September 9, 2008 at 6:12 pm
I am trying to count rows which have a column that matches a specific pattern. I have a "Status" column, and I want to count all of the "ERR%", "FAIL%", "PASS%", and "Other" values.
I have been looking at the string processing functions, and don't see anything that might help with this.
Any ideas?
Thanks,
Justin
My input data would look like:
DECLARE @raw TABLE (TestListID INT, Feature VARCHAR(256), RunStatus VARCHAR(64))
INSERT @raw
SELECT 1, 'Sample', 'FAIL_HW' UNION ALL
SELECT 2, 'trial', 'PASS' UNION ALL
SELECT 3, 'Stride', 'Error_SW' UNION ALL
SELECT 4, 'Regression', 'PASS_SW' UNION ALL
SELECT 5, 'Regression', 'PASS_HW' UNION ALL
SELECT 6, 'Setup', 'Waived' UNION ALL
SELECT 7, 'Setup', 'Defered' UNION ALL
SELECT 8, 'Setup', 'Error_HW' UNION ALL
SELECT 9, 'Regression', 'Error' UNION ALL
SELECT 10, 'Setup', 'Error' UNION ALL
SELECT 11, 'Unit', 'Ready' UNION ALL
SELECT 12, 'Unit', 'Ready' UNION ALL
SELECT 13, 'Unit', 'Ready' UNION ALL
SELECT 14, 'Unit', 'Ready'
And the output should look like:
Feature Status Count
------- ------ -----
Regression Err* 1
Regression Pass* 2
Sample Fail* 1
Setup Err* 2
Setup Other 2
Stride Err* 1
trial Pass* 1
Unit Other 4
September 9, 2008 at 9:09 pm
You would want to do a GROUP BY and COUNT to aggregate up the rows.
select fieldA, fieldB, count(fieldb)
from table
group by fieldA, fieldB
September 10, 2008 at 9:52 am
That will count all of the like items in fieldB, but I want to count all of the items in fieldB which match a specific pattern. For example, "Pass_HW" and "Pass_SW" should be considered the same and counted the same.
September 10, 2008 at 9:32 pm
SELECT Feature,
Status=CASE WHEN patindex('Err%',RunStatus)>0 THEN 'Err*'
WHEN patindex('Fail%',RunStatus)>0 THEN 'Fail*'
WHEN patindex('Pass%',RunStatus)>0 THEN 'Pass*'
ELSE 'Other'
END,
[Count]=count(1)
FROM @raw
GROUP BY Feature,CASE WHEN patindex('Err%',RunStatus)>0 THEN 'Err*'
WHEN patindex('Fail%',RunStatus)>0 THEN 'Fail*'
WHEN patindex('Pass%',RunStatus)>0 THEN 'Pass*'
ELSE 'Other'
END
ORDER BY Feature
September 12, 2008 at 10:49 am
That was perfect, thanks for the help!!
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