For my database, I have a list of company numbers where some of them start with two letters. I have created a regex which should eliminate these from a query and according to my tests, it should. But when executed, the result still contains the numbers with letters.
Here is my regex, which I've tested on https://www.regexpal.com
([^A-Z+|a-z+].*)
I've tested it against numerous variations such as SC08093, ZC000191 and NI232312 which shouldn't match and don't in the tests, which is fine.
My sql query looks like;
SELECT companyNumber FROM company_data
WHERE companyNumber ~ '([^A-Z+|a-z+].*)' order by companyNumber desc
To summerise, strings like SC08093 should not match as they start with letters.
I've read through the documentation for postgres but I couldn't seem to find anything regarding this. I'm not sure what I'm missing here. Thanks.
The ~ '([^A-Z+|a-z+].*)' does not work because this is a [^A-Z+|a-z+].* regex matching operation that returns true even upon a partial match (regex matching operation does not require full string match, and thus the pattern can match anywhere in the string). [^A-Z+|a-z+].* matches a letter from A to Z, +,|or a letter fromatoz`, and then any amount of any zero or more chars, anywhere inside a string.
You may use
WHERE companyNumber NOT SIMILAR TO '[A-Za-z]{2}%'
See the online demo
Here, NOT SIMILAR TO returns the inverse result of the SIMILAR TO operation. This SIMILAR TO operator accepts patterns that are almost regex patterns, but are also like regular wildcard patterns. NOT SIMILAR TO '[A-Za-z]{2}%' means all records that start with two ASCII letters ([A-Za-z]{2}) and having anything after (%) are NOT returned and all others will be returned. Note that SIMILAR TO requires a full string match, same as LIKE.
Your pattern: [^A-Z+|a-z+].* means "a string where at least some characters are not A-Z" - to extend that to the whole string you would need to use an anchored regex as shown by S-Man (the group defined with (..) isn't really necessary btw)
I would probably use a regex that specifies want the valid pattern is and then use !~ instead.
where company !~ '^[0-9].*$'
^[0-9].*$ means "only consists of numbers" and the !~ means "does not match"
or
where not (company ~ '^[0-9].*$')
Not start with a letter could be done with
WHERE company ~ '^[^A-Za-z].*'
demo: db<>fiddle
The first ^ marks the beginning. The [^A-Za-z] says "no letter" (including small and capital letters).
Edit: Changed [A-z] into the more precise [A-Za-z] (Why is this regex allowing a caret?)
Before posting, I tried the hive sentences function and did some search but couldn't get a clear understanding, my question is based on what delimiter hive sentences function breaks each sentence? hive manual says "appropriate boundary" what does that mean? Below is an example of my tries, I tried adding period (.) and exclamatory sign(!) at different points of the sentence. I'm getting different outputs, can someone explain on this?
with period (.)
select sentences('Tokenizes a string of natural language text into words and sentences. where each sentence is broken at the appropriate sentence boundary and returned as an array of words.') from dummytable
output - 1 array
[["Tokenizes","a","string","of","natural","language","text","into","words","and","sentences","where","each","sentence","is","broken","at","the","appropriate","sentence","boundary","and","returned","as","an","array","of","words"]]
with '!'
select sentences('Tokenizes a string of natural language text into words and sentences! where each sentence is broken at the appropriate sentence boundary and returned as an array of words.') from dummytable
output - 2 arrays
[["Tokenizes","a","string","of","natural","language","text","into","words","and","sentences"],["where","each","sentence","is","broken","at","the","appropriate","sentence","boundary","and","returned","as","an","array","of","words"]]
If you understand the functionality of sentences()..it clears your doubt.
Definition of sentences(str):
Splits str into arrays of sentences, where each sentence is an array
of words.
Example:
SELECT sentences('Hello there! I am a UDF.') FROM src LIMIT 1;
[ ["Hello", "there"], ["I", "am", "a", "UDF"] ]
SELECT sentences('review . language') FROM movies;
[["review","language"]]
An exclamation point is a type of punctuation mark that goes at the end of a sentence. Other examples of related punctuation marks include periods and question marks, which also go at the end of sentences.But as per the definition of sentences() ,Unnecessary punctuation, such as periods and commas in English, is automatically stripped.So,we are able to get two arrays of words with !. It completely involves java.util.Locale.java
I don't know the actual reason but observed after period(.) if you put space and next word first letter as capital then it is working.
Here I changed from where to Where it it worked. However this is not require for !
Tokenizes a string of natural language text into words and sentences. Where each sentence is broken at the appropriate sentence boundary and returned as an array of words.
And this is giving below output
[["Tokenizes","a","string","of","natural","language","text","into","words","and","sentences"],["Where","each","sentence","is","broken","at","the","appropriate","sentence","boundary","and","returned","as","an","array","of","words"]]
Unicode have categories of characters. Some are alpha numeric. Some are punctuation.
What about if I want to know whether a word belongs to keyword or not
For example,
A,a,b,c, tend to belong to words. So is Ƈ,Ǝ,ǟ, so are all chinese characters.
Sentences like
Hello World, I "like" (to) eat ƇƎǟ and 款开源 ©
Have keywords:
Hello
World
I
like
to
eat
ƇƎǟ
款
开
源
Here, , (),© are not word characters and hence should just be ignored and use.
© doesn't count as punctuation either. '©'.IsPunctuation returns false in vb.net but I want to get rid of that too.
Now I want to make a program that can split sentences into keywords. For that I need to know which characters are word characters and which one is not.
Is there a vb.net function for that?
Do it the other way round: use IsLetter for your test. Or better yet, use regular expressions to split your string by words:
Dim str = "Hello World, I ""like"" (to) eat ƇƎǟ and 款开源 ©"
Dim wordPattern As New Regex("\p{L}+")
For Each match in wordPattern.Matches(str))
Console.WriteLine(match)
Next
Here, \p{L} matches any word character. However, the above matches “款开源” in a single rather than in separate matches since there is no separator between the characters.
u need to deal with "keycodes"
like if u only want letters [a-z]
then
for(c>='a' && c<='z'){
}
or
for(c>=97 && C<=122){
}
I'm finding it hard to match strings using NSRegularExpression. Generic alpha characters are not a problem with [a-z] but if I need to match a word like 'import' I'm struggling to make it work. I'm sure I have to escape the word in some manner but I can't find any docs around this. A really basic example would be
{{import "hello"}}
where I want to get hold of the string: hello
edit: to clarify - 'hello' could be any string - it's the bit I want returned
This regular expression matches the text between the "-s in your example:
\{\{import "([^"]+)"\}\}
The match will be stored in the first match group.
I'm trying to implement stuff similar to spell check, but I need to get the word that is limited by a space. EX: "HI HOW R U", I need to collect HI, HOW and so on as they type. i.e. After user hits HI and space I need to collect HI and do a spell check.
Check the documentation for NSString Here. You want the message componentsSepeparatedByString:.
I don't know objective-C, but I'm fairly sure it'll have a Regexp library - although it'd be straightforward to code it without one.
Regexp: \b([^\s])*\b
\b = word boundary (whitespace, comma, dot, exclamation-mark, etc.)
\s = whitespace character
[...] = character set
[^...] = negated character set (any character(s) EXCEPT ...)
() = grouping construct
* = zero or more times
So the suggested expression would start matching at any word boundary, then match every subsequent character that is not a whitespace character, then match a word boundary.
Your stated case is so simple you may just want to look for spaces (one char at a time) and get the substring, but RegExp is very widely used across a range of languages and platforms, and so it's fairly easy to find an expression when you need to - and one often does for common stuff like checking if zip codes, phone numbers, email addresses and so on are syntactically correct. So it's worth learning in any case. :)