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Use only one of the characters in regular expression javascript

I guess that should be smth very easy, but I'm stuck with that for at least 2 hours and I think it's better to ask the question here.
So, I've got a reg expression /&t=(\d*)$/g and it works fine while it is not ?t instead of &t in url. I've tried different combinations like /\?|&t=(\d*)$/g ; /\?t=(\d*)$|/&t=(\d*)$/g ; /(&|\?)t=(\d*)$/g and various others. But haven't got the expected result which is /\?t=(\d*)$/g or /&t=(\d*)$/g url part (whatever is placed to input).
Thx for response. I think need to put some details here. I'm actually working on this peace of code
var formValue = $.trim($("#v").val());
var formValueTime = /&t=(\d*)$/g.exec(formValue);
if (formValueTime && formValueTime.length > 1) {
formValueTime = parseInt(formValueTime[1], 10);
formValue = formValue.replace(/&t=\d*$/g, "");
}
and I want to get the t value whether reference passed with &t or ?t in references like youtu.be/hTWKbfoikeg?t=82 or similar one youtu.be/hTWKbfoikeg&t=82

To replace, you may use
var formValue = "some?some=more&t=1234"; // $.trim($("#v").val());
var formValueTime;
formValue = formValue.replace(/[&?]t=(\d*)$/g, function($0,$1) {
formValueTime = parseInt($1,10);
return '';
});
console.log(formValueTime, formValue);
To grab the value, you may use
/[?&]t=(\d*)$/g.exec(formValue);
Pattern details
[?&] - a character class matching ? or &
t= - t= substring
(\d*) - Group 1 matching zero or more digits
$ - end of string

/\?t=(\d*)|\&t=(\d*)$/g
you inverted the escape character for the second RegEx.
http://regexr.com/3gcnu

I want to thank you all guys for trying to help. Special thanks to #Wiktor Stribiżew who gave the closest answer.
Now the piece of code I needed looks exactly like this:
/[?&]t=(\d*)$/g.exec(formValue);
So that's the [?&] part that solved the problem.
I use array later, so /\?t=(\d*)|\&t=(\d*)$/g doesn't help because I get an array like [t&=50,,50] when reference is & type and the correct answer [t?=50,50] when reference is ? type just because of the order of statements in RegExp.
Now, if you're looking for a piece of RegExp that picks either character in one place while the rest of RegExp remains the same you may use smth like this [?&] for the example where wanted characters are ? and &.

Matching and returning a regex between two values

I am trying to get the values from a string using regex, the value is that of the text between tt=" and "&
So, for example, "tt="Value"&" I would only want to get the word "Value" out of this.
So far I have this: /tt=.*&/ which gives me "tt=Value"&, Then, to get the value I am thinking to split the match on = and remove the 2 characters from the end. I feel though, that this would be an awful way to do this and would like to see if it could be done in the regex?

You're on the right track for matching the entire context inside of the string, but you want to use a capturing group to match/capture the value between the quotes instead of splitting on = and having to remove the two quote chars.
var r = 'tt="Value"&'.match(/tt="([^"]*)"/)[1];
if (r)
console.log(r); //=> "Value"

I know this isn't really the answer you are looking for since it doesn't involve regex but it's the way I usually do it.
strvariable = strvariable.Remove(0,strvariable.IndexOf("=") + 2);
strvariable = strvariable.Remove(strvariable.IndexOf("\""), strvariable.Length - strvariable.IndexOf("\""));
this would give you the result you were looking for which is Value in this instance.

regex search a string for contents between two strings

I am trying my upmost best to get my head around regex, however not having too much luck.
I am trying to search within a string for text, I know how the string starts, and i know how the string ends, I want to return ALL the text inbetween the string including the start and end.
Start search = [{"lx":
End search = }]
i.e
[{"lx":variablehere}]
So far I have tried
/^\[\{"lx":(*?)\}\]/;
and
/(\[\{"lx":)(*)(\}\])/;
But to no real avail... can anyone assist?
Many thanks

You're probably making the mistake of believing the * is a wildcard. Use the period (.) instead and you'll be fine.
Also, are you sure you want to stipulate zero or more? If there must be a value, use + (one or more).
Javascript:
'[{"lx":variablehere}]'.match(/^\[\{"lx":(.+?)\}\]/);

The * star character multiplies the preceding character. In your case there's no such character. You should either put ., which means "any character", or something more specific like \S, which means "any non whitespace character".

Possible solution:
var s = '[{"lx":variablehere}]';
var r = /\[\{"(.*?)":(.*?)\}\]/;
var m = s.match(r);
console.log(m);
Results to this array:
[ '[{"lx":variablehere}]',
'lx',
'variablehere',
index: 0,
input: '[{"lx":variablehere}]' ]

\[\{"lx"\:(.*)\}\]
This should work for you. You can reach the captured variable by \1 notation.

Try this:
^\[\{\"lx\"\:(.*)\}\]$
all text between [{"lx": and }] you will find in backreference variable (something like \$1 , depends on programming language).

how to extract this kind of data and put them into a nice array?

I got a string like this one:
var tweet ="#fadil good:))RT #finnyajja: what a nice day RT #fadielfirsta: how are you? #finnyajja yay";
what kind of code should work to extract any words with # character and also removing any special char at the end of the words? so it would an array like this :
(#fadil, #finnyajja, #fadielfirsta, #finnyajja);
i have tried the following code :
var users = $.grep(tweet.split(" "), function(a){return /^#/.test(a)});
it returns this:
(#fadil, #finnyajja:, #fadielfirsta:, #finnyajja)
there's still colon ':' character at the end of some words. What should I do? any solution guys? Thanks

Here is code that is more straightforward than trying to use split:
var tweet_text ="#fadil good:))RT #finnyajja: what a nice day RT #fadielfirsta: how are you? #finnyajja yay";
var result = tweet_text.match(/#\w+/g);

The easiest way without changing your current code too much would be to just remove all colons prior to calling split:
var users = $.grep(tweet_text.replace(":","").split(" "), function(a){return /^#/.test(a)});
You could also write a regex to do all the work for you using match. Something like this:
var regex = /#[a-z0-9]+/gi;
var matches = tweet.match(regex);
This assumes that you only want letters and numbers, if certain other characters are allowed, this regex will need to be modified.
http://jsfiddle.net/YHM87/

Using Regular Expressions with Javascript replace method

Friends,
I'm new to both Javascript and Regular Expressions and hope you can help!
Within a Javascript function I need to check to see if a comma(,) appears 1 or more times. If it does then there should be one or more numbers either side of it.
e.g.
1,000.00 is ok
1,000,00 is ok
,000.00 is not ok
1,,000.00 is not ok
If these conditions are met I want the comma to be removed so 1,000.00 becomes 1000.00
What I have tried so is:
var x = '1,000.00';
var regex = new RegExp("[0-9]+,[0-9]+", "g");
var y = x.replace(regex,"");
alert(y);
When run the alert shows ".00" Which is not what I was expecting or want!
Thanks in advance for any help provided.
strong text
Edit
strong text
Thanks all for the input so far and the 3 answers given. Unfortunately I don't think I explained my question well enough.
What I am trying to achieve is:
If there is a comma in the text and there are one or more numbers either side of it then remove the comma but leave the rest of the string as is.
If there is a comma in the text and there is not at least one number either side of it then do nothing.
So using my examples from above:
1,000.00 becomes 1000.00
1,000,00 becomes 100000
,000.00 is left as ,000.00
1,,000.00 is left as 1,,000.00
Apologies for the confusion!

Your regex isn't going to be very flexible with higher orders than 1000 and it has a problem with inputs which don't have the comma. More problematically you're also matching and replacing the part of the data you're interested in!
Better to have a regex which matches the forms which are a problem and remove them.
The following matches (in order) commas at the beginning of the input, at the end of the input, preceded by a number of non digits, or followed by a number of non digits.
var y = x.replace(/^,|,$|[^0-9]+,|,[^0-9]+/g,'');
As an aside, all of this is much easier if you happen to be able to do lookbehind but almost every JS implementation doesn't.
Edit based on question update:
Ok, I won't attempt to understand why your rules are as they are, but the regex gets simpler to solve it:
var y = x.replace(/(\d),(\d)/g, '$1$2');

I would use something like the following:
^[0-9]{1,3}(,[0-9]{3})*(\.[0-9]+)$
[0-9]{1,3}: 1 to 3 digits
(,[0-9]{3})*: [Optional] More digit triplets seperated by a comma
(\.[0-9]+): [Optional] Dot + more digits
If this regex matches, you know that your number is valid. Just replace all commas with the empty string afterwards.

It seems to me you have three error conditions
",1000"
"1000,"
"1,,000"
If any one of these is true then you should reject the field, If they are all false then you can strip the commas in the normal way and move on. This can be a simple alternation:
^,|,,|,$

I would just remove anything except digits and the decimal separator ([^0-9.]) and send the output through parseFloat():
var y = parseFloat(x.replace(/[^0-9.]+/g, ""));

// invalid cases:
// - standalone comma at the beginning of the string
// - comma next to another comma
// - standalone comma at the end of the string
var i,
inputs = ['1,000.00', '1,000,00', ',000.00', '1,,000.00'],
invalid_cases = /(^,)|(,,)|(,$)/;
for (i = 0; i < inputs.length; i++) {
if (inputs[i].match(invalid_cases) === null) {
// wipe out everything but decimal and dot
inputs[i] = inputs[i].replace(/[^\d.]+/g, '');
}
}
console.log(inputs); // ["1000.00", "100000", ",000.00", "1,,000.00"]