out why!
I assume my syntax is just slightly wrong, but I can't figure it out! Any ideas?
FYI - I know that the following will work (and have used this as a workaround in the meantime), but really want to understand why the former doesn't work.
Regular Expression: ([^\%C2])\%A3
Replace With: $1%C2%A3
TIA!

Why not just replace ((%C2)?%A3) with %C2%A3, making the prefix an optional part of the match? It means that you're "replacing" text with itself even when it's already right, but I don't foresee a performance issue.

Unfortunately, the (?!) syntax is negative lookahead. To the best of my knowledge, JavaScript does not support negative lookbehind.
What you could do is go forward with the replacement anyway, and end up with %C2%C2%A3 strings, but these could easily be converted in a second pass to the desired %C2%A3.

You could replace
(^.?.?|(?!%C2)...)%A3
with
$1%C2%A3

I would suggest you use the functional form of Javascript String.replace (see the section "Specifying a function as a parameter"). This lets you put arbitrary logic, including state if necessary, into a regexp-matching session. For your case, I'd use a simpler regexp that matches a superset of what you want, then in the function call you can test whether it meets your exact criteria, and if it doesn't then just return the matched string as is.
The only problem with this approach is that if you have overlapping potential matches, you have the possibility of missing the second match, since there's no way to return a value to tell the replace() method that it isn't really a match after all.