I have this table
id value
1 OK
2 xminimum
3 NO
4 YES
I want to sort this table by value where minimum is always first then the rest according to alphabetic order of value column
Meaning:
xminimum
NO
OK
YES
I wrote this query:
Select *
from table_a
order by case when value='xminimum' then 1 else ????? end
I don't know what to put in the else... conceptually it should be else value end so it means alphabetic order.. but I can not combine integer with text.
How do I fix it?
As requested, copied from my comment:
Select *
from table_a
order by case when value='xminimum' then 1 else 2 end, value
Another solution:
SELECT *
FROM table_a
ORDER BY value <> 'xminimum', value;
Do it like you have and add the value column as second column to sort by:
SELECT *
FROM table_a
ORDER BY CASE WHEN value='xminimum' THEN 1 ELSE 2 END, value
Related
id value
1 a
2 b
3 c
How do i add second value 'z' to id=1 (separated by comma)?
id value
1 a,z
2 b
3 c
and how to remove the 'z' now if i have that final table?
You can use update:
update t
set value = concat(value, ',z')
where id = 1;
To answer your secondary question, yes.
If you run Select value from table where id = 1 it will return a,z. that means that if you are going to use it again in queries, you will quite possibly need to utilize a Split() type function, dependent on what you're doing with it.
The best and simplest way to do this is the following query according to me :
update table1 set value = concat(value,'z') where id = 1
where : Table1 is the name of your table.
I have below data. The condition is that if the Id has two different types then take Long, such that there should not be any duplicate Id's
**id type**
1 Short
1 Long
2 Short
3 Short
3 Long
4 Short
And i need output like this.
**id type**
1 Long
2 Short
3 Long
4 Short
Does this work for you:
select id,
case when count(id) > 1 then 'Long' else 'Short' end as type
from tmp
group by id
You can simply take MIN from your Type column's value using GROUP BY on ID column. No CASE or COUNT statement is required. This following script will always work if you have specific value "short" and "long" in your column Type.
SELECT ID,MIN(Type) Type
FROM your_table
GROUP BY ID
You can do this:
Select id, case when count(id)>1 then 'Long' else min(Type) End as Type
from Tbl
group by id
Based on one column within my query results (Value), I am trying to write an if/else statement based on the value held which will display the result the in an additional row.
For example, if I have a record of 2 within the value field, but I want to check whether it is above < 5. If the value is less than 5 I basically want the additional column to display a hardcoded value of 5, else display actual value.
Any help would be appreciated.
Use a case statement
select a.*,
case
when a.TheField < 5 then 5
else a.TheField
end as NewField
from MyTable a
You can use a case
select value, case when value < 5
then 5
else value
end as calculated_column
from your_table
I have a simple query where I want to search for a value in a column, and if no match is found then I want to discard that where condition and select any random value from that column.
I have tried to achieve this using OR condition, using case, but to no avail
select top 10 *
from tblRFCWorkload
where (f1 ='mbb' or f1 = f1)
In the above query I want the result table to have all rows with 'mbb', if 'mbb' is not found then give any value instead of mbb. But instead it returns only two rows with mbb even though there are 10 matching rows
select top 10 *
from tblRFCWorkload
where 1 = case when f1 = 'mbb' then 1 else 1 end
This query also returns the same result as the 1st query
If I change it to
select top 10 *
from tblRFCWorkload
where 1 = case when f1 = 'mbb' then 1 else 0 end
then it only rows having the mbb value.
Need similar thing with between clause... search for one range if no results found then search for second range... how do I do that???
I think you want something like this:
select top 10 *
from tblRFCWorkload
order by (case when f1 = 'mbb' then 1 else 2 end);
This prioritizes the rows that you want, fetching the top 10 which will be 'mbb', if they are available.
You want only the results where f1 = 'mbb' if there are any, otherwise nothing? In that case you need to be testing whether there are any matching rows. Something like this
SELECT TOP 10 *
FROM tblRFCWorkload
WHERE f1 = 'mbb' --the match you need
OR NOT EXISTS --or there are no matches
(SELECT * FROM tblRFCWorkload WHERE f1 = 'mbb')
I'm trying to get an ordered list of rows out of my MYSQL database table based upon an integer value 'place'.
SELECT * FROM mytable
ORDER BY place;
This works okay, except that all rows with value place=0 should appear at the end of the table.
So if my table is:
name place
---- -----
John 1
Do 2
Eric 0
Pete 2
it should become:
name place
---- -----
John 1
Do 2
Pete 2
Eric 0
order by case when place = 0 then 1 else 0 end asc, place asc
that way you get all the non-zeroes ordered first.
SELECT *
FROM myTable
ORDER BY place>0 DESC, place
is a solution without CASE
SELECT *
FROM myTable
ORDER BY CASE place WHEN 0 THEN 9999 ELSE place END
This approach implies that we known that 9999 (or some other value) is bigger than all possible values in the place column.
Alternatively we can sort by two values as in:
ORDER BY CASE place WHEN 0 THEN 0 ELSE -1 END, place