How to parse for a word in text in JavaScript? - javascript

In the text page, I would like to examine each word. What is the best way to read each word at the time? It is easy to find words that are surrounded by space, but once you get into parsing out words in text it can get complicated.
Instead of defining my own way of parsing the words from text, is there something already built that parse out the words in regular expression or other methods?
Some example of words in text.
word word. word(word) word's word word' "word" .word. 'word' sub-word

You can use:
text = "word word. word(word) word's word word' \"word\" .word. 'word' sub-word";
words = text.match(/[-\w]+/g);
This will give you an array with all your words.
In regular expressions, \w means any character that is either a-z, A-Z, 0-9 or _. [-\w] means any character that is a \w or a -. [-\w]+ means any of these characters that appear 1 ore more times.
If you would like to define a word as being something more than the above expression, add the other characters that compose your words inside the [-\w] character class. For example, if you'd like words to also contain ( and ), make the character class be [-\w()].
For an introduction in regular expressions, check out the great tutorial at regular-expressions.info.

What you're talking about is Tokenisation. It's non-trivial to say the least, and a subject of intense reasearch at the major search engines. There are a number of open source tokenisation libraries in various server-side languages (e.g see the Stanford NLP and Lucene projects) but as far as I am aware there's nothing that would even come close to these in javascript. You may have to roll your own :) or perhaps do the processing server-side, and load the results via AJAX?

I support Richard's answer here - but to add to it - one of the easiest ways of building a tokeniser (imho) is Antlr; and some maniac has built a Javascript target for it; thus allowing you to run and execute a grammar in the web browser (look under 'runtime libraries' section here)
I won't pretend that there's not a learning curve there though.

Take a look at regular expressions - you can define almost any parsing algorithm you want.

Related

Regular expression matching multiple entries, spanning multiple lines [duplicate]

This question already has answers here:
Regular expression to stop at first match
(9 answers)
Closed 2 years ago.
I have this gigantic ugly string:
J0000000: Transaction A0001401 started on 8/22/2008 9:49:29 AM
J0000010: Project name: E:\foo.pf
J0000011: Job name: MBiek Direct Mail Test
J0000020: Document 1 - Completed successfully
I'm trying to extract pieces from it using regex. In this case, I want to grab everything after Project Name up to the part where it says J0000011: (the 11 is going to be a different number every time).
Here's the regex I've been playing with:
Project name:\s+(.*)\s+J[0-9]{7}:
The problem is that it doesn't stop until it hits the J0000020: at the end.
How do I make the regex stop at the first occurrence of J[0-9]{7}?
Make .* non-greedy by adding '?' after it:
Project name:\s+(.*?)\s+J[0-9]{7}:
Using non-greedy quantifiers here is probably the best solution, also because it is more efficient than the greedy alternative: Greedy matches generally go as far as they can (here, until the end of the text!) and then trace back character after character to try and match the part coming afterwards.
However, consider using a negative character class instead:
Project name:\s+(\S*)\s+J[0-9]{7}:
\S means “everything except a whitespace and this is exactly what you want.
Well, ".*" is a greedy selector. You make it non-greedy by using ".*?" When using the latter construct, the regex engine will, at every step it matches text into the "." attempt to match whatever make come after the ".*?". This means that if for instance nothing comes after the ".*?", then it matches nothing.
Here's what I used. s contains your original string. This code is .NET specific, but most flavors of regex will have something similar.
string m = Regex.Match(s, #"Project name: (?<name>.*?) J\d+").Groups["name"].Value;
I would also recommend you experiment with regular expressions using "Expresso" - it's a utility a great (and free) utility for regex editing and testing.
One of its upsides is that its UI exposes a lot of regex functionality that people unexprienced with regex might not be familiar with, in a way that it would be easy for them to learn these new concepts.
For example, when building your regex using the UI, and choosing "*", you have the ability to check the checkbox "As few as possible" and see the resulting regex, as well as test its behavior, even if you were unfamiliar with non-greedy expressions before.
Available for download at their site:
http://www.ultrapico.com/Expresso.htm
Express download:
http://www.ultrapico.com/ExpressoDownload.htm
(Project name:\s+[A-Z]:(?:\\w+)+.[a-zA-Z]+\s+J[0-9]{7})(?=:)
This will work for you.
Adding (?:\\w+)+.[a-zA-Z]+ will be more restrictive instead of .*

Regular expression: matching word boundaries [duplicate]

This question already has answers here:
Regular expression to stop at first match
(9 answers)
Closed 2 years ago.
I have this gigantic ugly string:
J0000000: Transaction A0001401 started on 8/22/2008 9:49:29 AM
J0000010: Project name: E:\foo.pf
J0000011: Job name: MBiek Direct Mail Test
J0000020: Document 1 - Completed successfully
I'm trying to extract pieces from it using regex. In this case, I want to grab everything after Project Name up to the part where it says J0000011: (the 11 is going to be a different number every time).
Here's the regex I've been playing with:
Project name:\s+(.*)\s+J[0-9]{7}:
The problem is that it doesn't stop until it hits the J0000020: at the end.
How do I make the regex stop at the first occurrence of J[0-9]{7}?
Make .* non-greedy by adding '?' after it:
Project name:\s+(.*?)\s+J[0-9]{7}:
Using non-greedy quantifiers here is probably the best solution, also because it is more efficient than the greedy alternative: Greedy matches generally go as far as they can (here, until the end of the text!) and then trace back character after character to try and match the part coming afterwards.
However, consider using a negative character class instead:
Project name:\s+(\S*)\s+J[0-9]{7}:
\S means “everything except a whitespace and this is exactly what you want.
Well, ".*" is a greedy selector. You make it non-greedy by using ".*?" When using the latter construct, the regex engine will, at every step it matches text into the "." attempt to match whatever make come after the ".*?". This means that if for instance nothing comes after the ".*?", then it matches nothing.
Here's what I used. s contains your original string. This code is .NET specific, but most flavors of regex will have something similar.
string m = Regex.Match(s, #"Project name: (?<name>.*?) J\d+").Groups["name"].Value;
I would also recommend you experiment with regular expressions using "Expresso" - it's a utility a great (and free) utility for regex editing and testing.
One of its upsides is that its UI exposes a lot of regex functionality that people unexprienced with regex might not be familiar with, in a way that it would be easy for them to learn these new concepts.
For example, when building your regex using the UI, and choosing "*", you have the ability to check the checkbox "As few as possible" and see the resulting regex, as well as test its behavior, even if you were unfamiliar with non-greedy expressions before.
Available for download at their site:
http://www.ultrapico.com/Expresso.htm
Express download:
http://www.ultrapico.com/ExpressoDownload.htm
(Project name:\s+[A-Z]:(?:\\w+)+.[a-zA-Z]+\s+J[0-9]{7})(?=:)
This will work for you.
Adding (?:\\w+)+.[a-zA-Z]+ will be more restrictive instead of .*

How do I match Unicode special alpha characters while NOT matching special characters

I have a dilemma here. I am trying to write a regex pattern that matches all alpha characters for eastern languages as well as western languages. One of the criteria is that no numbers can match (so José13) is not a match but (José) is, the other criteria is that special characters cannot match (ie: !##$% etc.)
I've played around with this in chrome's console, and I've gotten:
"a".match('[a-zA-z]');
to come back successfully, when I put in:
"a".match('[\p{L}]');
I get a null response, which I'm not quite understanding why. According to http://www.regular-expressions.info/unicode.html \p{L} is a match for any letter.
EDIT: the \p doesn't seem to work in my chrome console, so I'll try a different route. I have a chart of the unicode from Unifoundry. I'll match up the regex and attempt to make the range of characters invalid.
Any input would be greatly appreciated.
This works in the javascript console, but it seems like a hack:
.match('^[^\u0000-\u0040\u005B-\u0060\u007B-\u00BF\u00D7\u00F7]*');
However it does what I need it to do.
Referenced this post on SO: Javascript + Unicode regexes
Current Javascript implementations don't support such shortcuts, but you can specify a range, for example:
/[\u4E00-\u9FFF]+/g.test("漢字")

Validate field for all language characters through REGEX

i need to validate a field for empty. But it should allow English and the Foreign languages characters(UTF-8) but not the special characters. I'm not good at Regex. So any help on this would be great...
If you want to support a wide range of languages, you'll have to work by excluding only the characters you don't want, since specifying all of the ranges you do want will be difficult.
You'll need to look at the list of Unicode blocks and or the character database to identify the blocks you want to exclude (like, for instance, U+0000 through U+001F. This Wikipedia article may also help.
Then use a regular expression with character classes to look for what you want to exclude.
For example, this will check for the U+0000 through U+001F and the U+007F characters (obviously you'll be excluding more than just these):
if (/[\u0000-\u001F\u007F]/.exec(theString)) {
// Contains at least one invalid character
}
The [] identify a "character class" (list and/or range of characters to look for). That particular one says look for \u0000 through \u001F (inclusive) as well as \u007F.
It would have been nice if I could say "Just do /^\w+$/.test(word)", but...
See this answer for the current state of unicode support (or rather lack of) in JavaScript regular expressions.
You can either use the library he suggests, which might be slow or enlist the help of the server for this (which might be slower).
You can test for a unicode letter like this:
str.match(/\p{L}/u)
Or for the existence of a non-letter like this:
str.match(/[^\p{L}]/u)

How To Create This RegExp

I am looking to find this in a string: XXXX-XXX-XXX Where the X is any number.
I need to find this in a string using JavaScript so bonus points to those who can provide me the JavaScript too. I tried to create a regex and came out with this: ^[0-9]{4}\-[0-9]{3}\-[0-9]{3}$
Also, I would love to know of any cheat sheets or programs you guys use to create your regular expressions.
i suppose this is what you want:
\d{4}-\d{3}-\d{3}
in doubt? Google for "RegEx Testers"
With your attempt:
^[0-9]{4}\-[0-9]{3}\-[0-9]{3}$
Since the - is not a metacharacter, there is no need to escape it -- thus you are looking for explicit backslash characters.
Also, you've anchored the match at the beginning and end of the string -- this will match only strings that consist only of your number. (Well, assuming the rest were correct.)
I know most people like the {3} style of counting, but when the thing being matched is a single digit, I find this more legible:
\d\d\d\d-\d\d\d-\d\d\d
Obviously if you wanted to extend this to matching hexadecimal digits, extending this one would be horrible, but I think this is far more legible than alternatives:
\d{4}-\d{3}-\d{3}
[[:digit:]]{4}-[[:digit:]]{3}-[[:digit:]]{3}
[0-9]{4}-[0-9]{3}-[0-9]{3}
Go with whatever is easiest for you to read.
I tend to use the perlre(1) manpage as my main reference, knowing full well that it is far more featureful than many regexp engines. I'm prepared to handle the differences considering how conveniently available the perlre manpage is on most systems.
var result = (/\d{4}\-\d{3}\-\d{3}/).exec(myString);

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