Is it possible to perform a conditional if statement to collapse 3 fields into a single field? I am trying to achieve following. field1, field2 and field3 are are Int64 which is nullable. I could not find a way to do a null check so am checking if a positive value is assigned to any of the fields then set the a field to respective values. When syntax below I am getting error below:
case when field1 >= 0 then 0 end as field,
case when field2 >= 0 then 1 end as field,
case when field3 >= 0 then 2 end as field
Duplicate column names in the result are not supported. Found
duplicate(s): field
Is this what you want?
(case when field1 is not null then 0
when field2 is not null then 1
when field3 is not null then 2
end) as null_check
Or, if you want to turn this into a coded field:
concat(case when field1 is not null then 0 end,
case when field2 is not null then 1 end,
case when field3 is not null then 2 end,
) as null_check
This will list "0", "1", "2" in a string, depending on the values that are not NULL.
It sounds like perhaps you just want a CASE expression which can generate multiple values based on multiple conditions:
CASE WHEN field3 >= 0 THEN 2
WHEN field2 >= 0 THEN 1
WHEN field1 >= 0 THEN 0 END AS field
The logic here is that it will check field3 first, then field2, followed by field1, when determining which output to generate.
Related
Say that you need to query some data and that there are three fields like the following (this is part of a larger query):
Field1, Field2, Field3.
So you select them like this:
SELECT Field1=MyTable.Field1, Field2=MyTable.Field2, Field3=MyTable.Field3
FROM MyTable
I need to compare these values and return the variable Result that is computed as follows:
0 if they are all the same
1/2 if two are the same and one is different
1 if they are all different.
How should I restructure my query? I think I need a subquery but I am not sure how to structure it.
You can use case:
select (case when field1 = field2 and field1 = field3 then 0
when field1 in (field2, field3) or field2 = field3 then 0.5
else 1
end) as result
I've a table in my database for which I need to check if all rows have one field not null.
If there are no row or if there is at least 1 row with the field null => true
If there are rows and they are all with the field not null => False
Is there a way to do this in on simple query? Or I need to check if my table is empty first then if it's not check if I've a row with the field value empty ?
This will count how many NULL values you have in a field;
SELECT
SUM(CASE WHEN FieldName IS NULL THEN 1 ELSE 0 END) NullValues
FROM TableName
Will return 0 if there are no NULL values, and will return the number of NULLS if there are any present.
If you actually want to return a value as 'True' or 'False' then do this;
SELECT CASE
WHEN a.NullValues > 0
THEN 'True'
ELSE 'False'
END CheckField
FROM (
SELECT
SUM(CASE WHEN FieldName IS NULL
THEN 1
ELSE 0
END) NullValues
FROM TableName
) a
Use count(*) and count(field) and compare the two:
select
case when count(*) > 0 and count(*) = count(field) then 1 -- not empty and no nulls
else 0 end as isgood
from mytable;
Oracle SQL has no boolean data type , so I use 1 for true and 0 for false. You can replace this with whatever you like (e.g. 'true' instead of 1 and 'false' instead of 0).
As to turning this into a predicate (correlated to a main query), you'd use something along the lines of:
select ...
from main
where exists
(
select 1
from mytable
where mytable.colx = main.coly
having count(*) > 0 and count(*) = count(field)
);
You can do this with aggregation. However, it is difficult to understand what you are asking for. If you want to check that a field has no NULL values, you can do:
select (case when count(*) > 0 then 1 else 0 end) as HasNullValues
from t
where field is null;
Alternate way I found using max with putting null first:
select case when
max(field) keep (dense_rank first order by datfin desc nulls first) is null then 1
else 0 end as flag
from MYTABLE;
I have a SQL query which displays a list of results. Every row in my database has about
20 columns and not every column is mandatory. I would like the result of the SQL query to be
sorted by the number of filled in columns. The rows with the least empty columns at the top, the ones with the most empty columns at the bottom. Do any of you guys have an idea how to do this?
I thought about adding an extra column to the table which if updated every time the user edits their row, this number would indicate the number of empty columns and I could sort my list with that. This however, sounds like unnecessary troubles, but maybe there is no other way? I'm sure somebody on here will know!
Thanks,
Sander
You can do it in just about any database with a giant case statement:
order by ((case when col1 is not null then 1 else 0 end) +
(case when col2 is not null then 1 else 0 end) +
. . .
(case when col20 is not null then 1 else 0 end)
) desc
You could order by the amount of empty columns:
order by
case when col1 is null then 1 else 0 end +
case when col2 is null then 1 else 0 end +
case when col3 is null then 1 else 0 end +
...
case when col20 is null then 1 else 0 end
(Note the + at the end of the lines: it's only one column with the integer count of empty fields, sorted in ascending order.)
I have a table with 3 columns (containing Integer values that assume only values from 0 to 10). I want extract, with a single query, a table with 1 column. This column must assume a value based on the following logic:
If one of these three columns has value 0 ----> the value of column of table generated by query must be 0 too.
If none of the last three columns has value 0 ----> the value of column must assume the value 1.
You are looking for CASE construct or IF function:
SELECT CASE WHEN (t.field1 = 0 OR t.field2 = 0 OR t.field3 = 0) THEN 0
ELSE 1 END AS value
FROM t;
In this specific case you might also use the fact that any member being zero will zero the product:
SELECT CASE WHEN (t.field1*t.field2*t.field3 = 0) THEN 0 ELSE 1 END AS value
FROM t;
Or
SELECT IF((t.field1*t.field2*t.field3)=0, 0, 1) AS value FROM t;
This is a simple case statement. Assuming there are no NULL values, try this:
select (case when col1 = 0 or col2 = 0 or col3 = 0 then 0 else 1 end)
Try this
SELECT
CASE
WHEN column1 = 0 THEN 0
WHEN column2 = 0 THEN 0
WHEN column3 = 0 THEN 0
ELSE 1
END
FROM urtable
I have a table with 3 fields, let us say field1, field2, and field3. The value in field2 is either 0 or 1.
I am trying to fetch field1 in a way to always have all the rows where the value is 1 in field2 displayed first and arranged by field3, and then display the rest of data, also arranged by field3.
My research told me one can order by two fields, let us say 'Order by field2 Desc, field 3' but this really is not giving the expected result. Any idea?
One small change is required for MySQL:
ORDER BY field2 = 1 DESC, field3
Or for standard SQL:
ORDER BY CASE WHEN field2 = 1 THEN 1 ELSE 0 END DESC, field3
try this:
order by
case when field2=1 then 0 else 1 end,
field3