Oleg Netchaev (2/5/2010)

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JosephDMarsh (2/5/2010)

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I chose 4, but (as it turns out) I was guessing. (I thought I knew what was going on, chose an answer and clicked 'Submit' before actually studying the question -- yes, that's a problem I'm working on)

After re-reading the explanation, as well as all of the posts in this thread, I'm not clear on what is happening exactly.

As a Rookie, this is what I *think* is happening:

1. Temp table is created

2. 10 records are created in temp table

3. 4 More records are added to temp table (first part of @sql_str)

4. All records are deleted from temp table (second part of @sql_str)

Obviously, that is not the case. I can't find anything in BOL to help figger this out. Any help?

Thanks in advance,

- Joseph Marsh

Joseph,

You are correct, but step 3 needs tweaking and last important step is missing. Here is what is happening:

1. Temp table is created

2. 10 records are inserted into the temp table

3. The dynamic sql is examined and then executed by the engine. Since the first part of the dynamic sql selects 4 records, those are placed on the heap (saved in memory) in order to be returned back when needed. They cannot be returned back as of yet because the dynamic sql has second part.

4. All records (10 to be exact) are deleted from the temp table.

5. insert into part now kicks in and what the engine sees at this point is to execute the following: insert into temp table select whatever was selected and saved on the heap from the executing dynamic sql, which happens to be select first 4 records from temp table. Thus, the 4 originally selected records are inserted back into the temp table after every row has just been deleted from it. 4 records inserted into empty table make the table now have 4 records.

Hope this helps.

This is why I mentioned in my earlier post that if you were to replace the delete from temp table (second portion of the dynamic sql) with something like select 99, 99 then results of the first select (4 records) will be inserted and the results of the second select (one record) will be inserted as well and the result will then be 15 records in the temp table. And if you were to replace the delete from temp table part with the select of a different shape (something like select 1) then nothing will be inserted because the second select will return data not compatible with expected shape (2 columns per record returned).

Oleg

Oleg