We Keep Coding

Get integer number using regex in javascript

I am trying to write a regex to get only integer numbers e.g, 23, 234, 45, etc, and not select numbers with decimal points.
For Context :
I need it in a larger regex that I am writing to convert mixed fraction latex input
For example:
5\frac{7}{8}
But it should not select latex such as:
3.5\frac{7}{8}
The regex That I have so far is:
(^(.*)(?!(\.))(.*))\\frac{([^{}]+(?:{(?:[^{}]+)}|))}{([^{}]+(?:{(?:[^{}]+)}|))}
But it is for integer and decimal numbers alike. Need to change the regex for group1.

Maybe this will do it for you:
(?<!\d\.)\b(\d+)\\frac{([^{}]+(?:{(?:[^{}]+)}|))}{([^{}]+(?:{(?:[^{}]+)}|))}
It captures an integer expression before \fraq, unless it's preceded by a digit and a full stop.
(?<!\d\.) This ensures the number isn't preceded by a digit followed by a full stop, i.e.
the integer part of a floating.
\b Must be at the start of a number (to make sure we don't get a match with the end
of a multi digit number).
(\d+) Captures the integer number
\\frac Matches the string "\fraq"
The rest is the same as you original expression.
See it here at regex101.
Edit
Since there obviously are people out there, however unbelievable, that still haven't moved to a real browser ;) - the answer has to change to:
It depends of the syntax of latex, whether you can do it or not.
(And since I don't know that, I shouldn't have said anything in the first place ;)
The problem is that, without look behinds, you can't do it without matching characters outside the expression your interested in. In your regexr-example you clearly show that you want to be able to match expression not only at the beginning of strings, but also in the middle of the. Thus we need to be able to tell that the part before our match isn't the integer part of a decimal number. Now, doing this with matching isn't a problem. E.g.
(?:^|[^\d.])(\d+)\\frac{...
like Wiktor suggested in comments, will match if the expression is on the start of the line (^), or is preceded by something that isn't a decimal point or a digit ([^\d.]). That should do it. (Here at regex101.)
Well, as pointed out earlier, it depends on the syntax of latex. If two expressions can be directly adjacent, without any operators or stuff between them, you can't (as far as I can tell) do it with JS regex. Consider 1\fraq{0}{1}2\fraq{2}{3} (which I have no idea if it's syntactically correct). The first expression 1\fraq{0}{1} is a piece of a cake. But after that has been matched, we'd need to match a character before the second expression to verify the it doesn't start with a decimal number, but since the first expression already ate the characters, we can't. Because the test (?:^|[^\d.]) to verify that our expression doesn't start with a decimal number, would match one of the characters that actually belongs to our expression (the 2 in 2\fraq{2}{3}), thus making the match fail, because the remaining part doesn't start with the digit needed to satisfy the rest of the regex (\d+)\\frac{....
If, however, an expression always starts the string tested, or is preceded by and operator, or such, then it should be possible using
(?:^|[^\d.])(\d+)\\frac{([^{}]+(?:{(?:[^{}]+)})?)}{([^{}]+(?:{(?:[^{}]+)})?)}
Here at regex101.
(Sorry for my rambling)

Chaining multiple positive lookaheads in JavaScript regex

I'm new to learning Regular Expressions, and I came across this answer which uses positive lookahead to validate passwords.
The regular expression is - (/^(?=.*\d)(?=.*[a-z])(?=.*[A-Z])[0-9a-zA-Z]{8,}$/) and the breakdown provided by the user is -
(/^
(?=.*\d) //should contain at least one digit
(?=.*[a-z]) //should contain at least one lower case
(?=.*[A-Z]) //should contain at least one upper case
[a-zA-Z0-9]{8,} //should contain at least 8 from the mentioned characters
$/)
However, I'm not very clear on chaining multiple lookaheads together. From what I have learned, a positive lookahead checks if the expression is followed by what is specified in the lookahead. As an example, this answer says -
The regex is(?= all) matches the letters is, but only if they are immediately followed by the letters all
So, my question is how do the individual lookaheads work? If I break it down -
The first part is ^(?=.*\d). Does this indicate that at the starting of the string, look for zero or more occurrences of any character, followed by 1 digit (thereby checking the presence of 1 digit)?
If the first part is correct, then with the second part (?=.*[a-z]), does it check that after checking for Step 1 at the start of the string, look for zero or more occurrences of any character, followed by a lowercase letter? Or are the two lookaheads completely unrelated to each other?
Also, what is the use of the ( ) around every lookahead? Does it create a capturing group?
I have also looked at the Rexegg article on lookaheads, but it didn't help much.
Would appreciate any help.

As mentionned in the comments, the key point here are not the lookaheads but backtracking:
(?=.*\d) looks for a complete line (.*), then backtracks to find at least one number (\d).
This is repeated throughout the different lookaheads and could be optimized like so:
(/^
(?=\D*\d) // should contain at least one digit
(?=[^a-z]*[a-z]) // should contain at least one lower case
(?=[^A-Z]*[A-Z]) // should contain at least one upper case
[a-zA-Z0-9]{8,} // should contain at least 8 from the mentioned characters
$/)
Here, the principle of contrast applies.

Assertions are atomic, independent expressions with separate context
from the rest of the regex.
It is best visualized as: They exist between characters.
Yes, there is such a place.
Being independent though, they receive the current search position,
then they start moving through the string trying to match something.
They literally advance their private (local) copy of the search position
to do this.
They return a true or false, depending on if they matched something.
The caller of this assertion maintains it's own copy of the search position.
So, when the assertion returns, the callers search position has not changed.
Thus, you can weave in and out of places without having to worry about
the search position.
You can see this a little more dramatically, when assertions are nested:
Target1: Boy1 has a dog and a train
Target2: Boy2 has a dog
Regex: Boy\d(?= has a dog(?! and a train))
Objective: Find the Boy# that matches the regex.
Other noteworthy things about assertions:
They are atomic (ie: independent) in that they are immune to backtracking
from external forces.
Internally, they can backtrack just like anywhere else.
But, when it comes to the position they were given, that cannot change.
Also, inside assertions, it is possible to capture just like anywhere else.
Example ^(?=.*\b(\w+)\b) captures the last word in string, however the search position does not change.
Also, assertions are like a red light. The immediate expression that follows the assertion
must wait until it gets the green light.
This is the result the assertion passes back, true or false.

Since "a+?" is Lazy, Why does "a+?b" Match "aaab"?

While learning regular expressions in javascript using JavaScript: The Definitive Guide, I was confused by this passage:
But /a+?/ matches one or more occurrences of the letter a, matching as
few characters as necessary. When applied to the same string, this
pattern matches only the first letter a.
…
Now let’s use the nongreedy version: /a+?b/. This should match the
letter b preceded by the fewest number of a’s possible. When applied
to the same string “aaab”, you might expect it to match only one a and
the last letter b. In fact, however, this pattern matches the entire
string, just like the greedy version of the pattern.
Why is this so?
This is the explanation from the book:
This is because regular-expression pattern matching is done by finding
the first position in the string at which a match is possible. Since a
match is possible starting at the first character of the
string,shorter matches starting at subsequent characters are never
even considered.
I don't understand. Can anyone give me a more detailed explanation?

Okay, so you have your search space, "aaabc", and your pattern, /a+?b/
Does /a+?b/ match "a"? No.
Does /a+?b/ match "aa"? No.
Does /a+?b/ match "aaa"? No.
Does /a+?b/ match "aaab"? Yes.

Since you're matching literal characters and not any sort of wildcard, the regular expression a+?b is effectively the same as a+b anyway. The only type of sequence either one will match is a string of one or more a characters followed by a single b character. The non-greedy modifier makes no difference here, as the only thing an a can possibly match is an a.
The non-greedy qualifier becomes interesting when it's applied to something that can take on lots of different values, like .. (edit or cases where there's interesting stuff to the left of something like a+?)
edit — if you're expecting a+?b to match just the last a before the b in aaab, well that's not how it works. Searching for a pattern in a string implicitly means to search for the earliest occurrence of the pattern. Thus, though starting from the last a does give a substring that matches the pattern, it's not the first substring that matches.

The Engine Attempts a Match at the Beginning of the String
Can anyone give me a more detailed explanation?
Yes.
In short: .+? does not look for a shortest match globally, at the level of the entire string, but locally, from the position in the string where the engine is currently positioned.
How the Engine Works
When you try a regex against the string aaab, the engine first tries to find a match starting at the very first position in the string. That position is the position before the first a. If the engine cannot find a match at the first position, it moves on and tries again starting from the second position (between the first and second a)
So can a match be found by the regex a+?b at the first position? Yes.
a matches the first a
The +? quantifiers tells the engine to match the fewest number of a chars necessary. Since we are looking to return a match, necessary means that the following tokens (in this case) have to be allowed to match. In this case, the fewest number of a chars needed to allow the b to match is all the remaining a chars.
b matches
In the details the second point is a bit more complex (the engine tries to match b against the second a, fails, backtracks...) but you don't need to worry about that.

'?' after a+ means minimum number of characters to satisfy expression. /a+/ means one 'a' or as many as you can encounter before some other character. In order to satisfy /a+?/ (since it's nogreedy) it only needs single 'a'.
In order to satisfy /a+?b/, since we have 'b' at the end, in order to satisfy this expression it needs to match one or more 'a' before it hits 'b'. It has to hit that 'b'. /a+/ doesn't have to hit b because RegEx doesn't ask for that. /a+?b/ has to hit that 'b'.
Just think about it. What other meaning /a+?b/ could have?
Hope this helps

Find out the position where a regular expression failed

I'm trying to write a lexer in JavaScript for finding tokens of a simple domain-specific language. I started with a simple implementation which just tries to match subsequent regexps from the current position in a line to find out whether it matches some token format and accept it then.
The problem is that when something doesn't match inside such regexp, the whole regexp fails, so I don't know which character exactly caused it to fail.
Is there any way to find out the position in the string which caused the regular expression to fail?
INB4: I'm not asking about debugging my regexp and verifying its correctness. It is correct already, matches correct strings and drops incorrect ones. I just want to know programmatically where exactly the regexp stopped matching, to find out the position of a character which was incorrect in the user input, and how much of them were OK.
Is there some way to do it with just simple regexps instead of going on with implementing a full-blown finite state automaton?

Short answer
There is no such thing as a "position in the string that causes the
regular expression to fail".
However, I will show you an approach to answer the reverse question:
At which token in the regex did the engine become unable to match the
string?
Discussion
In my view, the question of the position in the string which caused the regular expression to fail is upside-down. As the engine moves down the string with the left hand and the pattern with the right hand, a regex token that matches six characters one moment can later, because of quantifiers and backtracking, be reduced to matching zero characters the next—or expanded to match ten.
In my view, a more proper question would be:
At which token in the regex did the engine become unable to match the
string?
For instance, consider the regex ^\w+\d+$ and the string abc132z.
The \w+ can actually match the entire string. Yet, the entire regex fails. Does it make sense to say that the regex fails at the end of the string? I don't think so. Consider this.
Initially, \w+ will match abc132z. Then the engine advances to the next token: \d+. At this stage, the engine backtracks in the string, gradually letting the \w+ give up the 2z (so that the \w+ now only corresponds to abc13), allowing the \d+ to match 2.
At this stage, the $ assertion fails as the z is left. The engine backtracks, letting the \w+, give up the 3 character, then the 1 (so that the \w+ now only corresponds to abc), eventually allowing the \d+ to match 132. At each step, the engine tries the $ assertion and fails. Depending on engine internals, more backtracking may occur: the \d+ will give up the 2 and the 3 once again, then the \w+ will give up the c and the b. When the engine finally gives up, the \w+ only matches the initial a. Can you say that the regex "fails on the "3"? On the "b"?
No. If you're looking at the regex pattern from left to right, you can argue that it fails on the $, because it's the first token we were not able to add to the match. Bear in mind that there are other ways to argue this.
Lower, I'll give you a screenshot to visualize this. But first, let's see if we can answer the other question.
The Other Question
Are there techniques that allow us to answer the other question:
At which token in the regex did the engine become unable to match the
string?
It depends on your regex. If you are able to slice your regex into clean components, then you can devise an expression with a series of optional lookaheads inside capture groups, allowing the match to always succeed. The first unset capture group is the one that caused the failure.
Javascript is a bit stingy on optional lookaheads, but you can write something like this:
^(?:(?=(\w+)))?(?:(?=(\w+\d+)))?(?:(?=(\w+\d+$)))?.
In PCRE, .NET, Python... you could write this more compactly:
^(?=(\w+))?(?=(\w+\d+))?(?=(\w+\d+$))?.
What happens here? Each lookahead builds incrementally on the last one, adding one token at a time. Therefore we can test each token separately. The dot at the end is an optional flourish for visual feedback: we can see in a debugger that at least one character is matched, but we don't care about that character, we only care about the capture groups.
Group 1 tests the \w+ token
Group 2 seems to test \w+\d+, therefore, incrementally, it tests the \d+ token
Group 3 seems to test \w+\d+$, therefore, incrementally, it tests the $ token
There are three capture groups. If all three are set, the match is a full success. If only Group 3 is not set (as with abc123a), you can say that the $ caused the failure. If Group 1 is set but not Group 2 (as with abc), you can say that the \d+ caused the failure.
For reference: Inside View of a Failure Path
For what it's worth, here is a view of the failure path from the RegexBuddy debugger.

You can use a negated character set RegExp,
[^xyz]
[^a-c]
A negated or complemented character set. That is, it matches anything
that is not enclosed in the brackets. You can specify a range of
characters by using a hyphen, but if the hyphen appears as the first
or last character enclosed in the square brackets it is taken as a
literal hyphen to be included in the character set as a normal
character.
index property of String.prototype.match()
The returned Array has an extra input property, which contains the
original string that was parsed. In addition, it has an index
property, which represents the zero-based index of the match in the
string.
For example to log index where digit is matched for RegExp /[^a-zA-z]/ in string aBcD7zYx
var re = /[^a-zA-Z]/;
var str = "aBcD7zYx";
var i = str.match(re).index;
console.log(i); // 4

Is there any way to find out the position in the string which caused the regular expression to fail?
No, there isn't. A Regex either matches or doesn't. Nothing in between.
Partial Expressions can match, but the whole pattern doesnt. So the engine always needs to evaluates the whole expression:
Take the String Hello my World and the Pattern /Hello World/. While each word will match individually, the whole Expression fails. You cannot tell whether Hello or World matched - independent, both do. Also the whitespace between them is available.

Javascript Regex for Javascript Regex and Digits

The title might seem a bit recursive, and indeed it is.
I am working on a Javascript which can highlight/color Javascript code displayed in HTML. Thus, in the Internet Browser, comments will be turned green, definitions (for, if, while, etc.) will be turned a dark blue and italic, numbers will be red, and so on for other elements. However, the coloring is not all that important.
I am trying to figure out two different regular expressions which have started to cause a minor headache.
1. Finding a regular expression using a regular expression
I want to find regular expressions within the script-tags of HTML using a Javascript, such as:
match(/findthis/i);
, where the regex part of course is "/findthis/i".
The rules are as follows:
Finding multiple occurrences (/g) is not important.
It must be on the same line (not /m).
Caseinsensitive (/i).
If a backward slash (ignore character) is followed directly by a forward slash, "/", the forward slash is part of the expression - not an escape character. E.g.: /itdoesntstop\/untilnow:/
Two forward slashes right next to each other (//) is: (A) At the beginning: Not a regex; it's a comment. (B) Later on: First slash is the end of the regex and the second slash is nothing but a character.
Regex continues until the line breaks or end of input (\n|$), or the escape character (second forward slash which complies with rule 4) is encountered. However, also as long as only alphabetic characters are encountered, following the second forward slash, they are considered part of the regex. E.g.: /aregex/allthisispartoftheregex
So far what I've got is this:
'\\/(?:[^\\/\\\\]|\\/\\*)*\\/([a-zA-Z]*)?'
However, it isn't consistent. Any suggestions?
2. Find digits (alphanumeric, floating) using a regular expression
Finding digits on their own is simple. However, finding floating numbers (with multiple periods) and letters including underscore is more of a challenge.
All of the below are considered numbers (a new number starts after each space):
3 3.1 3.1.4 3a 3.A 3.a1 3_.1
The rules:
Finding multiple occurrences (/g) is not important.
It must be on the same line (not /m).
Caseinsensitive (/i).
A number must begin with a digit. However, the number can be preceeded or followed by a non-word (\W) character. E.g.: "=9.9;" where "9.9" is the actual number. "a9" is not a number. A period before the number, ".9", is not considered part of the number and thus the actual number is "9".
Allowed characters: [a-zA-Z0-9_.]
What I've got:
'(^|\\W)\\d([a-zA-Z0-9_.]*?)(?=([^a-zA-Z0-9_.]|$))'
It doesn't work quite the way I want it.

For the first part, I think you are quite close. Here is what I would use (as a regex literal, to avoid all the double escapes):
/\/(?:[^\/\\\n\r]|\\.)+\/([a-z]*)/i
I don't know what you intended with your second alternative after the character class. But here the second alternative is used to consume backslashes and anything that follows them. The last part is important, so that you can recognize the regex ending in something like this: /backslash\\/. And the ? at the end of your regex was redundant. Otherwise this should be fine.
Test it here.
Your second regex is just fine for your specification. There are a few redundant elements though. The main thing you might want to do is capture everything but the possible first character:
/(?:^|\W)(\d[\w.]*)/i
Now the actual number (without the first character) will be in capturing group 1. Note that I removed the ungreediness and the lookahead, because greediness alone does exactly the same.
Test it here.