I use the LIKE statement like '%/_Description' but got the following results:
Description
Something/_Description
Using '%/[_]Description', "Description" is returned.
How can I restrict the result only to something like "Something/_Description"
Please note this is a standard SQL so supposed to be ran at both SQL Server/Oracle
Thanks,
You can escape the wildchar characters with the character defined in the ESCAPE clause:
match_expression [ NOT ] LIKE pattern [ ESCAPE escape_character ]
In your case (untested):
WHERE ... LIKE '%/\_Description' ESCAPE '\'
This syntax is shared by SQL Server and Oracle.
you need to understand how the LIKE works:
// return rows like text111, text222, text333 etc
select * from users where username like 'text%';
// return rows like 111text, 222text, 333text etc
select * from users where username like '%text';
// return rows like 111text111, 222text222, 333text333 etc
select * from users where username like '%text%';
You have to escape the _ properly as it's a placeholder for like too. Based on your DBMS it should be ! so your statement could look like %/!_Description
the _ char is a special jolly character in Oracle. That's why the rows returned are not as you expected. If this query is generated by a program you can have a substitute the _ with [_].
Alternatively, you can use the REPLACE function in oracle:
REPLACE( string1, string_to_replace [, replacement_string] )
in your case
REPLACE( string_var, '_', [_] )
be aware that this may be a bottleneck if your function is called too many times, so I suggest to use an escape character.
example:
select REPLACE( 'aaaa_bbb', '_', '[_]' ) from dual
Related
What I want to do is I want to filter from a table where it specifically starts with something but when I try to filter out 'BRA_' it should only give me fields that has BRA_ on it and not BRAzil BRAma which is what the Like % operator is giving me. I also tried using Like _ but that only took the underscore out completely. Is there a way to do this?
SELECT *
FROM MyTable
WHERE MyColumn Collate Latin1_General_CS_AS LIKE 'BRA?_' ESCAPE '?'
In the LIKE operator, ESCAPE create your own escape char that must be placed before the joker character that you would find as a litteral in the string.
I tryed some wildcards in Access in Where statement, but they don't works. For example:
This query SELECT staff.* FROM staff;returns:
I tryed to do a query with wildcard SELECT staff.* FROM staff WHERE (staff.s_name LIKE "A%");
but it returns an empty table:
What is the reason? My wildcard doesn't work
(s_name is the second column)
(look that "firstname" is the tag of "s_name" only for the view)
Wildcard character in Access is *, not % unlike in SQL Server.
See MSDN for details.
No, no, no, use '*', not '%'. Or, use 'Like'.
http://www.techrepublic.com/article/10-tips-for-using-wildcard-characters-in-microsoft-access-criteria-expressions/
https://www.techonthenet.com/access/queries/like.php
https://www.techonthenet.com/access/queries/like2007.php
For instance:
Like 'm*'
Result: all values that start with m
Like 'm'
Result: all values that contain m
Like '*m'
Result: all values that end with m
I have this situation on my azure database where I need to search for any rows that contains the _ character. This is a special character on the database so I try to escape it but I get every row as a result.
select * from table where fieldColumn like '%_%'
will return everything on the table
select * from table where fieldColumn like '%\_%'
returns nothing
select * from table where fieldColumn = '_'
works
so how can i get that row that has only one _ and all the other ones that may have the _ on the string?
You can set whatever escape character you want, like this:
select * from table where fieldColumn like '%!_%' ESCAPE '!'
Here I am using the ! as an escape character to tell SQL Server to treat the following character, the _ , as a string literal.
See the documentation for more info: http://technet.microsoft.com/en-us/library/ms179859.aspx
select * from table where fieldColumn like '%_%' escape '\';
LIKE:
escape_character
Is a character that is put in front of a wildcard character to indicate that the wildcard should be interpreted as a regular character and not as a wildcard. escape_character is a character expression that has no default and must evaluate to only one character.
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
I would like to select all records that have an underscore character in their 11th character,
so i try this:
SELECT * FROM "BOM_SUB_LEVEL" where TOP_CODE like '%%%%%%%%%%_%%%'
but this doesnt work as expected, can someone help?
Just use the "SUBSTRING" function :
SELECT * FROM "BOM_SUB_LEVEL" where SUBSTRING(TOP_CODE, 11, 1) = "_"
Marc
For a single character wildcard use _. For multiple characters wildcards, use %. To escape a "real" appearance of _, use \_ (thanks Bill!).
Try the following code:
SELECT * FROM "BOM_SUB_LEVEL" where TOP_CODE like '___________\_%'
To further elaborate following Dav's comment, note that '%%%' is exactly the same as '%', since by definition '%' covers multiple characters.
pervasive uses _ to match any single character and \_ to actually match an underscore.
so the select would be:
SELECT * FROM "BOM_SUB_LEVEL" where TOP_CODE like '___________\_%'
LIKE % can mean any number of characters, use LIKE _ to mean just one. Since you're looking for an underscore, you need to escape it with !.
SELECT * FROM BOM_SUB_LEVEL WHERE TOP_CODE LIKE '__________!_%'
The % is not a per character wildcard, its a beginning and end of string wild card.
i.e. if I want to find all rows that have "car" in them, I would do this:
Select * from myTable where myCol LIKE '%car%'
If I wanted just the rows that STARTED with car:
Select * from myTable where myCol LIKE 'car%'
and ended with car:
Select * from myTable where myCol LIKE '%car'
% is a wildcard and can replace an character, or combination of characters. Use ? instead which replaces a single character.
You can try something like: (play with the numbers, I don't have pervasive to test with)
SELECT *
FROM BOM_SUB_LEVEL
where SUBSTRING(TOP_CODE, 11,1) = '-'