on 10-22-2014 5:57 AM
Dear Friends,
I have a requiremnt to find out the number of students with noraml and abnormal eye sight based on 3 characteristics ( Left eye, right eye, and Use of Lens).
All thes characteristics have following values ( Normal, A-low, A-high, B-low, B-High, C-low etc)
Number of Students with Normal eye sight - If my 3 characteristics have value "Normal" then the student is normal.
To achieve this, In columns I have created a selection restricting these 3 charcteistics with value "Normal" and able to find out the result.
Number of Students with Abnormal eye sight - If any one of 3 characteristics has value other than normal than the student is abnormal.
How to find out this..?
I tried restricting my 3 charcteristics with other than "normal" values but it doesnt work. it works only when all the 3 charcteriscs have values not equal to "normal". But the condition i want is if any one carcteristics has value other than "normal" it shud count as a abnormal student,
please guide me.
thanks
Vijay
Hi Vijay,
What Loed Despuig replied is right. You can put it in different way.
In the report there should be 2 KF in Display Mode.
KF_TESTED = No of students tested = 1
Add 3 more KF and Hide those.
RKF_LEFT_NORMAL = KF_TESTED - restrict on Char LEFT_EYE = NORMAL
RKF_RIGHT_NORMAL = KF_TESTED - restrict on Char RIGHT_EYE = NORMAL
RKF_LENSE_NORMAL = KF_TESTED - restrict on Char USE_OF_LENS = NORMAL
Add Formula columns as below
KF_NORMAL = (RKF_LEFT_NORMAL+RKF_RIGHT_NORMAL+RKF_LENSE_NORMAL == 3) * 1
KF_ABNORMAL = (RKF_LEFT_NORMAL+RKF_RIGHT_NORMAL+RKF_LENSE_NORMAL <> 3) * 1
Note: Exc Aggre should set to Student_ID
Dept No of students tested No of studnts normal No of students Abnormal
Mech 4 1 3
Regards,
Sucheta
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Hi Vijay,
Instead of inccluding characterstics other than normal for abnormal key-figure why don't you use the exclude option and exclude 'normal' only.
This should help you in achieving the desired result.
Hope this helps.
Regards,
Amit
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Hi Amith,
I tried already but it doesnt working. The condition satisfy only when all the 3 characteristics have values other than "Normal"
ex student1 will not be considered as his left eye is normal
Student2 will be counted because all 3 characteristics hav values other than normal.
hope you understand.
We dont want to disturb the dataflow, shud be acheived at front end.
thanks
Hi Vijay,
Ok now i get it.
Create 3 different key-figures for each of the characterstic and restrict them with value other than 'normal'.
Now create a calculated key-figure and use these 3 key-figures and write a condition to check if any of them returns a value using 'if' condition.
If yes display abnormal.
Regards,
Amit
Hi,
if conditions not satisfy check with the applying the exception aggregation.
create the all diffrent KFvalues with resctricting the Normal, A-low, A-high, B-low, B-High, C-low etc) and abnormal.
then apply the exception aggregation -> create the new formula -> insert the resctricted KF -> select exception aggregation -> summation/total -> put the reference char -> student id
check the out put.
Thanks,
Phani.
Hi Vijay,
Can you post here your desired report layout?
Right now, the solution I can think is to have a DUMMY KF in the transformation level and make a FIELD ROUTINE for the three (3) characteristics (Left eye, right eye, and Use of Lens)..
Routine Code:
if LEFT_EYE = "NORMAL" and RIGHT_EYE = "NORMAL" and USE_OF_LENS = "NORMAL".
result = 1.
else.
result = 0.
endif.
Then in BEx level, just use a condition or filter in the report the normal and abnormal students using the DUMMY KF (which is 1 will be NORMAL and 0 for ABNORMAL)..
Thanks!
Loed
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Hi Loed
Rows : Test Year, Department
Columns : Counter( dummy key figure - hidden),
CKF1 ( No of students Tested) acheived
Selection - ( No of Students with Normal Vision) - Achieved by restricting characteristics Zleft-eye, ZRighteye, Zlens with value "Normal"
Now how to calculate no of students with abnormal vision.
ex data in cube:
StudentID Dept Left eye Right eye Use of Lens
1 Mech Normal A-low No
2 Mech B-low A-low Yes
3 Mech A-high Normal Yes
4 Mech Normal Normal No
For my above example, report layout is below
Dept No of students tested No of studnts normal No of students Abnormal
Mech 4 1 3
Thanks
Vijay
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