Sorting by value with ORDER BY?

For clarification, are you able to use MySQL this way to sort?

ORDER BY Comp开发者_Python百科anyID = XXX DESC

What I am trying to do is use one sql query to sort everything where X = Y, in this case, where CompanyID = XXX. All values where CompanyID is not XXX should come after all the results where CompanyID = XXX.

I don't want to limit my query but I do want to sort a particular company above other listings.

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ORDER BY FIELD(CompanyID, 'google', 'apple', 'microsoft') ASC

Google = first
Microsoft = third
The rest = last

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You can use a case in an order by, like:

order by case when CompanyID = XXX then 1 else 2 end

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Your query is fine. CompanyID = xxx yields 1 on a match, and 0 otherwise. Just add the CompanyID column as the secondary order column.

ORDER BY CompanyID = XXX DESC, CompanyID ASC

This way, you get records where the CompanyID matches your constant first, then the rest are in order.

Alternatively, you could do this:

ORDER BY CompanyID <> XXX ASC, CompanyID ASC

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You might want to try using a CASE statement in your order by expression. When the company id matches your preferred company, return 0, else return 1. Then order by company name like usual in a secondary order by column.

 order by case when CompanyID = XXX then 0 else 1 end, CompanyName

That way you have your preferred company at the top, then all of the rest order alphabetically by name (or however you want it ordered).

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You could probably fake it out

SELECT CompanyID, IF(CompanyID = XXX, -1, CompanyID) AS sortby
FROM companies
ORDER BY sortby DESC