如何仅在第一个数字是从0到9的数字的位置选择行？

Question

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.

Answer 1

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.

Answer 2

SELECT * FROM table WHERE field REGEXP '^[0-9]' 

Answer 3

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.

Answer 4

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

Answer 5

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!