SQL Server Not Recognizing Leading Spaces in Select Query - sql-server-2016

Question

I get a TXT file from one of our sources systems each night. It's basically a terminal report with headers, footers, titles, column headers, sub-totals, etc. I'm trying to scrape discrete data elements from the file using SQL Server. The file is being FTP'd to a Windows file share. The source system is AIX and the file's encoding is UTF-8, with an EOL marker of LF. I'm using SSIS to import the raw text report into a single column table with each report row being one record in my table. The column I'm storing the rows in is a VARCHAR(240) and I'm using SQL Server 2016.
For the report rows that I want to use, the one thing they have in common is that they all start with two spaces. Here's an example record from a text report I've loaded to SQL:
COLUMN_NAME
AD DEPT 0 0 0 0 0 0 0 0 0 0.0 0 0 0.00 0.00 0 0 0 0.0
When I try to select the record using:
SELECT *
FROM TABLE_NAME
WHERE COLUMN_NAME LIKE ' %';
No rows are returned in my result set. However, REPLACE seems to recognize the the row starts with two spaces.
So this:
SELECT REPLACE([COLUMN_NAME],' ','$')
FROM TABLE_NAME
Returns this:
COLUMN_NAME
$$AD$DEPT$$$$$$0$$$$$0$$$$$0$$$0$$$$$0$$$$$0$$$$0$$$$$0$$$$0$$$0.0$$$0$$$$$$0$$0.00$$0.00$$$$$$0$$$$$$0$$$$$$0$$$0.0
Can someone help me understand why REPLACE sees that there are two leading spaces in the row but the plain SELECT does not?

Answer 1

If you know that you have 2 spaces as the start of the column then you can use the single character wildcards in your LIKE expression.
For example:
CREATE TABLE testString (
sampleImport varchar(20)
)
GO
INSERT testString
VALUES (' AD DEPT 0 0 0'), ('BD DEPT 0 0 0')
GO
SELECT *
FROM testString
WHERE sampleImport LIKE '[ ][ ]%'
GO
SELECT *
FROM testString
WHERE sampleImport NOT LIKE '[ ][ ]%'
GO
The [] is used to signify a single character - the specific character is the one enclosed in the brackets. So placing a single space within the brackets allows you to match the spaces.
I also noted that your criteria had only a single space before the % character. Although I cannot see it documented as such, I suspect that your version is failing as it is not seeing a leading space as a valid character for the wildcard (although by definition it should). When using LIKE ' %' it works with my test data.