I've looked at lots of examples for TRIM and REPLACE on the internet and for some reason I keep getting errors when I try.
I need to strip suffixes from my Netsuite item record names in a saved item search. There are three possible suffixes: -T, -D, -S. So I need to turn 24335-D into 24335, and 24335-S into 24335, and 24335-T into 24335.
Here's what I've tried and the errors I get:
Can you help me please? Note: I can't assume a specific character length of the starting string.
Use case: We already have a field on item records called Nickname with the suffixes stripped. But I've ran into cases where Nickname is incorrect compared to Name. Ex: Name is 24335-D but Nickname is 24331-D. I'm trying to build a saved search alert that tells me any time the Nickname does not equal suffix-stripped Name.
PS: is there anywhere I can pay for quick a la carte Netsuite saved search questions like this? I feel bad relying on free technical internet advice but I greatly appreciate any help you can give me!
You are including too much SQL - a formulae is like a single result field expression not a full statement so no FROM or AS. There is another place to set the result column/field name. One option here is Regex_replace().
REGEXP_REPLACE({name},'\-[TDS]$', '')
Regex meaning:
\- : a literal -
[TDS] : one of T D or S
$ : end of line/string
To compare fields a Formulae (Numeric) using a CASE statement can be useful as it makes it easy to compare the result to a number in a filter. A simple equal to 1 for example.
CASE WHEN {custitem_nickname} <> REGEXP_REPLACE({name},'\-[TDS]$', '') then 1 else 0 end
You are getting an error because TRIM can trim only one character : see oracle doc
https://docs.oracle.com/javadb/10.8.3.0/ref/rreftrimfunc.html (last example).
So try using something like this
TRIM(TRAILING '-' FROM TRIM(TRAILING 'D' FROM {entityid}))
And always keep in mind that saved searches are running as Oracle SQL queries so Oracle SQL documentation can help you understand how to use the available functions.
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"]]
I'm using PostgreSQL database with VB.NET and ODBC (Windows).
I'm searching sentences for whole words by combining SELECT with a regular expression, like this:
"SELECT dtbl_id, name
FROM mytable
WHERE name ~*'" + "( |^)" + TextBox1.Text + "([^A-z]|$)"
This searches well in some cases but because of syntax errors in text (or other reasons) it sometimes fails. For example, if I have the sentence
BILLY IDOL: WHITE WEDDING
the word "white" will be found. But if I have
CLASH-WHITE RIOT
then "white" will not be found, because there is no space between start of word "white".
The simplest solution would be to temporarily change or replace characters in the sentences :,.\/-= etc to spaces.
Is this possible to do in single SELECT line to be suitable for use with .NET/ODBC? Maybe inside the same regular expression?
If it is, how?
Try this:
SELECT 'CLASH-WHITE RIOT' ~ '[[:<:]]WHITE[[:>:]]';
[[:<:]] and [[:>:]] simply mean beginning and end of a word respectively
more info you can find at: http://www.postgresql.org/docs/9.1/static/functions-matching.html#FUNCTIONS-POSIX-REGEXP
OK So i am confused (obviously)
I'm trying to return rows (from Oracle) where a text field contains a complete word, not just the substring.
a simple example is the word 'I'.
Show me all rows where the string contains the word 'I', but not simply where 'I' is a substring somewhere as in '%I%'
so I wrote what i thought would be a simple regex:
select REGEXP_INSTR(upper(description), '\bI\b') from mytab;
expecting that I should be detected with word boundaries. I get no results (or rather the result 0 for each row.
what i expect:
'I am the Administrator' -> 1
'I'm the administrator' -> 0
'Am I the administrator' -> 1
'It is the infamous administrator' -> 0
'The adminisrtrator, tis I' -> 1
isn't the /b supposed to find the contained string by word boundary?
tia
I believe that \b is not supported by your flavor of regex :
http://download.oracle.com/docs/cd/B19306_01/appdev.102/b14251/adfns_regexp.htm#i1007670
Therefore you could do something like :
(^|\s)word(\s|$)
To at least ensure that your "word" is separated by some whitespace or it's the whole string.
Oracle doesn't support word boundary anchors, but even if it did, you wouldn't get the desired result: \b matches between an alphanumeric character and a non-alphanumeric character. The exact definition of what an alnum is differs between implementations, but in most flavors, it's [A-Za-z0-9_] (.NET also considers Unicode letters/digits).
So there are two boundaries around the I in %I%.
If you define your word boundary as "whitespace before/after the word", then you could use
(^|\s)I(\s|$)
which would also work at the start/end of the string.
Oracle native regex support is limited. \b or < cannot be used as word delimiters. You may want Oracle Text for word search.
How can I find words like and, or, to, a, no, with, for etc. in a sentence using VB.NET and remove them. Also where can I find all words list like above.
Note that unless you use Regex word boundaries you risk falling afoul of the Scunthorpe (Sfannythorpe) problem.
string pattern = #"\band\b";
Regex re = new Regex(pattern);
string input = "a band loves and its fans";
string output = re.Replace(input, ""); // a band loves its fans
Notice the 'and' in 'band' is untouched.
You can indeed replace your list of words using the .Replace function (as colithium described) ...
myString.Replace("and", "")
Edit:
... but indeed, a nicer way is to use Regular Expressions (as edg suggested) to avoid replacing parts of words.
As your question suggests that you would like to clean-up a sentence to keep meaningfull words, you have to do more than just remove two- and three letter words.
What you need is a list of stop-words:
http://en.wikipedia.org/wiki/Stop_word
A comma seperated list of stop-words for the English language can be found here:
http://www.textfixer.com/resources/common-english-words.txt
The easiest way is:
myString.Replace("and", "")
You'd loop over your word list and have a statement like the above. Google for a list of common English words?
List of English 2 Letter Words
List of English 3 Letter Words
You can match the words and remove them using regular expressions.