How to exclude all rows with the same field value if one of them doesn't match?

Question

MySQL

For example, there is a table
id | objid | status
1 | 1 | 1
2 | 1 | 0
3 | 1 | 0
4 | 1 | 1
5 | 2 | 0
6 | 2 | 0
7 | 2 | 0
8 | 2 | 0
9 | 3 | 1
10 | 3 | 1
In this case, I only need objid = 2 because all values ​​have status=0. At the same time, objid 1 and 3 are not suitable, because for 3 status=1 and for 1 the status is 1 for some, and 0 for some.
Yes, in theory, you can take those that are equal to 1 and exclude these IDs from the sample. But let's imagine that there are about 100 thousand such records from 1, and it will be expensive to exclude them in a subquery.
Is it possible to tell mysql not to need all objid where status is not 0? Or, to rephrase, exclude all objids if they have at least one value equal to 1?
Let's even assume that we got a list of those where there is at least one zero - how to remove from this list all entries with the same objid where there is at least one unit?
PS. Statuses can be not 2 but more. For example 5 statuses. You need to get those objid where there is status 0 and 4 but no status 1 2 and 3.

Answer 1

Clumsy, but it works
sqlfiddle.com/#!9/cb6ea/1
Another option
sqlfiddle.com/#!9/cb6ea/2
Or even like this:
sqlfiddle.com/#!9/cb6ea/3
For good, you need to look explain, think and/or test against the data, but I'll leave that to you.

Answer 2

Select * from mytable t1
Where not exists (select * from mytable t2
Where t2.objid=t1.objid and t2.status=1)

Answer 3

SELECT `objid`
    FROM `table`
    GROUP BY `objid`
    HAVING MAX(`status`) = 0

SELECT DISTINCT `t1`.`objid`
    FROM `table` AS `t1`
    LEFT JOIN (
        SELECT DISTINCT `objid`
            FROM `table`
            WHERE `status` = 1
    ) AS `t2` USING(`objid`)
    WHERE `t2`.`objid` IS NULL