Can you advise if it is possible, to select a count for numerous substrings in a query
so if I have a message field which contains for example, text messages and I could do
SELECT COUNT(1)
FROM MESSAGES
WHERE MESSAGE_BODY LIKE '%hello%'
but what I want to do is more:
SELECT STRING, COUNT(1)
FROM MESSAGES
WHERE MESSAGE_BODY IN (list of strings with wild card)
is this possible?
to break down example:
ID | Message_Body
1 | Hello, How Are You?
2 | Hi, Great Thanks
3 | Hello, How is things?
4 | Ciao
Output wanted:
hello , 2
ciao, 1
SELECT (input strings), COUNT(1)
FROM TABLE
WHERE (input strings) IN ('%hello%','%ciao%')
If I understood you correctly, you can try something like this:
SELECT t.string,
CASE WHEN t.MESSAGE_BODY LIKE '%laptop%' then 1 else 0 END +
CASE WHEN t.MESSAGE_BODY LIKE '%one%' then 1 else 0 END +
CASE WHEN t.MESSAGE_BODY LIKE '%two%' then 1 else 0 END as count_col
FROM YourTable t
If you just want multiple LIKE comaparison, use REGEXP_LIKE() :
SELECT STRING, COUNT(1)
FROM MESSAGES
where regexp_like(MESSAGE_BODY, 'one|two|laptop')
EDIT: You can use a derived table containing all strings you are intrested on and left join to the original table for count:
SELECT t.wrd,COUNT(s.id) as cnt
FROM (
SELECT 'hello' as wrd FROM DUAL
UNION ALL
SELECT 'ciao' as wrd FROM DUAL) t
LEFT OUTER JOIN messages s
ON(s.message_body LIKE '%' || t.wrd || '%')
GROUP BY t.wrd
Here is with looking for whole words:
SELECT a.word, COUNT (message.message_body)
FROM ( SELECT REGEXP_SUBSTR ('hello,ciao', '[^,]+', 1, LEVEL) word
FROM DUAL
CONNECT BY REGEXP_SUBSTR ('hello,ciao', '[^,]+', 1, LEVEL) IS NOT NULL) a
LEFT OUTER JOIN MESSAGES ON REGEXP_INSTR (MESSAGE_BODY, '(^|\s)' || a.word || '(\s|$)', 1, 1, 0, 'i') > 0
GROUP BY a.word
Related
How can I search for a specific number in an array using REGEXP?
I have an array and need to verify if it has a specific number.
Ex: [5,2,1,4,6,19] and I am looking for number 1, but just the number 1 and not any number that contain the digit 1.
I had to do this:
case when REGEXP_INSTR(JSON_QUERY(MY_JSON_COLUMN,'$.path') , '[[]{1}[1][,]')<>0
or REGEXP_INSTR(JSON_QUERY(MY_JSON_COLUMN,'$.path') , '[,]{1}[1][,]{1}')<>0
or REGEXP_INSTR(JSON_QUERY(MY_JSON_COLUMN,'$.path') , '[,]{1}[1][]]')<>0
or REGEXP_INSTR(JSON_QUERY(MY_JSON_COLUMN,'$.path') , '[[]{1}[1][]]') <>0
then 'DIGIT_ONE' else 'NO_DIGIT_ONE'
end
Is there anything simpler?
You can use
(^|\D)1(\D|$)
This will seach for 1 not enclosed with other digits.
See this regex demo.
Details
(^|\D) - start of string or non-digit
1 - a 1 char
(\D|$) - non-digit or end of string.
Do NOT use regular expressions, use a proper JSON parser and then filter for the number you want:
SELECT my_json_column,
CASE
WHEN JSON_EXISTS( my_json_column, '$?(#.path[*] == 1)' )
THEN 'DIGIT ONE'
ELSE 'NO DIGIT ONE'
END AS has_one
FROM table_name;
or (if you are using Oracle 12.1 and cannot use path filter expressions with JSON_EXISTS, which is only available from Oracle 12.2):
SELECT my_json_column,
CASE
WHEN EXISTS(
SELECT 'X'
FROM JSON_TABLE(
t.my_json_column,
'$.path[*]'
COLUMNS (
value NUMBER PATH '$'
)
)
WHERE value = 1
)
THEN 'DIGIT ONE'
ELSE 'NO DIGIT ONE'
END
FROM table_name t;
Which, for the sample data:
CREATE TABLE table_name (
my_json_column CHECK ( my_json_column IS JSON )
) AS
SELECT '{"path":[5,2,1,4,6,19],"not_this_path":[1,2,3,4,5]}' FROM DUAL UNION ALL
SELECT '{"path":[5,2,4,6,19],"not_this_path":[1,2,3,4,5]}' FROM DUAL UNION ALL
SELECT '{"path":[11],"not_this_path":[1]}' FROM DUAL UNION ALL
SELECT '{"path":[2],"not_this_path":[1]}' FROM DUAL UNION ALL
SELECT '{"path":[1,11]}' FROM DUAL;
Both output:
MY_JSON_COLUMN | HAS_ONE
:-------------------------------------------------- | :-----------
{"path":[5,2,1,4,6,19],"not_this_path":[1,2,3,4,5]} | DIGIT ONE
{"path":[5,2,4,6,19],"not_this_path":[1,2,3,4,5]} | NO DIGIT ONE
{"path":[11],"not_this_path":[1]} | NO DIGIT ONE
{"path":[2],"not_this_path":[1]} | NO DIGIT ONE
{"path":[1,11]} | DIGIT ONE
db<>fiddle here
Alternatively, with a little bit more typing (a little bit? Am I kidding?!), splitting the string into rows and comparing values to the search string:
SQL> with test (col) as
2 (select '[5,2,1,4,6,19]' from dual)
3 select t.col,
4 case when '&par_search_string' in
5 (select regexp_substr(substr(col, 2, length(col) - 1), '[^,]+', 1, level) val
6 from test
7 connect by level <= regexp_count(col, ',') + 1
8 )
9 then 'Search string exists'
10 else 'Search string does not exist'
11 end result
12 from test t;
Enter value for par_search_string: 1
COL RESULT
-------------- ----------------------------
[5,2,1,4,6,19] Search string exists
SQL> /
Enter value for par_search_string: 24
COL RESULT
-------------- ----------------------------
[5,2,1,4,6,19] Search string does not exist
SQL>
I have a column that contains multiple values in a delimited(comma-separated) format -
id | code
------------
1 11,19,21
2 55,87,33
3 3,11
4 11
I want to be able to compare to each value inside the 'code' column as below -
SELECT id FROM myTbl WHERE code = '11'
This should return -
1
3
4
I've tried the solution below but it does not work for all cases -
SELECT id FROM myTbl WHERE POSITION('11' IN code) <> 0
This will work with a 2 digit number like '11' as it will return a value that is <> 0 if it finds a match. But it will fail when searching for say '3' because rows with 'id' 2 and 3 both will be returned.
Here is link that talks about the POSITION function in REDSHIFT.
Any other approach that will solve this problem?
you can get the count of this string
SELECT id FROM myTbl WHERE regexp_count(user_action, '[11]') > 0
I think we can use regexp_substr() as follow.
select tb .id from myTbl tb where '11' in (
select regexp_substr( (select code from myTbl where id=tb.id),'[^,]+', 1, LEVEL) from dual
connect by regexp_substr((select code from myTbl where id=tb.id) , '[^,]+', 1, LEVEL) is not null);
just try this.
Use split_part() function
SELECT distinct id
FROM myTbl
WHERE '11' in ( split_part( code||',' , ',', 1 ),
split_part( code||',' , ',', 2 ),
split_part( code||',' , ',', 3 ) )
This is a very, very bad data model. You should be storing this information in a junction/association table, with one row per value.
But, if you have no choice, you can use like:
SELECT id
FROM myTbl
WHERE ',' || code || ',' LIKE '%,11,%';
I have a master table that contains a list of strings to search for. it returns TRUE/FALSE if any string in the cell contains text from the master lookup table. Currently I use excel's
=SUMPRODUCT(--ISNUMBER(SEARCH(masterTable,[#searchString])))>0
is there a way to do something like this in SQL? LEFT JOIN or OUTER APPLY would be simple solutions if the strings were equal; but they need be contains..
SELECT *
FROM t
WHERE col1 contains(lookupString,lookupColumn)
--that 2nd table could be maintained and referenced from multiple queries
hop
bell
PRS
2017
My desired results would be a column that shows TRUE/FALSE if the row contains any string from the lookup table
SEARCH_STRING Contained_in_lookup_column
hopping TRUE
root FALSE
Job2017 TRUE
PRS_tool TRUE
hand FALSE
Sorry i dont have access to the DB now to confirm the syntax, but should be something like this:
SELECT t.name,
case when (select count(1) from data_table where data_col like '%' || t.name || '%' > 0) then 'TRUE' else 'FALSE' end
FROM t;
or
SELECT t.name,
case when exists(select null from data_table where data_col like '%' || t.name || '%') then 'TRUE' else 'FALSE' end
FROM t;
Sérgio
You can use a combination of % wildcards with LIKE and EXISTS.
Example (using Oracle syntax) - we have a v_data table containing the data and a v_queries table containing the query terms:
with v_data (pk, value) as (
select 1, 'The quick brown fox jumps over the lazy dog' from dual union all
select 2, 'Yabba dabba doo' from dual union all
select 3, 'forty-two' from dual
),
v_queries (text) as (
select 'quick' from dual union all
select 'forty' from dual
)
select * from v_data d
where exists (
select null
from v_queries q
where d.value like '%' || q.text || '%');
I have a table EmployeeTable.
If I want only that records where employeename have character of 1 to 5
will be palindrome and there also condition like total character is more then 10 then 4 to 8 if character less then 7 then 2 to 5 and if character less then 5 then all char will be checked and there that are palindrome then only display.
Examples :- neen will be display
neetan not selected
kiratitamara will be selected
I try this something on string function like FOR first case like name less then 5 character long
SELECT SUBSTRING(EmployeeName,1,5),* from EmaployeeTable where
REVERSE (SUBSTRING(EmployeeName,1,5))=SUBSTRING(EmployeeName,1,5)
I want to do that without string functions,
Can anyone help me on this?
You need at least SUBSTRING(), I have a solution like this:
(In SQL Server)
DECLARE #txt varchar(max) = 'abcba'
;WITH CTE (cNo, cChar) AS (
SELECT 1, SUBSTRING(#txt, 1, 1)
UNION ALL
SELECT cNo + 1, SUBSTRING(#txt, cNo + 1, 1)
FROM CTE
WHERE SUBSTRING(#txt, cNo + 1, 1) <> ''
)
SELECT COUNT(*)
FROM (
SELECT *, ROW_NUMBER() OVER (ORDER BY cNo DESC) as cRevNo
FROM CTE t1 CROSS JOIN
(SELECT Max(cNo) AS strLength FROM CTE) t2) dt
WHERE
dt.cNo <= dt.strLength / 2
AND
dt.cChar <> (SELECT dti.cChar FROM CTE dti WHERE dti.cNo = cRevNo)
The result will shows the count of differences and 0 means no differences.
Note :
Current solution is Non-Case-Sensitive for change it to a Case-Sensitive you need to check the strings in a case-sensitive collation like Latin1_General_BIN
You can use this solution as a SVF or something like that.
I dont realy understand why you dont want to use string functions in your query, but here is one solution. Compute everything beforehand:
Add Column:
ALTER TABLE EmployeeTable
ADD SubString AS
SUBSTRING(EmployeeName,
(
CASE WHEN LEN(EmployeeName)>10
THEN 4
WHEN LEN(EmployeeName)>7
THEN 2
ELSE 1 END
)
,
(
CASE WHEN LEN(EmployeeName)>10
THEN 8
WHEN LEN(EmployeeName)>7
THEN 5
ELSE 5 END
)
PERSISTED
GO
ALTER TABLE EmployeeTable
ADD Palindrome AS
REVERSE(SUBSTRING(EmployeeName,
(
CASE WHEN LEN(EmployeeName)>10
THEN 4
WHEN LEN(EmployeeName)>7
THEN 2
ELSE 1 END
)
,
(
CASE WHEN LEN(EmployeeName)>10
THEN 8
WHEN LEN(EmployeeName)>7
THEN 5
ELSE 5 END
)) PERSISTED
GO
Then your query will looks like:
SELECT * from EmaployeeTable
where Palindrome = SubString
BUT!
This is not a good idea. Please tell us, why you dont want to use string functios.
You could do it building a list of palindrome words using a recursive query that generates palindrome words till a length o n characters and then selects employees with the name matching a palindrome word. This may be a really inefficient way, but it does the trick
This is a sample query for Oracle, PostgreSQL should support this feature as well with little differences on syntax. I don't know about other RDBMS.
with EmployeeTable AS (
SELECT 'ADA' AS employeename
FROM DUAL
UNION ALL
SELECT 'IDA' AS employeename
FROM DUAL
UNION ALL
SELECT 'JACK' AS employeename
FROM DUAL
), letters as (
select chr(ascii('A') + rownum - 1) as letter
from dual
connect by ascii('A') + rownum - 1 <= ascii('Z')
), palindromes(word, len ) as (
SELECT WORD, LEN
FROM (
select CAST(NULL AS VARCHAR2(100)) as word, 0 as len
from DUAL
union all
select letter as word, 1 as len
from letters
)
union all
select l.letter||p.word||l.letter AS WORD, len + 1 AS LEN
from palindromes p
cross join letters l
where len <= 4
)
SEARCH BREADTH FIRST BY word SET order1
CYCLE word SET is_cycle TO 'Y' DEFAULT 'N'
select *
from EmployeeTable
WHERE employeename IN (
SELECT WORD
FROM palindromes
)
DECLARE #cPalindrome VARCHAR(100) = 'SUBI NO ONIBUS'
SET #cPalindrome = REPLACE(#cPalindrome, ' ', '')
;WITH tPalindromo (iNo) AS (
SELECT 1
WHERE SUBSTRING(#cPalindrome, 1, 1) = SUBSTRING(#cPalindrome, LEN(#cPalindrome), 1)
UNION ALL
SELECT iNo + 1
FROM tPalindromo
WHERE SUBSTRING(#cPalindrome, iNo + 1, 1) = SUBSTRING(#cPalindrome, LEN(#cPalindrome) - iNo, 1)
AND LEN(#cPalindrome) > iNo
)
SELECT IIF(MAX(iNo) = LEN(#cPalindrome), 'PALINDROME', 'NOT PALINDROME')
FROM tPalindromo
I need to check if a partial name matches full name. For example:
Partial_Name | Full_Name
--------------------------------------
John,Smith | Smith William John
Eglid,Timothy | Timothy M Eglid
I have no clue how to approach this type of matching.
Another thing is that name and last name may come in the wrong order, making it harder.
I could do something like this, but this only works if names are in the same order and 100% match
decode(LOWER(REGEXP_REPLACE(Partial_Name,'[^a-zA-Z'']','')), LOWER(REGEXP_REPLACE(Full_Name,'[^a-zA-Z'']','')), 'Same', 'Different')
you could use this pattern on the text provided - works for most engines
([^ ,]+),([^ ,]+)(?=.*\b\1\b)(?=.*\b\2\b)
Demo
WITH
/*
tab AS
(
SELECT 'Smith William John' Full_Name, 'John,Smith' Partial_Name FROM dual
UNION ALL SELECT 'Timothy M Eglid', 'Eglid,timothy' FROM dual
UNION ALL SELECT 'Tim M Egli', 'Egli,Tim,M2' FROM dual
UNION ALL SELECT 'Timot M Eg', 'Eg' FROM dual
),
*/
tmp AS (
SELECT Full_Name, Partial_Name,
trim(CASE WHEN instr(Partial_Name, ',') = 0 THEN Partial_Name
ELSE regexp_substr(Partial_Name, '[^,]+', 1, lvl+1)
END) token
FROM tab t CROSS JOIN (SELECT lvl FROM (SELECT LEVEL-1 lvl FROM dual
CONNECT BY LEVEL <= (SELECT MAX(LENGTH(Partial_Name) - LENGTH(REPLACE(Partial_Name, ',')))+1 FROM tab)))
WHERE LENGTH(Partial_Name) - LENGTH(REPLACE(Partial_Name, ',')) >= lvl
)
SELECT Full_Name, Partial_Name
FROM tmp
GROUP BY Full_Name, Partial_Name
HAVING count(DISTINCT token)
= count(DISTINCT CASE WHEN REGEXP_LIKE(Full_Name, token, 'i')
THEN token ELSE NULL END);
In the tmp each partial_name is splitted on tokens (separated by comma)
The resulting query retrieves only those rows which full_name matches all the corresponding tokens.
This query works with the dynamic number of commas in partial_name. If there can be only zero or one commas then the query will be much easier:
SELECT * FROM tab
WHERE instr(Partial_Name, ',') > 0
AND REGEXP_LIKE(full_name, substr(Partial_Name, 1, instr(Partial_Name, ',')-1), 'ix')
AND REGEXP_LIKE(full_name, substr(Partial_Name,instr(Partial_Name, ',')+1), 'ix')
OR instr(Partial_Name, ',') = 0
AND REGEXP_LIKE(full_name, Partial_Name, 'ix');
This is what I ended up doing... Not sure if this is the best approach.
I split partials by comma and check if first name present in full name and last name present in full name. If both are present then match.
CASE
WHEN
instr(trim(lower(Full_Name)),
trim(lower(REGEXP_SUBSTR(Partial_Name, '[^,]+', 1, 1)))) > 0
AND
instr(trim(lower(Full_Name)),
trim(lower(REGEXP_SUBSTR(Partial_Name, '[^,]+', 1, 2)))) > 0
THEN 'Y'
ELSE 'N'
END AS MATCHING_NAMES