hello experts,
i have a small issue...i have 3 numbers...suppose 10 14 and 18 and i got one more number 15 which is to be compared with the first 3 values...this 15 dont match with any of the 3 values(10,14 and 18)...in this case i got to print the nearest value i.e., 14....
plz tell me the logic how i can do that
PARAMETERS znum TYPE i.
data: znum1 TYPE i, znum2 TYPE i, znum3 TYPE i.
data:diff1 type i, diff2 type i, diff3 type i.
znum1 = 10.
znum2 = 14.
znum3 = 18.
diff1 = abs( znum1 - znum ).
diff2 = abs( znum2 - znum ).
diff3 = abs( znum3 - znum ).
if diff1 lt diff2 and diff1 lt diff3.
WRITE / znum1.
ELSEIF diff2 lt diff1 and diff2 lt diff3.
WRITE / znum2.
elseif diff3 lt diff1 and diff3 lt diff2.
WRITE / znum3.
endif.
plz reward points if dis helps
Message was edited by:
abapuser
Hi vinil,
data: x1 type i value 10,
x2 type i value 20,
x3 type i value 30,
x4 type i value 23.
if x4 BETWEEN X1 AND X2.
WRITE X1.
ELSEIF X4 BETWEEN X2 AND X3.
WRITE X2.
ELSE.
WRITE X4.
ENDIF.
Reward points and close the thread...if it is answered.
Hello Vinil,
If you have n numbers... Just put those into an internal table with one filed.
data: begin of itab occurs 0,
num type i,
end of itab.
itab-num = 10. append itab. clear itab.
itab-num = 14. append itab. clear itab.
itab-num = 18. append itab. clear itab.
itab-num = 15. append itab. clear itab.
now sort the itab. store the last entered value.
now loop at the internal table... if last entered value occurs just print sy-tabix - 1 th record. this is the nearest value of last enterd value.
Reward If Useful.
Regards
--
Sasidhar Reddy Matli.
Hi,
try this logic this will be dynamic
PARAMETERS znum TYPE i.
DATA: BEGIN OF itab OCCURS 0,
num TYPE i,
END OF itab.
DATA: diff TYPE i.
itab-num = '10'.
APPEND itab.
itab-num = '20'.
APPEND itab.
itab-num = '12'.
APPEND itab.
itab-num = '13'.
APPEND itab.
SORT itab BY num.
data: diffc type i.
data: begin of itab1 occurs 0,
number type i,
differ type i,
end of itab1.
loop at itab.
if itab-num > znum.
diff = itab-num - znum.
else.
diff = znum - itab-num.
endif.
itab1-number = itab-num.
itab1-differ = diff.
clear: diff.
append itab1.
endloop.
sort itab1.
read table itab1 index 1.
if sy-subrc eq 0.
write:/ itab1-number.
endif.
regards,
Venkatesh
hi,
try like this,
data: no1 type i,
no2 type i,
no3 type i,
no type i,
diff1 type i,
diff2 type i,
diff 3 type .
diff1 = abs(no1 - no).
diff2 = abs(no2 -no).
diff3 = abs(no3 - no).
if diff1 = diff2 or diff1 =diff3.
write:/10 no,' IS IN THE RANGE OF',no1,no2,no3.
else.
if diff2 = diff3.
write:/10 no,' IS IN THE RANGE OF',no1,no2,no3.
else.
write:/10 no-1.
endif.
REWARD ALL USEFUL ANSWERS.
WITH REGARDS,
SURESH ALURI.
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