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Get list of parameter names from native sql expression (regex)

I'm having trouble getting list of all parameters in SQL query using Regex.
Example of the query:
SELECT ... WHERE col1 = :user AND col2 = 'HELLO' OR col3 = :language
To obtain parameters, I use following regex pattern:
Pattern.compile(":([\\w.$]+|\"[^\"]+\"|'[^']+')", Pattern.MULTILINE)
The pattern returns list of parameters correctly:
:user
:language
The problem is with another type of query, where literals might contain character ':'
WHERE col1 = :user AND some_date > '2022-09-26T10:22:55'
The list of parameters for this case is:
:user
:22
:55
Is there any better approach that will not consider contents of literals as parameters?

You could simplify your problem by assuming that a named param in sql is just a word with prefix : and always follows after a space (this is actually not a requirement or always true but might be just good enough to get you acceptable results with as simple of regex as possible)
Pattern.compile(" :\\w+", Pattern.MULTILINE)
--
summary of the comments:
had to match
- foo = :param AND :param = bar AND foo=:param AND :param=bar
- AND FUNC(:param) OR FUNC(0, :param) OR FUNC(:param, 0)
finally this regex with fixed length lookahead and variable length lookbehind was helpful:
Pattern.compile("(?<=[=(])\\s*:[\\w_.]+|:[\\w_.]+(?=\s*[=)])", Pattern.MULTILINE)

How to use CriteriaBuilder to call db function

I have a string that might eventually contain spaces. I would like to replace those spaces with a regex that matches against \t,\r,\n. After replacing I would like to call regexp_like (an oracle function) to match a field against this string.
I know it is possible to call db functions using criteria builder as described for example in this link
I am not very familiar with the difference between regex in java versus oracle or how to cobble this together (I have never called functions from criteriabuilder). Here are my tentative steps with places where I am stuck in the comments
// first replace all spaces with regex for \s,\r,\t, value is the original string
value.replaceAll(" +", "[\\t\\n\\r]+")
// build the db function call expression, seems I cant do table.<String>get(field) and cant pass value as a string
Expression<String> regExp = cb.function("regexp_like", String.class, table.<String>get(field), value);
// now create a predicate not sure how
Predicate fieldMatch = cb.equal(...)
Is this possible?

Its possible. You only need to do few small changes.
Extends your Oracle dialog
public class Oracle10gCustomDialect extends Oracle10gDialect {
public Oracle10gCustomDialect() {
super();
registerFunction("regexp_like", new SQLFunctionTemplate(StandardBasicTypes.BOOLEAN,
"(case when (regexp_like(?1, ?2)) then 1 else 0 end)"));
}
}
Then use this path at your application.properties or hibernate properties config.
Then in your specification, need to do something like this.
Expression<Boolean> regExprLike = criteriaBuilder.function("regexp_like", Boolean.class, root.get("yourColumn"), criteriaBuilder.literal("<your regexp value>"));
predicates.add(criteriaBuilder.isTrue(regExprLike));
...
And thats all!

JPA Select query not returning results with one letter word

I have a query that when given a word that starts with a one-letter word followed by space character and then another word (ex: "T Distribution"), does not return results. While given "Distribution" alone returns results including the results for "T Distribution". It is the same behavior with all search terms beginning with a one-letter word followed by space character and then another word.
The problem appears when the search term is of this pattern:
"[one-letter][space][letter/word]". example: "o ring".
What would be the problem that the LIKE operator not working correctly in this case?
Here is my query:
#Cacheable(value = "filteredConcept")
#Query("SELECT NEW sina.backend.data.model.ConceptSummaryVer04(s.id, s.arabicGloss, s.englishGloss, s.example, s.dataSourceId,
s.synsetFrequnecy, s.arabicWordsCache, s.englishWordsCache, s.superId, s.categoryId, s.dataSourceCacheAr, s.dataSourceCacheEn,
s.superTypeCasheAr, s.superTypeCasheEn, s.area, s.era, s.rank, s.undiacritizedArabicWordsCache, s.normalizedEnglishWordsCache,
s.isTranslation, s.isGloss, s.arabicSynonymsCount, s.englishSynonymsCount) FROM Concept s
where s.undiacritizedArabicWordsCache LIKE %:searchTerm% AND data_source_id != 200 AND data_source_id != 31")
List<ConceptSummaryVer04> findByArabicWordsCacheAndNotConcept(#Param("searchTerm") String searchTerm, Sort sort);
the result of the query on the database itself:
link to screenshot
results on the database are returned no matter the letters case:
link to screenshot

I solved this problem.
It was due to the default configuration of the Full-text index on mysql database which is by default set to 2 (ft_min_word_len = 2).
I changed that and rebuilt the index. Then, one-letter words were returned by the query.
12.9.6 Fine-Tuning MySQL Full-Text Search

Use some quotes:
LIKE '%:searchTerm%';

Set searchTerm="%your_word%" and use it on query like this :
... s.undiacritizedArabicWordsCache LIKE :searchTerm ...

String split function with hql

I have following hql query,
from Channe where ip='1.11.6.0';
But in the db the IP is saving as 1.11.6.0:8080 .
So I need to modify the query in a way that, split the ip with a delimiter ':' and take the firstcome value. I do not wish to modify the search with value 1.11.6.0:8080.

See this page in the Hibernate docs. On the page below there is a section called 14.10. Expressions
http://docs.jboss.org/hibernate/orm/3.3/reference/en/html/queryhql.html
It says, among other things:
string concatenation ...||... or concat(...,...) current_date(),
...
Any function or operator defined by EJB-QL 3.0: substring(), trim(), lower(), upper(),
length(), locate(), abs(), sqrt(), bit_length(), mod()
But you are actually better off doing as #Hansraj suggests in the comments and appending a wildcard to your search term
String query = "from Channe where ip like :term";
entityManager.createQuery(query).setParameter("term",ipString + "%");
This assumes that your data type is string, of course.

Try the following:
Say variable ip had the address
ip = "10.131.56.40:8080";
var ipSplit = ip.Split(':');
var ipStart = ipSplit[0];
ipStart will store only 10.131.56.40
This could solve your problem

Try this:
SPLIT(".", FIELDNAME)

java, Morphia how to compare strings when using the Morphia Find Methods

Im new to this so here goes.
Trying to get a user called "Bob" from the MongoDb.
I have the:
UserData ud = MonConMan.instance().getDb().find(UserData.class, "name","bob").get();
The "bob" cannot be found if it has capital "Bob".
I understand i can get a List and do equalsIgnoreCase but are
there some Operators i can use?
I have users logging on and must test to see if they are registered. A user can type his name anyway he likes so must find a way to equalsIgnoreCase. Yea this is a problem, i cannot get all names and do equalsIgnoreCase, if there are like 10,000. One could of course initially save all user names in lowercase but that would destroy the visual appearance of the name.
looking at the wiki but cannot see any..
http://code.google.com/p/morphia/wiki/Query

Use java regex, like this.
String name = "bob";
Pattern pattern = Pattern.compile("^" + bob + "$", Pattern.CASE_INSENSITIVE);//This line will create a pattern to match words starts with "b", ends with "b" and its case insensitive too.
Query<UserData> query = createQuery().field("name").equal(pattern).retrievedFields(true, "id");//Replace `id` with what ever name you use in UserData for '_id'
UserData user = query.get();
if(user!=null){
//he is already registered
}
else{
//He is a new guy
}
(I am not good at regex, so you may have read about$&^somewhere. )
You should be sure that the user names you are using to validate a new user should be unique across your system.

Ended up keeping two fields like
- lowercaseusername
- originalusername
This way i could search for a user using the lowercaseusername

You can make find a name of a UserData using this code :
Query<UserData> query = createQuery().filter("name","bob");
find(query);
In my application, this code return all UserData that haves a field name with "bob" value.
The code can be this way too :
Query<UserData> query = createQuery().field("name").equal("bob");
find(query);
These codes will be in a UserDataDao that extends BasicDao, and receives in the construtor the datastore from morphia.