append logic

[Former Member]

int_a with values:

order item good qty rej qty

200 1 10 0

200 2 0 2

201 1 20 0

201 2 0 6

sort int_a by order item

loop at int_a

v_good = v_good + int_a-good

v_rej = v_rej + int_a-rej

at new order.

v_index = 1.

endat.

if v_index = 1.

move-corresponding int_a to int_b.

int_b-good = v_good.

int_b_rej = v_rej.

append int_b.

clear: v_index,int_b,v_good,v_rej.

endif.

endloop.

int_b-rej values coming topsy turvy.. i tried at end of ord,, still no use..

can any one tell me how to append the 2 line item values(good and rej) of int_a into int_b in single line?

[Former Member]

sort int_a by order item

clear: v_index,int_b,v_good,v_rej.

loop at int_a

v_good = v_good + int_a-good

v_rej = v_rej + int_a-rej

<b>at end of order.</b> " not at new

v_index = 1.

endat.

if v_index = 1.

move-corresponding int_a to int_b.

int_b-good = v_good.

int_b_rej = v_rej.

append int_b.

clear: v_index,int_b,v_good,v_rej.

endif.

endloop.

regards

shiba dutta

[Former Member]

Hi Shiba,

thanks a lot for quick reply,,

i was using at end of before. but when i use that the item value is getting as 2..i want item as 1 in my Int_b.

does the <b>clear: v_index,int_b,v_good,v_rej before looping will work?</b>

Sorry for asking silly questions, but my mind got jammed..:)

[Former Member]

just change this part

if v_index = 1.

move-corresponding int_a to int_b.

<b>int_b-item = v_index.</b> " or int_b-item = 1.

int_b-good = v_good.

int_b_rej = v_rej.

append int_b.

clear: v_index,int_b,v_good,v_rej.

endif.

regards

shiba dutta

[Former Member]

thanks a ton Shiba ,,,it worked

[Former Member]

Hi Nihi

you can modify code like this , using collect statement to sum the data .

sort int_a by order.

loop at int_a.

move-corresponding int_a to int_b. " no item in int_b

collect int_b " you can collect

endloop.

Regards

Wiboon

[Former Member]

int_a with values:

order item good qty rej qty

200 1 10 0

200 2 0 2

201 1 20 0

201 2 0 6

sort int_a by order item

loop at int_a

v_good = v_good + int_a-good

v_rej = v_rej + int_a-rej

at end order.

move-corresponding int_a to int_b.

int_b-good = v_good.

int_b_rej = v_rej.

append int_b.

clear: v_index,int_b,v_good,v_rej.

endat.

endloop.

[Former Member]

Try this, this will surely resolve your requirement.

sort int_a by order item.

loop at int_a.

v_good = v_good + int_a-good.

v_rej = v_rej + int_a-rej.

at END OF order.

CLEAR V_INDEX.

v_index = 1.

endat.

if v_index = 1.

move-corresponding int_a to int_b..

move int_a-ORDER to int_b-ORDER.

move int_a-ITEM to int_b-ITEM.

int_b-good = v_good.

int_b-rej = v_rej.

append int_b.

clear: v_index,int_b,v_good,v_rej.

endif.

endloop.

Regards

Amit Singla