2

I have a workbook of 12 sheet and every sheet has some value in column A. All the useful information starts from row 9 column 0 of every sheet. Once the cell (AX,0) has no data it should move to the next sheet. Here num is the number of sheet in the workbook. The code is below:

workbook = xlrd.open_workbook('example.xlsx')
row = 9
num = 0
col = 0

while True:
    worksheet = workbook.sheet_by_index(num)
    if worksheet.cell(row, col).value != xlrd.empty_cell.value:
        worksheet.cell(row, col).value
        row+=1
    else:
        if num != 12:
            num = num + 1
            row = 9

I am getting the below error:

File "", line 4, in if worksheet.cell(row, col).value != xlrd.empty_cell.value: File "C:\Python27\lib\site-packages\xlrd\sheet.py", line 412, in cell self._cell_types[rowx][colx], IndexError: list index out of range

Help me!!!

FBruzzesi
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Rakesh Sharma
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  • Welcome! You can quickly and easily format code in posts using backticks. You can see a few examples [here](https://stackoverflow.com/editing-help#syntax-highlighting). – edd May 02 '20 at 08:25

2 Answers2

0

I think you should add a break in the While loop, so you dont loop infinitely.

if num != 12:
    num = num + 1
    row = 9
else:
    break

Edit

Another problem is with your if statement of checking empty cell. Based in Openpyxl check for empty cell The problem is, that the "if" is true all the time, so row is unlimitely incremented and you get index out of range error.

If you want to check for not empty cell use this 

if worksheet.cell(row, col).value != None:
    # Do stuff
Blue Print
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0

I think the problem is that if the program enter the "else" block will be stock there because there in no way out so I suggest to either use ( break) or (return)

Ahlam
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