Using LIKE to match floats when Rounding is not wanted

Question

So here is what I am trying to accomplish. I have coordinate data in two tables, which I am trying to link together with an Association table. It is a one to one association, so if I have 40 records in table A, 40 records in table B, then I should have 40 records in the association table.

Now the data in these two tables is approximately the same, but never idential, in fact they rarely even have the same precision. One table(we'll say A) always has 6 decimal places, whereas table B may have no decimal places or up to 6 decimal places.

So let's say we're just matching up one pair of data, say, 12.345678 in table A against table B, with 12.34.

So I have a foreach in my asp.net code that forces zeros onto the end of the table B data, so we first compare 12.345678 against 12.340000.

Then 12.34567 against, 12.34000. Then 12.3456 against, 12.3400 Then 12.345 against,12.340

Then 12.34, against,12.34.

So long as an association record doesn't already exist, containing a reference to 12.345678 in table A, or 12.34 in table B, a new association record is created.

Now you may be asking, Joe, so how are you comparing data in Table A against data in Table B? I saved this part for last because it's the weirdest.

I am using LIKE, which I am sure will upset some people because you're already thinking, "Why in the hell are you using LIKE, which is meant for string matches for floats?"

Well because it works the best so far, about 95% of the time. The majority of that other 5% is just because the data is too different, but there is a very odd subset that most absolutely should be matching.

So before I insert a record, I check for the match and so long as I have only one match, I create the association record.

SELECT COUNT(*) FROM dbo.StartCoord 

WHERE StartLatitude LIKE '12.817%' 
AND StartLongitude LIKE '12.819%'

Now I am looking right now at the record where the 12.817 and 12.819 came from, and the full values are actually 12.8179 and 12.8199. So it shoould work, and for 95% of the time it does work.

Now for the weird part, maybe, using LIKE(which should be only for string matching) is causing SQL Server to do rounding in the background. My above stmt does not work, but if I throw it in Microsoft SQL Server Management, and change it to...

SELECT COUNT(*) FROM dbo.StartCoord 

WHERE StartLatitude LIKE '12.817%' --trying to string match 12.8179
AND StartLongitude LIKE '12.82%'   --trying to string match 12.8199

...it Works!

I am assuming someone is going to say that it's not actually the LIKE, but the fact that I am comparing the LIKE '12.817%' to a float and that float is causing SQL Server to enact some rounding mechanism.

However, if that was the case, why would LIKE '12.817%' match the original 12.8179? Should it not have rounded as well, and only matched in the case of 12.82?

After reading this, if anyone has a better title I could use for anyone else in the future having the same problem, that would be great.

Thanks.

EDIT: So I completely forgot to mention why this approach was taken. In one table, actual truth data is stored up to six decimal places, which I think I consistently used as the Table A example. However, the data in Table B, that varies from no decimal places to six is sometimes rounded, sometimes not.

So in Table A we may have 12.123456 and in somes cases, they give us Table B data that could be 12.1234 or sometimes it could be 12.1235. How they give us data is not consistent, which is why I am working around it this way. Using rounding or casting(numeric) to handle both of these cases results in less Associations being created, but I've only started experimenting with that. I also found a STR() function I am interested to look at.

Answer 1

If you do not want to change from using LIKE. You could cast the float to a decimal and then to an nvarchar, this should stop the rounding issue.

SELECT COUNT(*) FROM dbo.StartCoord 
WHERE CAST(CAST(StartLatitude  as DECIMAL(12,6)) as nvarchar(20)) LIKE '12.817%' 
AND   CAST(CAST(StartLongitude as DECIMAL(12,6)) as nvarchar(20)) LIKE '12.819%'

I have assumed 6 decimal places is all you need.