I am trying to getting just the first two words on sql query, I am using the match: ^\w{2}- but with no success because nothing is coming to me, I need to get those values
BA, CE, DF, ES, GO, I don't know how can I do that, below some data example.
SC&Tipo=FM
SC&Tipo=Web
SC&Tipo=Comunitaria
RS&Tipo=Todas
RS&Tipo=AM
RS&Tipo=FM
RS&Tipo=Web
RS&Tipo=Comunitaria
BA-Salvador&Tipo=12horas
CE-Fortaleza&Tipo=12horas
CE-Interior&Tipo=12horas
DF-Brasilia&Tipo=12horas
ES-Interior&Tipo=12horas
ES-Vitoria&Tipo=12horas
GO-Goiania&Tipo=12horas
MG-ZonaDaMata/LestedeMinas&Tipo=12horas
MG-AltoParanaiba&Tipo=12horas
MG-BeloHorizonte&Tipo=12horas
MG-CentroOestedeMinas&Tipo=12horas
Query: SELECT * FROM tabel WHERE filter REGEXP '^\w{2}-'
EDIT SOLVED:
To solve the query should be:
SELECT SUBSTRING(column, 1, 2) AS column FROM table WHERE column REGEXP '^[[:alnum:]_]{2}-'
MySQL doesn't support the character class \w or \d. Instead of \w you have to use [[:alnum:]]. You can find all the supported character classes on the official MySQL documentation.
So you can use the following solution using REGEXP:
SELECT *
FROM table_name
WHERE filter REGEXP '^[[:alnum:]]{2}-'
You can use the following to get the result with regular expression too, using REGEXP_SUBSTR:
SELECT REGEXP_SUBSTR(filter, '^[[:alnum:]]{2}-')
FROM table_name
WHERE filter REGEXP '^[[:alnum:]]{2}-';
Or another solution using HAVING to filter the result:
SELECT REGEXP_SUBSTR(filter, '^[[:alnum:]]{2}-') AS colResult
FROM table_name
HAVING colResult IS NOT NULL;
To get the value before MySQL 8.0 you can use the following with LEFT:
SELECT LEFT(filter, 3)
FROM table_name
WHERE filter REGEXP '^[[:alnum:]]{2}-';
demo: https://www.db-fiddle.com/f/7mJEmCkEiYhCYK3PcEZTNE/0
Using SUBSTRING(<column>, 1, 2) should also work..
More or less like below
SELECT
<column>
, SUBSTRING(<column>, 1, 2)
FROM
<table>
WHERE
SUBSTRING(<column>, 1, 2) IN ('BA' [,<value>..])
Some things are BNF (Backus-Naur form) in the SQL code.
<..> means replace with what you need.
[, ..] means optional unlimited repeat the comma in there is part off SQL syntax
Related
I have a table with a varchar column that represents a path. I want to search for rows that have a path that follow a pattern like name.name[*] where name can be anything. I am looking for repeated strings contained anywhere in the path column that are separated by a period and have a square bracket after them.
This seems to call for Regexp, so through python I have something like https://regex101.com/r/apS20a/4
However, trying to implement this with MySQL Regexp is not working. I have been able to translate the shorthand into REGEXP '([A-Za-z_]+).(\1[[0-9]+])', but it seems that MySql Regex does not support capture groups. Is there a way to accomplish what I am trying to do with mysql regexp? Thank you
I don't think that MySQL supports capture groups. But if you only have one example of .name[ in the string between the first . and the first [, you can hack your way around it. This is not a general solution, just a specific approach in this case.
You can get the name with:
select substring_index(substring_index(url, '[', 1), '.', -1) as name
And then incorporate this into a regular expression:
select t.*
from (select t.*,
substring_index(substring_index(url, '[', 1), '.', -1) as name
from t
) t
where url like concat('%', name, '.', name, '[%');
This just uses like instead of regexp, because [ and . are regular expression wildcards. Of course, this assumes that name does not have _ or %.
EDIT:
Here is a method that actually identifies when this occurs -- and works even if there are multiple patterns.
The idea is to construct the regular expression based on what happens between the . and [ -- and then to apply it. Delightfully self-referential:
select t.*,
(url regexp regex)
from (select t.*,
substr(regexp_replace(url, '[^.]*[.]([^\\[]*)\\[[^.]*', '|$1[.]$1\\\\['), 2) as regex
from (select 'abcde.de[12345.345[ABC' as url union all
select 'abcdefdef[[[[..123.124['
) t
) t;
Here is the above in a db<>fiddle.
I am a beginner so please help me.
There are 2 things you need to combine in this case.
Because you didn't provide enough information in your question we have to guess what you mean by name. I'm going to assume that you have a single name column, but that would be unusual.
With strings, to match a character column that is not an exact match, you need to use LIKE which allows for wildcards.
You also need to negate the match, or in other words show things that are NOT (something).
First to match names that START with 'A'.
SELECT * FROM table_name WHERE name LIKE 'A%';
This should get you all the PEOPLE who have names that "Start with A".
Some databases are case sensitive. I'm not going to deal with that issue. If you were using MySQL that is not an issue. Case sensitivity is not universal. In some RDBMS like Oracle you have to take some steps to deal with mixed case in a column.
Now to deal with what you actually want, which is NOT (starting with A).
SELECT * FROM table_name WHERE name NOT LIKE 'A%';
your question should have more detail however you can use the substr function
SELECT name FROM yourtable
WHERE SUBSTR(name,1,1) <> 'A'
complete list of mysql string functions here
mysql docs
NOT REGXP operator
MySQL NOT REGXP is used to perform a pattern match of a string expression expr against a pattern pat. The pattern can be an extended regular expression.
Syntax:
expr NOT REGEXP pat
Query:
SELECT * FROM emp_table WHERE emp_name NOT REGEXP '^[a]';
or
SELECT * FROM emp_table WHERE emp_name NOT REGEXP '^a';
I want to to create a regex to find all columns that only have a single character ([A-Z]) as name, like N or M but not NM.
I've tried:
SELECT * FROM 'table' WHERE Name REGEXP '^[A-Z]'
But it's not displaying the expected result.
Try ^[A-Z]$.
You then state that this character is first and also last character of the value.
The regex functions in Oracle work only on one column. So, to search for just one character in a column, you would do the following:
select * from yourTable where REGEXP_LIKE (col1, '^[A-z]$');
Now, to search all the char/varchar columns on your table, you'll need to chain the regex expressions together, like so:
select * from yourTable where REGEXP_LIKE (col1, '^[A-z]$') or REGEXP_LIKE (col3, '^[A-z]$');
SQL solution:
where name in ('N','M')
I have a code which I recently discovered :) and it does do its job and well done I might add. But, I want to check all columns instead of checking it by column. Is it possible
Check my code below:
SELECT column_name
FROM table_name
WHERE column_name REGEXP '[[.DLE.]-[.US.]]'
Now, what I want is something like this but it won't work
SELECT *
FROM table_name
WHERE REGEXP '[[.DLE.]-[.US.]]'
Kindly advice and I apologize for asking many questions :)
REGEXP is a binary operator which means you have to have a left operand and a right operand.
Like most arithmetic operators.
You could check all columns like this:
SELECT *
FROM table_name
WHERE CONCAT(a, b, c, d, ...) REGEXP '[[.DLE.]-[.US.]]'
I'm using ... for the sake of the example, but you'd need to name all your columns explicitly. There's no option to use a wildcard for the columns inside an expression.
You can't set "all columns" in a single SET clause. You'd need to do something like the following:
UPDATE table_name SET
a = REPLACE(a,char(16),''),
b = REPLACE(b,char(16),''),
c = REPLACE(c,char(16),''),
d = REPLACE(d,char(16),''),
...similar for other columns;
If you think this is an unexpected omission in the SQL language, then I wonder if you can name any other programming language that lets you compare to or assign a value to "all variables" in a single expression?
I have the following strings in the following pattern in a table in my db:
this_is_my_string_tester1
this_is_my_string_mystring2
this_is_my_string_greatstring
I am trying to match all strings that start with a specific pattern split by underscores i.e.this_is_my_string_ and then a wildcard final section
Unfortunately there is an added complication where some strings like the following:
this_is_my_string_tester1_yet_more_text
this_is_my_string_mystring2_more_text
this_is_my_string_greatstring_more
Therefore taking the following as examples:
this_is_my_string_tester1
this_is_my_string_mystring2
this_is_my_string_greatstring
this_is_my_string_tester1_yet_more_text
this_is_my_string_mystring2_more_text
this_is_my_string_greatstring_more
I am trying to have returned:
this_is_my_string_tester1
this_is_my_string_mystring2
this_is_my_string_greatstring
I have no idea how to do this with a like statement. Is this possible if so how?
EDIT
There is one final complication:
this_is_my_string
needs to be supplied as a list i.e in
(this_is_my_string, this_is_my_amazing_string, this_is_another_amazing_string)
SELECT * FROM atable WHERE afield REGEXP 'this_is_my_string_[a-z]+'
It might be faster if you have an index on afield and do
SELECT * FROM atable WHERE afield REGEXP 'this_is_my_string_[a-z]+'
AND afield LIKE 'this_is_my_string_%'
After edit of question:
Either
SELECT * FROM atable
WHERE afield REGEXP '(this_is_my_string|this_is_my_amazing_string)_[a-z]+'
or maybe you want something like having a table with the prefixes:
SELECT *
FROM atable AS t,
prefixes AS p
WHERE afield REGEXP CONCAT(p.prefix, '_[a-z]+')
As by the reference documentation this should not be possible, as a pattern (string literal) is required. Give it a try nevertheless.
There the answer of #KayNelson with LIKE (?) and INSTR might do instead of REGEXP.
try this
SELECT * FROM db.table WHERE strings LIKE 'this_is_my_string_%' AND instr(Replace(strings,"this_is_my_string_",""),"_") = 0;
It checks if more _ occurs after replacing the standard "this_is_my_string_"