As far as I know I can do something like:
"SELECT *
FROM my_table
WHERE my_field LIKE '0%'
OR my_field LIKE '1%'
OR my_field LIKE '2%' ";
Is there a way to achieve this with a regular expression or something like this:
"SELECT *
FROM my_table
WHERE my_field LIKE [only first char is 0-9]"??
EDIT: The field is not numeric and it can be something like "1 People", "211 Pies" and so on.
Try this
SELECT *
FROM BadlyDesignedTable
WHERE AnswerColumn RLIKE '^[0-9]+'
I was wondering if it was even possible to regex in where, found it on google in 30 seconds.
SELECT *
FROM my_table
WHERE left(my_field,1) REGEXP '^[0-9]+';
If you have MSSQL you can use
SELECT *
FROM my_table
WHERE isNumeric(left(my_field,1)) = 1
SELECT * FROM table WHERE field REGEXP '^[0-9]'
Select * From my_table Where (REGEXP_LIKE(my_field, '[[:digit:]]%'))
The (REGEXP_LIKE(Source_String, '[[:character class:]]')) is a function you can use for numerous issues such as the one you have. Simply use it to tell it to do that for the first digit.
http://docs.oracle.com/cd/B14117_01/server.101/b10759/conditions018.htm
EDIT: Select * From my_table Where (SUBSTR(my_field,1,1) = '[[:digit:]]') Tried this in a similar query of mine, and it works.
A simple one without regular expressions:
SELECT *, SUBSTR(my_field, 1, 1) AS mf FROM my_table HAVING mf BETWEEN '0' AND '9';
The quotes in the between clause are important!