I'm trying to find clobs in an oracle database that contain specific unicode whitespace characters, but I'm not sure how to the select statement can be written.
Trying the following query, but not sure if this is the correct way to go about it:
select * from mytable where my_clob like '%'|| n'0xEF' || '%';
I'm not sure what character you're wanting to search for, but I think it's the UNISTR command you're looking for. Also, LIKE will do an implicit conversion to VARCHAR2 (I think), so you're only searching in the first 4000 characters. Try this to search for the non-breaking space:
SELECT *
FROM mytable
WHERE dbms_lob.instr( my_clob, UNISTR( '\00A0' )) > 0;
https://docs.oracle.com/cd/E11882_01/server.112/e41084/functions224.htm#SQLRF06154
If you know the character code of the character you're looking for, you can use CHR() to build an expression to look for the specific code;
SELECT *
FROM mytable
WHERE my_clob LIKE '%' || CHR(15052183) || '%';
An SQLfiddle to test with.
Related
I have a sql server database that has medical descriptions in it. I've created a full text index on it, but I'm still figuring out how this works.
The easiest example to give is if there is a description of Hypertensive heart disease
Now they would like to be able to type hyp hea as a search term and have it return that.
So from what I've read it seems like my query needs to be something like
DECLARE #Term VARCHAR(100)
SET #Term = 'NEAR(''Hyper*'',''hea*'')'
SELECT * FROM Icd10Codes WHERE CONTAINS(Description, #Term)
If I take the wild card out for Hypertensive and heart, and type out the full words it works, but adding the wild card in returns nothing.
If it makes any difference I'm using Sql Server 2017
So it was a weird syntax issue that didn't cause an error, but stopped the search from working.
I changed it to
SELECT * FROM Icd10Codes where CONTAINS(description, '"hyper*" NEAR "hea*"')
The key here being I needed double quotes " and not to single quotes. I assumed it was two single quotes, the first to escape the second, but it was actually double quotes. The above query returns the results exactly as expected.
this will work:
SELECT * FROM Icd10Codes where SOUNDEX(description)=soundex('Hyp');
SELECT * FROM Icd10Codes where DIFFERENCE(description,'hyp hea')>=2;
You could try a like statement. You can find a thorough explanation here.
Like so:
SELECT * FROM Icd10Codes WHERE Icd10Codes LIKE '%hyp hea%';
And then instead of putting the String in there just use a variable.
If you need to search for separated partial words, as in an array of search terms, it gets a bit tricky, since you need to dynamically build the SQL statement.
MSSQL provides a few features for full text search. You can find those here. One of them is the CONTAINS keyword:
SELECT column FROM table WHERE CONTAINS (column , 'string1 string2 string3');
For me - this had more mileage.
create a calculated row with fields as full text search.
fullname / company / lastname all searchable.
ALTER TABLE profiles ADD COLUMN fts tsvector generated always as (to_tsvector('english', coalesce(profiles.company, '') || ' ' || coalesce(profiles.broker, '') || ' ' || coalesce(profiles.firstname, '') || ' ' || coalesce(profiles.lastname, '') || ' ' )) stored;
let { data, error } = await supabase.from('profiles')
.select()
.textSearch('fts',str)
I am after all cells containing the 'LINE SEPARATOR' (U+2028) unicode point. Normally this is encoded as \u+2028 or something similar. However googling how this translates to SQL has given various options none of which seem to work ((N'2028'), set #hexstring = '2028';, vchar(2028))
SELECT * FROM myTable WHERE desc LIKE '% [SOME WAY TO ESCAPE U+2028 ] %'
ANSI SQL answer, may or may not work with Postgresql.
SELECT * FROM myTable WHERE desc LIKE U&'%\2028%'
Alternative solution using regexp:
select * from mytable where desc ~ '\x2028';
I've executed the query on DB2:
SELECT * FROM MYTABLE WHERE MYFIELD LIKE '%B'
Although I know there are records that end with 'B' in the database, the query returned no results. After some research, it seems that DB2 is not recognizing LIKE expressions that don't end with '%'. So the following query would work:
SELECT * FROM MYTABLE WHERE MYFIELD LIKE '%B%'
but naturally not as expected, because it will return only the rows, where MYFIELD contains 'B', but doesn't end with it.
How to go around that strage, hmmm, feature? How to match the text on the end of the word in LIKE-like expressions?
DB2 can match patterns at the end of the string. The problem is probably that there are other characters.
You can try:
WHERE rtrim(MYFIELD) LIKE '%B'
You can also look at the lengths of the field and delimit the string value to see if there are other characters:
select length(MyField), '|' || MyField || '|'
from mytable
where MyField like '%B%';
If 'like' operator does not work for you, you can try regular expressions (with xQuery)
I'm trying to select some rows from an Oracle database like so:
select * from water_level where bore_id in ('85570', '112205','6011','SP068253');
This used to work fine but a recent update has meant that bore_id in water_level has had a bunch of whitespace added to the end for each row. So instead of '6011' it is now '6011 '. The number of space characters added to the end varies from 5 to 11.
Is there a way to edit my query to capture the bore_id in my list, taking account that trialling whitespace should be ignored?
I tried:
select * from water_level where bore_id in ('85570%', '112205%','6011%','SP068253%');
which returns more rows than I want, and
select * from water_level where bore_id in ('85570\s*', '112205\s*','6011\s*', 'SP068253\s*');
which didn't return anything?
Thanks
JP
You should RTRIM the WHERE clause
select * from water_level where RTRIM(bore_id) in ('85570', '112205','6011');
To add to that, RTRIM has an overload which you can pass a second parameter of what to trim, so if the trailing characters weren't spaces, you could remove them. For example if the data looked like 85570xxx, you could use:
select * from water_level where RTRIM(bore_id, 'x') IN ('85570','112205', '6011');
You could use the replace function to remove the spaces
select * from water_level where replace(bore_id, ' ', '') in ('85570', '112205', '6011', 'SP068253');
Although, a better option would be to remove the spaces from the data if they are not supposed to be there or create a view.
I'm guessing bore_id is VARCHAR or VARCHAR2. If it were CHAR, Oracle would use (SQL-standard) blank-padded comparison semantics, which regards 'foo' and 'foo ' as equivalent.
So, another approach is to force comparison as CHARs:
SELECT *
FROM water_level
WHERE CAST(bore_id AS CHAR(16)) IN ('85570', '112205', '6011', 'SP068253');
I have an SQL query as below.
Select * from table
where name like '%' + search_criteria + '%'
If search_criteria = 'abc', it will return data containing xxxabcxxxx which is fine.
But if my search_criteria = 'abc%', it will still return data containing xxxabcxxx, which should not be the case.
How do I handle this situation?
If you want a % symbol in search_criteria to be treated as a literal character rather than as a wildcard, escape it to [%]
... where name like '%' + replace(search_criteria, '%', '[%]') + '%'
Use an escape clause:
select *
from (select '123abc456' AS result from dual
union all
select '123abc%456' AS result from dual
)
WHERE result LIKE '%abc\%%' escape '\'
Result
123abc%456
You can set your escape character to whatever you want. In this case, the default '\'. The escaped '\%' becomes a literal, the second '%' is not escaped, so again wild card.
See List of special characters for SQL LIKE clause
The easiest solution is to dispense with "like" altogether:
Select *
from table
where charindex(search_criteria, name) > 0
I prefer charindex over like. Historically, it had better performance, but I'm not sure if it makes much of difference now.
To escape a character in sql you can use !:
EXAMPLE - USING ESCAPE CHARACTERS
It is important to understand how to "Escape Characters" when pattern matching. These examples deal specifically with escaping characters in Oracle.
Let's say you wanted to search for a % or a _ character in the SQL LIKE condition. You can do this using an Escape character.
Please note that you can only define an escape character as a single character (length of 1).
For example:
SELECT *
FROM suppliers
WHERE supplier_name LIKE '!%' escape '!';
This SQL LIKE condition example identifies the ! character as an escape character. This statement will return all suppliers whose name is %.
Here is another more complicated example using escape characters in the SQL LIKE condition.
SELECT *
FROM suppliers
WHERE supplier_name LIKE 'H%!%' escape '!';
This SQL LIKE condition example returns all suppliers whose name starts with H and ends in %. For example, it would return a value such as 'Hello%'.
You can also use the escape character with the _ character in the SQL LIKE condition.
For example:
SELECT *
FROM suppliers
WHERE supplier_name LIKE 'H%!_' escape '!';
This SQL LIKE condition example returns all suppliers whose name starts with H and ends in _ . For example, it would return a value such as 'Hello_'.
Reference: sql/like
Select * from table where name like search_criteria
if you are expecting the user to add their own wildcards...
You need to escape it: on many databases this is done by preceding it with backslash, \%.
So abc becomes abc\%.
Your programming language will have a database-specific function to do this for you. For example, PHP has mysql_escape_string() for the MySQL database.
Escape the percent sign \% to make it part of your comparison value.
May be this one help :)
DECLARE #SearchCriteria VARCHAR(25)
SET #SearchCriteria = 'employee'
IF CHARINDEX('%', #SearchCriteria) = 0
BEGIN
SET #SearchCriteria = '%' + #SearchCriteria + '%'
END
SELECT *
FROM Employee
WHERE Name LIKE #SearchCriteria