Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
1.2k views
in Technique[技术] by (71.8m points)

excel - Worksheet.UsedRange.Rows.Count returns wrong result

In VB.NET I want to get the used rows so I wrote that:

Dim objWorksheet As Excel.Worksheet = workbook.Sheets(2)
Dim lastRow = objWorksheet.UsedRange.Rows.Count

The number of lastRow is less than used rows. I searched the site and someone suggested:

Dim range As Excel.Range = objWorkSheet.UsedRange
Dim lastRow = range.Rows.Count

This returns less than actual used rows.

The solution is in the image:
img1

I found this question it is an overall understanding of the last used row.

UsedRange.Rows.Count means from the first rows that has data to the last rows that has data, which means you can have empty rows between the first non-empty rows to the last non-empty rows which not affect the result, but if the first row is empty, the result will be one less the actual non-empty row, and if the first two row is empty the result will be two less, so the link question said never use UsedRange.Rows.Count to get the last row.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

Try to avoid UsedRange. It can be misleading. For instance, if you fill range A1:B5 and clear the contents of column B, then UsedRange.Columns.Count will return 2 - because Excel remembers cells with formatting and includes them into UsedRange.

UPDATE

To get real last column and row, use following code:

lRealLastRow = _
    Cells.Find("*", Range("A1"), xlFormulas, , xlByRows, xlPrevious).Row
lRealLastColumn = _
    Cells.Find("*", Range("A1"), xlFormulas, , xlByColumns, xlPrevious).Column

UPDATE 2 img2


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...