I have two tables. Main table is "CompleteEmailListJuly11" and the second table is "CurrentCustomersEmailJuly11". I want to delete rows in CompleteEmailListJuly11 table that CurrentCustomersEmailJuly11 has based off email.
I've tried this following Delete example, but it doesn't do anything close to what I'm trying to do. This only shows me the ones that EXIST in the database, it doesn't show me the the list of emails that AREN'T matching.
DELETE * FROM CompleteEmailListJuly11 AS i
WHERE EXISTS (
SELECT 1 FROM CurrentCustomersEmailJuly11
WHERE CurrentCustomersEmailJuly11.email = i.EmailAddress
)
Help is greatly appreciated.
This is the query I think you need:
DELETE FROM CompleteEmailListJuly11
WHERE EmailAddress IN (SELECT email FROM CurrentCustomersEmailJuly11)
Ps: The DELETE query does not delete individual fields, only entire rows, so the * is not necessary, you will also need to "Execute" this query rather than "Previewing" or "Exporting"
If you're building your DELETE
query in Access' query designer, notice there are two different modes of operation which seem similar to "go ahead and do this".
- Datasheet View (represented by the grid icon labeled "View" on the "Design" section of the ribbon). That view enables you to preview the affected records, but does not actually delete them.
- The "Run" icon (represented by a red exclamation point). "Run" will actually execute the query and delete the affected records.
If you already know this, my description may seem insulting. Sorry. However, it seems that folks new to Access can easily overlook the distinction between them.
You can use something like this adapted to delete
SELECT ... // complete
EXCEPT
SELECT ... // current
I am not sure exactly how it maps to delete but take a look at that.
I fond it in a similar question: How do I 'subtract' sql tables?
We can use Correlated Query to resolve the issue like
DELETE FROM COMPLETE C
WHERE EMAIL = (SELECT EMAIL FROM CURR CU WHERE CU.EMAIL=C.EMAIL);
来源:https://stackoverflow.com/questions/11437677/how-to-remove-row-that-exists-in-another-table