如何从sql中的列获取特定的子字符串

Question

I have a Orders table that has one of the columns called "details" as:

Contact ID: A18YTX7GWEJRU8    City/Site and Site Name: Orlando - Orlando (UFL4)    Date of Call (MM/DD/YYYY): 01/23/2017    Time of Call (Local Time): 16:44    Order ID(s): 112-0654231-9637802    Call Summary: Cx did not receive. Order marked as delivered to doorstep at 16:27      created by flx-cstech on behalf of sssmiley.

There are different cell values in that column. Also could be like:

Short Description: Dry Ice Retrieval    Please enter the following information for the site ops to pick up the dry ice from the customer:    Contact ID:AD3R60PA1QCCF    Order ID:112-6254812-3186644

Or anything else.

I just want to extract the Order ID(s): 112-0654231-9637802 part from it. How do I do that?

Answer 1

SELECT REGEXP_SUBSTR(
         your_column,
         'Order\s+ID(\s*\(s\))?:\s*\d{3}-\d{7}-\d{7}'
       )
FROM   your_table

To just get the number you can wrap the number in a capture group:

SELECT REGEXP_SUBSTR(
         your_column,
         'Order\s+ID(\s*\(s\))?:\s*(\d{3}-\d{7}-\d{7})',
         1,                                              -- Start from the 1st character
         1,                                              -- Get the 1st match
         NULL,                                           -- Apply default flags
         2                                               -- Get the 2nd capture group
       )
FROM   your_table

Or, if you do not have anything else with the same 3-digit, dash, 7-digit, dash, 7-digit format:

SELECT REGEXP_SUBSTR(
         your_column,
         '\d{3}-\d{7}-\d{7}',
       )
FROM   your_table

Answer 2

你的字符串看起来像一个固定的格式字符串，所以最简单的方法是：

select substr(detail, 160, 31)

Answer 3

https://docs.oracle.com/cd/B12037_01/appdev.101/b10795/adfns_re.htm

REGEXP_LIKE This function searches a character column for a pattern. Use this function in the WHERE clause of a query to return rows matching the regular expression you specify.

and similar functions