I want to select names from a table where the 'name' column contains '%' anywhere in the value. For example, I want to retrieve the name 'Approval for 20 % discount for parts'.
SELECT NAME FROM TABLE WHERE NAME ... ?
You can use like with escape. The default is a backslash in some databases (but not in Oracle), so:
select name
from table
where name like '%\%%' ESCAPE '\'
This is standard, and works in most databases. The Oracle documentation is here.
Of course, you could also use instr():
where instr(name, '%') > 0
One way to do it is using replace with an empty string and checking to see if the difference in length of the original string and modified string is > 0.
select name
from table
where length(name) - length(replace(name,'%','')) > 0
Make life easy on yourselves and just use REGEXP_LIKE( )!
SQL> with tbl(name) as (
select 'ABC' from dual
union
select 'E%FS' from dual
)
select name
from tbl
where regexp_like(name, '%');
NAME
----
E%FS
SQL>
I read the documentation mentioned by Gordon. The relevent sentence is:
An underscore (_) in the pattern matches exactly one character (as opposed to one byte in a multibyte character set) in the value
Here was my test:
select c
from (
select 'a%be' c
from dual) d
where c like '_%'
The value a%be was returned.
While the suggestions of using instr() or length in the other two answers will lead to the correct answer, they will do so slowly. Filtering on function results simply take longer than filtering on fields.
Related
I need to find (exclude in fact) any results that contain '%' sign, wherever in a string field. That would mean ... WHERE string LIKE '%%%'. Googling about escaping gave me the following ideas. The first throws syntax error, the second returns rows but there are records actually contain '%'.
1st:
SELECT * FROM table
WHERE string NOT LIKE '%!%%' ESCAPE '!'
///tried with different escape characters
2nd:
SELECT * FROM table
WHERE string NOT LIKE '%[%]%'
Trying on GCP BigQuery.
Try:
SELECT *
FROM table
WHERE string NOT LIKE '%!%%' {ESCAPE '!'}
With curly braces as shown in microsoft sql server docs
Or also:
WITH indata(s) AS (
SELECT 'not excluded'
UNION ALL SELECT '%excluded'
UNION ALL SELECT 'Ex%cluded'
UNION ALL SELECT 'Excluded%'
)
SELECT * FROM indata WHERE INSTR(s,'%') = 0;
-- out s
-- out --------------
-- out not excluded
find (exclude in fact) any results that contain '%'
Consider below simple approach
select *
from your_table
where not regexp_contains(string , '%')
I would like to keep rows only with Numeric Character i.e. 0-9. My source data can have any type of character e.g. 2,%,( .
Input (postcode)
3453gds sdg3
454232
sdg(*d^
452
Expected Output (postcode)
454232
452
I have tried using WHERE REGEXP_LIKE(postcode, '^[[:digit:]]+$');
however in my version of Oracle I get an error saying
function regexp_like(character varying, "unknown") does not exist
You want regexp_like() and your version should work:
select t.*
from t
where regexp_like(t.postcode, '^[0-9]+$');
However, your error looks more like a Postgres error, so perhaps this will work:
t.postcode ~ '^[0-9]+$'
For Oracle 10 or higher you can use regexp functions. In earlier versions translate function will help you :
SELECT postcode
FROM table_name
WHERE length(translate(postcode,'0123456789','1')) is null
AND postcode IS NOT NULL;
OR
SELECT translate(postcode, '0123456789' || translate(postcode,'x123456789','x'),'0123456789') nums
FROM table_name ;
the above answer also works for me
SELECT translate('1234bsdfs3#23##PU', '0123456789' || translate('1234bsdfs3#23##PU','x123456789','x'),'0123456789') nums
FROM dual ;
Nums:
1234323
For an alternative to the Gordon Linoff answer, we can try using REGEXP_REPLACE:
SELECT *
FROM yourTable
WHERE REGEXP_REPLACE(postcode, '[0-9]+', '') IS NULL;
The idea here is to strip away all digit characters, and then assert that nothing were left behind. For a mixed digit-letter value, the regex replacement would result in a non-empty string.
Table DATA
----------------------------
Name
ABC:000
DEF:0
ABD:000
FFF:00
GGG:000
I need only those names which contains only 3 characters post the semicolon.
In the event that the field is stored as a char() and varying, then use trim():
where trim(name) like '%:___'
with
table_name ( name ) as (
select 'ABC:000' from dual union all
select 'DEF:0' from dual union all
select 'ABD:000' from dual union all
select 'FFF:00' from dual union all
select 'GGG:000' from dual
)
-- End of SIMULATED inputs (not part of the SQL query).
-- Solution begins BELOW THIS LINE. Use your actual table and column names.
select name
from table_name
where name like '%:___'
;
NAME
-------
ABC:000
ABD:000
GGG:000
Explanation: like is a comparison operator for strings. % stands for any sequence of characters, of any length (including of length zero). : stands for itself. Underscore stands for exactly one character - ANY character. The comparison string is one % sign, one : semicolon, and three underscores.
I have names in my dataset and they include parentheses. But, I am trying to clean up the names to exclude those parentheses.
Example: ABC Company (Somewhere, WY)
What I want to turn it into is: ABC Company
I'm using standard SQL with google big query.
I've done some research and I know big query has left(), but I do not know the equivalent of find(). My plan was to do something that finds the ( and then gives me everything to the left of -1 characters from the (.
My plan was to do something that finds the ( and then gives me everything to the left of -1 characters from the (.
Good plan! In BigQuery Standard SQL - equivalent of LEFT is SUBSTR(value, position[, length]) and equivalent of FIND is STRPOS(value1, value2)
With this in mind your query can look like (which is exactly as you planned)
#standardSQL
WITH names AS (
SELECT 'ABC Company (Somewhere, WY)' AS name
)
SELECT SUBSTR(name, 1, STRPOS(name, '(') - 1) AS clean_name
FROM names
Usually, string functions are less expensive than regular expression functions, so if you have pattern as in your example - you should go with above version
But in more generic cases, when pattern to clean is more dynamic like in Graham's answer - you should go with solution in Graham's answer
Just use REGEXP_REPLACE + TRIM. This will work with all variants (just not nested parentheses):
#standardSQL
WITH
names AS (
SELECT
'ABC Company (Somewhere, WY)' AS name
UNION ALL
SELECT
'(Somewhere, WY) ABC Company' AS name
UNION ALL
SELECT
'ABC (Somewhere, WY) Company' AS name)
SELECT
TRIM(REGEXP_REPLACE(name,r'\(.*?\)',''), ' ') AS cleaned
FROM
names
Use REGEXP_EXTRACT:
SELECT
RTRIM(REGEXP_EXTRACT(names, r'([^(]*)')) AS new_name
FROM yourTable
The regex used here will greedily consume and match everything up until hitting an opening parenthesis. I used RTRIM to remove any unwanted whitespace picked up by the regex.
Note that this approach is robust with respect to the edge case of an address record not having any term with parentheses. In this case, the above query would just return the entire original value.
I can't test this solution at the moment, but you can combine SUBSTR and INSTR. Like this:
SELECT CASE WHEN INSTR(name, '(') > 0 THEN SUBSTR( name, 1, INSTR(name, '(') ) ELSE name END as name FROM table;
I have (and don't own, so I can't change) a table with a layout similar to this.
ID | CATEGORIES
---------------
1 | c1
2 | c2,c3
3 | c3,c2
4 | c3
5 | c4,c8,c5,c100
I need to return the rows that contain a specific category id. I starting by writing the queries with LIKE statements, because the values can be anywhere in the string
SELECT id FROM table WHERE categories LIKE '%c2%';
Would return rows 2 and 3
SELECT id FROM table WHERE categories LIKE '%c3%' and categories LIKE '%c2%'; Would again get me rows 2 and 3, but not row 4
SELECT id FROM table WHERE categories LIKE '%c3%' or categories LIKE '%c2%'; Would again get me rows 2, 3, and 4
I don't like all the LIKE statements. I've found FIND_IN_SET() in the Oracle documentation but it doesn't seem to work in 10g. I get the following error:
ORA-00904: "FIND_IN_SET": invalid identifier
00904. 00000 - "%s: invalid identifier"
when running this query: SELECT id FROM table WHERE FIND_IN_SET('c2', categories); (example from the docs) or this query: SELECT id FROM table WHERE FIND_IN_SET('c2', categories) <> 0; (example from Google)
I would expect it to return rows 2 and 3.
Is there a better way to write these queries instead of using a ton of LIKE statements?
You can, using LIKE. You don't want to match for partial values, so you'll have to include the commas in your search. That also means that you'll have to provide an extra comma to search for values at the beginning or end of your text:
select
*
from
YourTable
where
',' || CommaSeparatedValueColumn || ',' LIKE '%,SearchValue,%'
But this query will be slow, as will all queries using LIKE, especially with a leading wildcard.
And there's always a risk. If there are spaces around the values, or values can contain commas themselves in which case they are surrounded by quotes (like in csv files), this query won't work and you'll have to add even more logic, slowing down your query even more.
A better solution would be to add a child table for these categories. Or rather even a separate table for the catagories, and a table that cross links them to YourTable.
You can write a PIPELINED table function which return a 1 column table. Each row is a value from the comma separated string. Use something like this to pop a string from the list and put it as a row into the table:
PIPE ROW(ltrim(rtrim(substr(l_list, 1, l_idx - 1),' '),' '));
Usage:
SELECT * FROM MyTable
WHERE 'c2' IN TABLE(Util_Pkg.split_string(categories));
See more here: Oracle docs
Yes and No...
"Yes":
Normalize the data (strongly recommended) - i.e. split the categorie column so that you have each categorie in a separate... then you can just query it in a normal faschion...
"No":
As long as you keep this "pseudo-structure" there will be several issues (performance and others) and you will have to do something similar to:
SELECT * FROM MyTable WHERE categories LIKE 'c2,%' OR categories = 'c2' OR categories LIKE '%,c2,%' OR categories LIKE '%,c2'
IF you absolutely must you could define a function which is named FIND_IN_SET like the following:
CREATE OR REPLACE Function FIND_IN_SET
( vSET IN varchar2, vToFind IN VARCHAR2 )
RETURN number
IS
rRESULT number;
BEGIN
rRESULT := -1;
SELECT COUNT(*) INTO rRESULT FROM DUAL WHERE vSET LIKE ( vToFine || ',%' ) OR vSET = vToFind OR vSET LIKE ('%,' || vToFind || ',%') OR vSET LIKE ('%,' || vToFind);
RETURN rRESULT;
END;
You can then use that function like:
SELECT * FROM MyTable WHERE FIND_IN_SET (categories, 'c2' ) > 0;
For the sake of future searchers, don't forget the regular expression way:
with tbl as (
select 1 ID, 'c1' CATEGORIES from dual
union
select 2 ID, 'c2,c3' CATEGORIES from dual
union
select 3 ID, 'c3,c2' CATEGORIES from dual
union
select 4 ID, 'c3' CATEGORIES from dual
union
select 5 ID, 'c4,c8,c5,c100' CATEGORIES from dual
)
select *
from tbl
where regexp_like(CATEGORIES, '(^|\W)c3(\W|$)');
ID CATEGORIES
---------- -------------
2 c2,c3
3 c3,c2
4 c3
This matches on a word boundary, so even if the comma was followed by a space it would still work. If you want to be more strict and match only where a comma separates values, replace the '\W' with a comma. At any rate, read the regular expression as:
match a group of either the beginning of the line or a word boundary, followed by the target search value, followed by a group of either a word boundary or the end of the line.
As long as the comma-delimited list is 512 characters or less, you can also use a regular expression in this instance (Oracle's regular expression functions, e.g., REGEXP_LIKE(), are limited to 512 characters):
SELECT id, categories
FROM mytable
WHERE REGEXP_LIKE('c2', '^(' || REPLACE(categories, ',', '|') || ')$', 'i');
In the above I'm replacing the commas with the regular expression alternation operator |. If your list of delimited values is already |-delimited, so much the better.