Hello I have made a dummy table that I am practicing with and I am trying to get the lasts name first letter for example. Aba Kadabra and Alfa Kadabra the last letter of their last name is 'K' so when I was testing some queries such as...
select * from employees
where full_name like 'K%'
select * from employees
where full_name like 'K%'
Neither of these worked. Can anyone tell me the best way to accomplish this?
Because % works that way. See here
So, 'K%' just brings all full_name that start with K.
and '%K' brings all full_name that end with K.
What you need is '% K%', test it please.
MySQL LIKE operator checks whether a specific character string matches
a specified pattern.
The LIKE operator does a pattern matching comparison. The operand to
the right of the LIKE operator contains the pattern and the left hand
operand contains the string to match against the pattern. A percent
symbol ("%") in the LIKE pattern matches any sequence of zero or more
characters in the string. An underscore ("_") in the LIKE pattern
matches any single character in the string. Any other character
matches itself or its lower/upper case equivalent (i.e.
case-insensitive matching). (A bug: SQLite only understands
upper/lower case for ASCII characters by default. The LIKE operator is
case sensitive by default for unicode characters that are beyond the
ASCII range. For example, the expression 'a' LIKE 'A' is TRUE but 'æ'
LIKE 'Æ' is FALSE.)
You can use below query:
select * from table where full_name like '% K%'
Related
I try to find query to find a string that 2nd character and 2nd last character both are letter m.
SELECT last_name
FROM employees
WHERE (last_name LIKE '_m%m_' AND LENGTH(last_name) >= '3');
Thanks in advance :)
Why not just OR instead of AND? I don't see the point of AND when your LIKE operator allready rules out names below three characters. You don't need to use regex nor a check for length:
SELECT last_name FROM employees
WHERE last_name LIKE '_m_'
OR last_name LIKE '_m%m_';
The use of OR and LIKE does catch any string that has at least 3 characters.
If you must use regex, try REGEXP operator:
SELECT last_name FROM employees WHERE last_name REGEXP '^.m(.*m)?.$';
Where the pattern means:
^.m - Start-line anchor with a single character and a literal 'm';
(.*m)? - Optional capture group to match 0+ characters upto a literal 'm';
.$ - A single character with end-line anchor.
The benefit of REGEXP is that it's a bit less verbose if you need case-insensitive matching using pattern: '^.[Mm](.*[Mm])?.$'. See an online demo.
If you need all record with second and last character is m you can use the following query:
select * from <table> where <column> like '_m%m'
the _ in the query is a placeholder for one character and % for many characters
I am trying to pull a product code from a long set of string formatted like a URL address. The pattern is always 3 letters followed by 3 or 4 numbers (ex. ???### or ???####). I have tried using REGEXP and LIKE syntax, but my results are off for both/I am not sure which operators to use.
The first select statement is close to trimming the URL to show just the code, but oftentimes will show a random string of numbers it may find in the URL string.
The second select statement is more rudimentary, but I am unsure which operators to use.
Which would be the quickest solution?
SELECT columnName, SUBSTR(columnName, LOCATE(columnName REGEXP "[^=\-][a-zA-Z]{3}[\d]{3,4}", columnName), LENGTH(columnName) - LOCATE(columnName REGEXP "[^=\-][a-zA-Z]{3}[\d]{3,4}", REVERSE(columnName))) AS extractedData FROM tableName
SELECT columnName FROM tableName WHERE columnName LIKE '%___###%' OR columnName LIKE '%___####%'
-- Will take a substring of this result as well
Example Data:
randomwebsite.com/3982356923abcd1ab?random_code=12480712_ABC_DEF_ANOTHER_CODE-xyz123&hello_world=us&etc_etc
In this case, the desired string is "xyz123" and the location of said pattern is variable based on each entry.
EDIT
SELECT column, LOCATE(column REGEXP "([a-zA-Z]{3}[0-9]{3,4}$)", column), SUBSTR(column, LOCATE(column REGEXP "([a-zA-Z]{3}[0-9]{3,4}$)", column), LENGTH(column) - LOCATE(column REGEXP "^.*[a-zA-Z]{3}[0-9]{3,4}", REVERSE(column))) AS extractData From mainTable
This expression is still not grabbing the right data, but I feel like it may get me closer.
I suggest using
REGEXP_SUBSTR(column, '(?<=[&?]random_code=[^&#]{0,256}-)[a-zA-Z]{3}[0-9]{3,4}(?![^&#])')
Details:
(?<=[&?]random_code=[^&#]{0,256}-) - immediately on the left, there must be & or &, random_code=, and then zero to 256 chars other than & and # followed with a - char
[a-zA-Z]{3} - three ASCII letters
[0-9]{3,4} - three to four ASCII digits
(?![^&#]) - that are followed either with &, # or end of string.
See the online demo:
WITH cte AS ( SELECT 'randomwebsite.com/3982356923abcd1ab?random_code=12480712_ABC_DEF_ANOTHER_CODE-xyz123&hello_world=us&etc_etc' val
UNION ALL
SELECT 'randomwebsite.com/3982356923abcd1ab?random_code=12480712_ABC_DEF_ANOTHER_CODE-xyz4567&hello_world=us&etc_etc'
UNION ALL
SELECT 'randomwebsite.com/3982356923abcd1ab?random_code=12480712_ABC_DEF_ANOTHER_CODE-xyz89&hello_world=us&etc_etc'
UNION ALL
SELECT 'randomwebsite.com/3982356923abcd1ab?random_code=12480712_ABC_DEF_ANOTHER_CODE-xyz00000&hello_world=us&etc_etc'
UNION ALL
SELECT 'randomwebsite.com/3982356923abcd1ab?random_code=12480712_ABC_DEF_ANOTHER_CODE-aaaaa11111&hello_world=us&etc_etc')
SELECT REGEXP_SUBSTR(val,'(?<=[&?]random_code=[^&#]{0,256}-)[a-zA-Z]{3}[0-9]{3,4}(?![^&#])') output
FROM cte
Output:
I'd make use of capture groups:
(?<=[=\-\\])([a-zA-Z]{3}[\d]{3,4})(?=[&])
I assume with [^=\-] you wanted to capture string with "-","\" or "=" in front but not include those chars in the result. To do that use "positive lookbehind" (?<=.
I also added a lookahead (?= for "&".
If you'd like to fidget more with regex I recommend RegExr
I am trying to select all client names without vowels from a table (should therefore return an empty list) using the setminus operator with regular expressions, but it is simply returning the entire column. The same happens if I try to select all client names without 'a' or 'e' or any other vowel.
This is the query I'm using:
select client_name from client
where client_name regexp '[^aeiou]';
If I try doing a condition like below, then the inside caret actually does take every character other than 'a'. I'm not sure why it doesn't work by itself though.
select client_name from client
where client_name regexp '^[^a]'
Expected - empty output
Actual Results - whole column is returned
The regular expression can match anywhere in the name. So it will match any name that has any non-vowel character, not where all the characters are not vowels. You need to anchor it and quantify it:
WHERE client_name REGEXP '^[^aeiou]*$'
This tests all the characters in the name.
Or you can negate the test:
WHERE client_name NOT REGEXP '[aeiou]'
The regexp matches a vowel anywhere in the name. Then using NOT makes this return the names that don't match.
I need a SELECT query in MYSQL that will retrieve all rows in one table witch field values contain "?" char with one condition: the char is not the last character
Example:
ID Field
1 123??see
2 12?
3 45??78??
Returning rows would then be those from ID 1 and 3 that match the condition given
The only statement I have is:
SELECT *
FROM table
WHERE Field LIKE '%?%'
But, the MySQL query does not solve my problem..
The LIKE expressions also support a wildcard "_" which matches exactly one character.
So you can write an expression like the example below, and know that your "?" will not be the last character in the string. There must be at least one more character.
WHERE intrebare LIKE '%?_%'
Re comment from #JohnRuddell,
Yes, that's true, this will match the string "??" because a "?" exists in a position that is not the last character.
It depends whether the OP means for that to be a match or not. The OP says the string "45??78??" is a match, but it's not clear if they would intend that "4578??" to be a match.
An alternative is to use a regular expression, but this is a little more tricky because you have to escape a literal "?", so it won't be interpreted as a regexp metacharacter. Then also escape the escape character.
WHERE intrebare REGEXP '\\?[^?]'
you can just add an additional where where the last character is not a ?
SELECT *
FROM intrebari
WHERE intrebare LIKE '%?%' AND intrebare NOT LIKE '%?'
you could also do it like this
SELECT *
FROM intrebari
WHERE intrebare LIKE '%?%' AND RIGHT(intrebare,1) <> '?'
DEMO
example:
SELECT country
FROM data
WHERE city LIKE
(SELECT LEFT ('jakartada',7));
it's working nicely but if i give delimiter with value:4 ,i must give wildcard
like this ---> "%string%" ,where i can give wildcard in the query?
With LIKE you can use the following two wildcard characters in the pattern.
Character Description
% Matches any number of characters, even zero characters
_ Matches exactly one character
\% Matches one “%” character
\_ Matches one “_” character
does this query help you?
SELECT country
FROM data
WHERE city LIKE CONCAT ('%', (SELECT LEFT ('jakartada',7)), '%');
why you using sub query?
SELECT * FROM `data` WHERE city LIKE 'jakar%'