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')
Related
I suck at REGEX, but I need to pull all the records from a table column that stats with AST, and the rest only contains numbers after. I am assuming this can be done with just REGEX and not LIKE but I'm not sure.
For instance AST000001
and not AST99XXH011
SELECT * FROM table WHERE column LIKE 'AST%' AND column REGEXP '[0-9]$'
You can use REGEXP/RLIKE on the whole column value (using start-of-string (^) and end-of-string ($) anchors to ensure you match the entire column):
SELECT *
FROM `table`
WHERE `column` REGEXP '^AST[0-9]+$'
Demo on dbfiddle
I have a table, one of the columns contains a text values, some of which are comma separated string, like this:
Downtown, Market District, Warehouse District
I need to modify my query to see is a given value matches this column. I decided that using IN() is the best choice.
SELECT *
FROM t1
WHERE myValue IN (t1.nighborhood)
I am getting spotty results - sometimes I return records and sometimes not. If there's a value in t1.nighborhood that matches myValue, I do get data.
I checked and there are no MySQL errors. What am I missing?
You can use FIND_IN_SET() to search a comma-delimited list:
SELECT *
FROM t1
WHERE FIND_IN_SET(myValue, REPLACE(t1.nighborhood, ', ', ','));
The REPLACE() is necessary to remove the extra spaces.
Another solution is to use regex to match your search value surrounded by commas if necessary:
SELECT *
FROM t1
WHERE t1.nighborhood REGEXP CONCAT('(^|, )', myValue, '(, |$)');
In general, it's bad design to store distinct values in a single column. The data should be normalized into a related table with a foreign key.
I'm trying to select all rows that contain only alphanumeric characters in MySQL using:
SELECT * FROM table WHERE column REGEXP '[A-Za-z0-9]';
However, it's returning all rows, regardless of the fact that they contain non-alphanumeric characters.
Try this code:
SELECT * FROM table WHERE column REGEXP '^[A-Za-z0-9]+$'
This makes sure that all characters match.
Your statement matches any string that contains a letter or digit anywhere, even if it contains other non-alphanumeric characters. Try this:
SELECT * FROM table WHERE column REGEXP '^[A-Za-z0-9]+$';
^ and $ require the entire string to match rather than just any portion of it, and + looks for 1 or more alphanumberic characters.
You could also use a named character class if you prefer:
SELECT * FROM table WHERE column REGEXP '^[[:alnum:]]+$';
Try this:
REGEXP '^[a-z0-9]+$'
As regexp is not case sensitive except for binary fields.
There is also this:
select m from table where not regexp_like(m, '^[0-9]\d+$')
which selects the rows that contains characters from the column you want (which is m in the example but you can change).
Most of the combinations don't work properly in Oracle platforms but this does. Sharing for future reference.
Try this
select count(*) from table where cast(col as double) is null;
Change the REGEXP to Like
SELECT * FROM table_name WHERE column_name like '%[^a-zA-Z0-9]%'
this one works fine
I have a table, such as
create table table1(
name varchar(32),
);
And there's some data in it. When I select like this:
select * from table1 where name like 'Jack2%';
there will be Jack2.
But if I select like this:
select * from table1 where name like 'Jack[0-9]%';
there will be nothing;
And I also tried regexp to subsitute like, but it also didn't work!
What's wrong?
You've confused two different pattern-matching mechanisms. SQL LIKE uses % to match anything and _ to match any single character; it does not have anything like [0-9] to match a digit. That looks like a character class from a regular expression.
Standard SQL has no support for regular expressions at all, but MySQL does - you just have to use RLIKE (or REGEXP, but that doesn't read as nicely IMO) instead of LIKE. But that means that you have to replace the % with the regular-expression equivalent .*, too.
SELECT * FROM table1 WHERE name RLIKE 'Jack[0-9].*';
Fiddle
MySQL REGEX
select * from Table1 where `name` REGEXP 'Jack[0-9]'
You can use RLIKE instead
SELECT * FROM table1 WHERE name RLIKE 'Jack[0-9].*';
And please note the the '%' operator won't work with RLIKE, you have to use a regular expression pattern like '.*' instead.
I want to use the LIKE operator to match possible values in a column.
If the value begins with "CU" followed by a digit (e.g. "3") followed by anything else, I would like to return it. There only seems to be a wildcard for any single character using underscore, however I need to make sure it is a digit and not a-z.
I have tried these to no avail:
select name from table1 where name like 'CU[0-9]%'
select name from table1 where name like 'CU#%'
Preferably this could be case sensitive i.e. if cu or Cu or cU then this would not be a match.
You need to use regexp:
select name
from table1
where name regexp binary '^CU[0-9]'
The documentation for regexp is here.
EDIT: binary is required to ensure case-sensitive matching
The like operator only have the % and _ wildcards in MySQL, but you can use a regular expression with the rlike operator:
select name from table1 where name rlike '^CU[0-9]'
You can use REGEXP operator, see http://dev.mysql.com/doc/refman/5.1/en/regexp.html#operator_regexp
so your query would be:
select name from table where name regexp 'CU[0-9].*';
Have you tried with:
select name from table where name between 'CU0' and 'CU9'