Select only half the records

霸气de小男生 提交于 2021-02-05 02:35:58

问题


I am trying to figure out how to select half the records where an ID is null. I want half because I am going to use that result set to update another ID field. Then I am going to update the rest with another value for that ID field.

So essentially I want to update half the records someFieldID with one number and the rest with another number splitting the update basically between two values for someFieldID the field I want to update.


回答1:


In oracle you can use the ROWNUM psuedocolumn. I believe in sql server you can use TOP. Example:

select TOP 50 PERCENT * from table



回答2:


You can select by percent:

 SELECT TOP 50 PERCENT *fields* FROM YourTable WHERE ...



回答3:


update x set id=@value from (select top 50 percent * from table where id is null) x



回答4:


The following SQL will return the col_ids of the first half of the table.

SELECT col_id FROM table
WHERE rownum <= (SELECT count(col_id)/2 FROM table);

If the total number of col_ids is an odd number then you will get the first half - 1. This is because, for instance, we have 51 total records, the count(col_id)/2 returns 25.5, and since there is no rownum equal to this result, we get everything equal to 25 and below. That means the other 26 are not returned.

However, I have not seen the reverse statement working:

SELECT col_id FROM table
WHERE rownum > (SELECT count(col_id)/2 FROM table);

So if you want the other half of the table, you could just store the first results into a temp table, lets call it TABLE_A. Then just do MINUS on the original table from this table:

SELECT col_id FROM table
MINUS
SELECT col_id FROM table_a

Hopefully this helps someone.



来源:https://stackoverflow.com/questions/686648/select-only-half-the-records

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