Excel Sumif, Sumifs with partial strings in multip

问题:

So this is the simplified question I broke down from a former question I had here: Excel help on combination of Index - match and sumifs? .

For this one, I have Table1 (the black-gray one) with two or more columns for adjustments for various order numbers. See this image below:

What I want to achieve is to have total adjustments for those order numbers that contain the numbers in Total Adjustment column in the blue table, each of which will depend on the cell beside it.

Example: Order number 17051 has two products: 17051A (Apple) and 17051B (Orange).

Now what I want to achieve in cell C10 is the sum of adjustment for both 17051A and 17051B, which will be: Apple Adjustment (5000) + Orange Adjustment (4500) = 9500.

The formula I used below (and in the image) kept giving me error messages, and this happens even before I add the adjustment for Orange.

=SUMIF(Text(LEFT(Table1[Order Number],5),"00000"),text(B10,"00000"),Table1[Apple Adjustment])

I have spent the whole day looking for a solution for this and didn’t even come close to find any. Any suggestion is appreciated.

回答1:

Assuming your headers always have the text "adjustment" in them, you could use:

=SUMPRODUCT((LEFT($B$4:$B$7,5)=B10&"")*(RIGHT($C$3:$F$3,10)="adjustment")*$C$4:$F$7)

回答2:

In C10 you could add two sumproducts. This assumes that products are always 5 numbers long at the start. If not swop the 5 to use the length of the product reference part you are matching on.

=SUMPRODUCT(--(1*LEFT($B$4:$B$7,5)=$B10),$D$4:$D$7)+SUMPRODUCT(--(1*LEFT($B$4:$B$7,5)=$B10),$F$4:$F$7)

Which with table syntax is:

=SUMPRODUCT(--(1*LEFT(Table1[Order Number],5)=$B10),Table1[Apple Adjustment])+SUMPRODUCT(--(1*LEFT(Table1[Order Number],5)=$B10),Table1[Orange Adjustment])

Using LEN

=SUMPRODUCT(--(1*LEFT(Table1[Order Number],LEN($B10))=$B10),Table1[Apple Adjustment])+SUMPRODUCT(--(1*LEFT(Table1[Order Number],LEN($B10))=$B10),Table1[Orange Adjustment])

I am multiplying by 1 to ensure Left, 5 becomes numeric.