I want my query to return the rows of a table in groups where a column contains specific values. After I got the rows ordered in the groups I want to be able to order them by name.
Example Table
- Id - Name - Group
- 1 George Group_2_1
- 2 Alfred Group_2_2
- 3 Eric Group_3
- 4 Mary Group_1_2
- 5 Jon Group_1_1
I want them ordered by their group and after that ordered by their name
- Id - Name - Group
- 1 Jon Group_1_1
- 2 Mary Group_1_2
- 3 Alfred Group_2_2
- 4 George Group_2_1
- 5 Eric Group_3
I found this SQL-Query-Snippet
ORDER BY CASE WHEN Group LIKE '%Group_1%' THEN 1 ELSE 2 END, Group
but it is not enough. The result is only grouped by the first group (obviously) but I can't extend it to order the second group because it is in the same column.
Please don't get confused by the example.
I just want to be able to group certain rows and put them in front of the results. I want a result that has all rows containing group 1 in the top, containing group 2 in the middle and containing group 3 in the bottom.
The values are not "Group_1_1" or something like that. They are just some strings and I want certain strings to be always in the first row (group 1) and some always below group 1
The problem here seems to be that some of your group names have an extra underscore, otherwise you could just order by the Group and all would be good. You could probably do something like this to work around this?
WITH Data AS (
SELECT 'Group1_1' AS Value
UNION
SELECT 'Group_3_2' AS Value
UNION
SELECT 'Group_2_2' AS Value
UNION
SELECT 'Group_3_1' AS Value
)
SELECT * FROM Data ORDER BY CASE WHEN Value LIKE 'Group_%' THEN SUBSTRING(Value, 7, 10) ELSE SUBSTRING(Value, 6, 10) END;
Results:
Value
Group1_1
Group_2_2
Group_3_1
Group_3_2
---- EDIT ----
Okay, seeing as your example isn't really an "example" it sounds like you are going to need a really, REALLY long case statement. You could do something like this (using the original Group_1_1, Group_2_2 codes) that would extend to different values. The key is that a CASE statement works from left to right and a value is assigned to the first case that matches:
ORDER BY
CASE
WHEN [Group] = 'Group_1_1' THEN 1
WHEN [Group] = 'Group_1_2' THEN 2
WHEN [Group] LIKE 'Group_1_%' THEN 3
WHEN [Group] = 'Group_2_1' THEN 4
WHEN [Group] = 'Group_2_2' THEN 5
WHEN [Group] LIKE 'Group_2_%' THEN 6
etc.
END;
Obviously that's very generic and depends on what the actual values are in your database.
Edits for mssql
If there is ANY instance of 3 underscores then the following simply won't work. However if there is the possibility of Group_12_6 or Group_21_1 then this approach may be worth trying.
It removes Group_ or Group from the string, leaving 1_1 or 12_6 or 21_1 then it replaces the remaining underscore with . giving 1.1 or 12.6 or 21.1 and casts this to decimal.
All utterly dependent of the consistency of those group names.
SELECT
id
, name
, [Group]
FROM YourData
ORDER BY
CAST(REPLACE(REPLACE(REPLACE([Group], 'Group_', ''), 'Group', ''), '_', '.') AS decimal(12,3))
, name
I'm really hoping you do not have a column called [Group] but if you do it has to be referenced as [Group] or "Group". Test result:
| ID | NAME | GROUP |
|----|--------|-----------|
| 1 | Jon | Group_1_1 |
| 2 | Mary | Group_1_2 |
| 4 | George | Group_2_1 |
| 3 | Alfred | Group_2_2 |
| 5 | Eric | Group_3 |
see http://sqlfiddle.com/#!3/e95b07/1
Related
I have a table like so:
id | value
---+------
1 | 10
2 | 5
3 | 11
4 | 8
5 | 9
6 | 7
The data in this table is really pairs of values, which I need to take the average of, which should result in:
pair_id | pair_avg
--------+---------
1 | 7.5
2 | 9.5
3 | 8
I have got some other information (a pair of flags) which could also help to pair them, though they still have to be in id order. I cannot really change how the data comes to me.
As I'm more used to arrays than SQL, all I can think is that I need to loop through the table and sum the pairs. But this doesn't strike me as very SQL-ish.
Update
In making this minimal example, I have apparently over simplified.
As the table I am working with is the result of several selects, the IDs will not be quite so clean, apologies for not specifying this.
The table looks a lot more like:
id | value
----------
1 | 10
4 | 5
6 | 11
7 | 8
10 | 9
15 | 7
The results will be used to create a second table, I don't care about the index on this new table, it can provide its own, therefore giving the result already indicated above.
If your data is as clean as the question makes it seem: no NULL values, no gaps, pairs have consecutive positive numbers, starting with 1, and assuming id is type integer, it can be as simple as:
SELECT (id+1)/2 AS pair_id, avg(value) AS pair_avg
FROM tbl
GROUP BY 1
ORDER BY 1;
Integer division truncates the result and thus takes care of grouping pairs automatically this way.
If your id numbers are not as regular but at least strictly monotonically increasing like your update suggests (still no NULL or missing values), you can use a surrogate ID generated with row_number() instead:
SELECT id/2 AS pair_id, avg(value) AS pair_avg
FROM (SELECT row_number() OVER (ORDER BY id) + 1 AS id, value FROM tbl) t
GROUP BY 1
ORDER BY 1;
db<>fiddle here
I think you can just use group by with arithmetic:
select row_number() over (order by min(id)), min(id), max(id), avg(id)
from t
group by floor( (id - 1) / 2 );
I'm not sure why you would want to renumber the ids after aggregation. The original ids seem more useful.
You may use ceil function by appliying division by 2 to id column as in the following select statement :
with t(id,value) as
(
select 1 , 10 union all
select 2 , 5 union all
select 3 , 11 union all
select 4 , 8 union all
select 5 , 9 union all
select 6 , 7
)
select ceil(id/2::numeric) as "ID", avg(t.value) as "pair_avg"
from t
group by "ID"
order by "ID";
id | pair_avg
-------------
1 | 7.5
2 | 9.5
3 | 8
I am trying to get 3% of total membership which the code below does, but the results are bringing me back two rows one has the % and the other is "0" not sure why or how to get rid of it ...
select
sum(Diabetes_FLAG) * 100 / (select round(count(medicaid_no) * 0.03) as percent
from membership) AS PERCENT_OF_Dia
from
prefinal
group by
Diabetes_Flag
Not sure why it brought back a second row I only need the % not the second row .
Not sure what I am doing wrong
Output:
PERCENT_OF_DIA
1 11.1111111111111
2 0
SELECT sum(Diabetes_FLAG)*100 / (SELECT round(count(medicaid_no)*0.03) as percentt
FROM membership) AS PERCENT_OF_Dia
FROM prefinal
WHERE Diabetes_FLAG = 1
# GROUP BY Diabetes_Flag # as you're limiting by the flag in the where clause, this isn't needed.
Remove the group by if you want one row:
select sum(Diabetes_FLAG)*100/( SELECT round(count(medicaid_no)*0.03) as percentt
from membership) AS PERCENT_OF_Dia
from prefinal;
When you include group by Diabetes_FLAG, it creates a separate row for each value of Diabetes_FLAG. Based on your results, I'm guessing that it takes on the values 0 and 1.
Not sure why it brought back a second row
This is how GROUP BY query works. The group by clause group data by a given column, that is - it collects all values of this column, makes a distinct set of these values and displays one row for each individual value.
Please consider this simple demo: http://sqlfiddle.com/#!9/3a38df/1
SELECT * FROM prefinal;
| Diabetes_Flag |
|---------------|
| 1 |
| 1 |
| 5 |
Usually GROUP BY column is listed in in SELECT clause too, in this way:
SELECT Diabetes_Flag, sum(Diabetes_Flag)
FROM prefinal
GROUP BY Diabetes_Flag;
| Diabetes_Flag | sum(Diabetes_Flag) |
|---------------|--------------------|
| 1 | 2 |
| 5 | 5 |
As you see, GROUP BY display two rows - one row for each unique value of Diabetes_Flag column.
If you remove Diabetes_Flag colum from SELECT clause, you will get the same result as above, but without this column:
SELECT sum(Diabetes_Flag)
FROM prefinal
GROUP BY Diabetes_Flag;
| sum(Diabetes_Flag) |
|--------------------|
| 2 |
| 5 |
So the reason that you get 2 rows is that Diabetes_Flag has 2 distict values in the table.
How can I group by a comma delineated list within a row?
Situation:
I have a view that shows me information on support tickets. Each ticket is assigned to an indefinite number of resources. It might have one name in the resource list, it might have 5.
I would like to aggregate by individual names, so:
| Ticket ID | resource list
+-----------+----------
| 1 | Smith, Fred, Joe
| 2 | Fred
| 3 | Smith, Joe
| 4 | Joe, Fred
Would become:
| Name | # of Tickets
+-----------+----------
| Fred | 3
| Smith | 2
| Joe | 3
I did not design the database, so I am stuck with this awkward resource list column.
I've tried something like this:
SELECT DISTINCT resource_list
, Count(*) AS '# of Tickets'
FROM IEG.vServiceIEG
GROUP BY resource_list
ORDER BY '# of Tickets' DESC
...which gives me ticket counts based on particular combinations, but I'm having trouble getting this one step further to separate that out.
I also have access to a list of these individual names that I could do a join from, but I'm not sure how I would make that work. Previously in reports, I've used WHERE resource_list LIKE '%' + #tech + '%', but I'm not sure how I would iterate through this for all names.
EDIT:
This is my final query that gave me the information I was looking for:
select b.Item, Count(*) AS 'Ticket Count'
from IEG.vServiceIEG a
cross apply (Select * from dbo.Split(REPLACE(a.resource_list, ' ', ''),',')) b
Group by b.Item
order by 2 desc
Check this Post (Function Definition by Romil) for splitting strings into a table:
How to split string and insert values into table in SQL Server
Use it this way :
select b.Item, Count(*) from IEG.vServiceIEG a
cross apply (
Select * from dbo.Split (a.resource_list,',')
) b
Group by b.Item
order by 2 desc
I realize that this has probably been asked a billion times, and I could swear I've done this in the past, but tonight I've got brain block or something and can't figure it out...
I have a database table ("t1") where I need to be able to retrieve only the first row where a particular value appears in a particular column.
Here's a sample of the data:
id | qID | Name
---------------------
1 | 1 | Bob
2 | 3 | Fred
3 | 1 | George
4 | 1 | Jack
What I want as a result is:
id | qID | Name
---------------------
1 | 1 | Bob
2 | 3 | Fred
The only column I actually need to get out of the query is the first one, but that's not where the duplicates need to be eliminated, and I thought it might be confusing not to show the entire row.
I've tried using this:
select id, qID, ROW_NUMBER() over(partition by qID order by qID) as zxy
from t1 where zxy = 1
But it gives me this error:
Msg 207, Level 16, State 1, Line 14
Invalid column name 'zxy'.
If I remove the where part of the query, the rest of it works fine. I've tried different variable names, using single or double quotes around 'zxy' but it seems to make no difference. And try as I might, I can't find the part of the SQL Server documentation where it discusses assigning a variable name to an expression, as in the "as zxy" part of the above query... if anybody has a link for that, that's quite useful.
Needless to say, I've tried other variable names besides "zxy" but that makes no difference.
Help!
WHERE clause is applied earlier in the process than SELECT. Therefore the calculated column zxy is not available in WHERE. In order to achieve your goal you need to put your original query in a subquery or CTE.
select id, qid
from
(
select id, qID, ROW_NUMBER() over(partition by qID order by qID) as zxy
from t1
) q
where zxy = 1
Output:
| id | qid |
|----|-----|
| 1 | 1 |
| 2 | 3 |
Here is SQLFiddle demo
Logical Processing Order of the SELECT statement
1 FROM
2 ON
3 JOIN
4 WHERE
5 GROUP BY
6 WITH CUBE or WITH ROLLUP
7 HAVING
8 SELECT
9 DISTINCT
10 ORDER BY
11 TOP
Where Clause Execute Before Select Clause so You can not find ZXY in Where cluase
with cte as
(
select id, qID, ROW_NUMBER() over(partition by qID order by qID) as zxy
from t1
)
select * from cte where zxy = 1
Here is my blog it might help you http://sqlearth.blogspot.in/2015/05/how-sql-select-statement-logically-works.html
I have the following table in my database.
# select * FROM matches;
name | prop | rank
------+------+-------
carl | 1 | 4
carl | 2 | 3
carl | 3 | 9
alex | 1 | 8
alex | 2 | 5
alex | 3 | 6
alex | 3 | 8
alex | 2 | 11
anna | 3 | 8
anna | 3 | 13
anna | 2 | 14
(11 rows)
Each person is ranked at work by different properties/criterias called 'prop' and the performance is called 'rank'. The table contains multiple values of (name, prop) as the example shows. I want to get the best candidate following from some requirements. E.g. I need a candidate that have (prop=1 AND rank > 5) and (prop=3 AND rank >= 8). Then we must be able to sort the candidates by their rankings to get the best candidate.
EDIT: Each person must fulfill ALL requirements
How can I do this in SQL?
select x.name, max(x.rank)
from matches x
join (
select name from matches where prop = 1 AND rank > 5
intersect
select name from matches where prop = 3 AND rank >= 8
) y
on x.name = y.name
group by x.name
order by max(rank);
Filtering the data to match your criteria here is quite simple (as shown by both Amir and sternze):
SELECT *
FROM matches
WHERE prop=1 AND rank>5) OR (prop=3 AND rank>=8
The problem is how to aggregate this data so as to have just one row per candidate.
I suggest you do something like this:
SELECT m.name,
MAX(DeltaRank1) AS MaxDeltaRank1,
MAX(DeltaRank3) AS MaxDeltaRank3
FROM (
SELECT name,
(CASE WHEN prop=1 THEN rank-6 ELSE 0 END) AS DeltaRank1,
(CASE WHEN prop=3 THEN rank-8 ELSE 0 END) AS DeltaRank3,
FROM matches
) m
GROUP BY m.name
HAVING MaxDeltaRank1>0 AND MaxDeltaRank3>0
SORT BY MaxDeltaRank1+MaxDeltaRank3 DESC;
This will order the candidates by the sum of how much they exceeded the target rank in prop1 and prop3. You could use different logic to indicate which is best though.
In the case above, this should be the result:
name | MaxDeltaRank1 | MaxDeltaRank3
------+---------------+--------------
alex | 3 | 0
... because neither anna nor carl reach both the required ranks.
A typical case of relational division. We assembled a whole arsenal of techniques under this related question:
How to filter SQL results in a has-many-through relation
Assuming you want the minimum rank of a person, I might solve your particular case with LEAST():
SELECT m1.name, LEAST(m1.rank, m2.rank, ...) AS best_rank
FROM matches m1
JOIN matches m2 USING (name)
...
WHERE m1.prop = 1 AND m1.rank > 5
AND m2.prop = 3 AND m2.rank >= 8
...
ORDER BY best_rank;
Also assuming name to be unique per individual person. You'd probably use some kind of foreign key to a pk column of a person table in reality.
And if you have such a person table like you should, the best rank would be stored in a column there ...
If I understand you question, then you just need to execute the following operation:
SELECT * FROM matches where (prop = 1 AND rank > 5) OR (prop = 3 AND rank >= 8) ORDER BY rank
It gives you the canidates that either have prop=1 and rank > 5 or prop=3 and rank >= 8 sorted by their rankings.