如何打印下一行每次函数运行?(openpyxl)

Question

我正在做的整个项目是，我正在使用openxyl从excel文档中的特定行和列中获取值，并在网页上填写表单。我填写表单，然后保存它，然后用excel文档中的下一行值填写下一个表单。

Excel是这样的:

First Name         Last Name         DOB
George             Hill              9/9/99
Sam                Genius            8/8/88
Bill               Smith             7/7/77

我想运行下面的函数三次，每次它运行时打印出下一行:

def AddFirstName():
for i in range(3):
for row in ws.iter_rows(min_row=1, min_col=1, max_row=3,max_col=1):
for cell in row:
print(cell.value)

break
break  # Tried to use breaks to break the loop so it only prints once for each loop
AddFirstName()
AddFirstName()
AddFirstName()

或者像这样在函数外运行:

for i in range(3):
for row in ws.iter_rows(min_row=1, min_col=1, max_row=3,max_col=1):
for cell in row:
print(cell.value)

break
break  # Tried to use breaks to break the loop so it only prints once for each loop

这个打印:

George
George
George

如何解决这个问题，使它打印:

George             
Sam                
Bill 

Answer 1

我在这里找到了答案

基本上，它将根据运行函数的次数打印出我想要挂起的行数。

listofnames = [
['George'],
['Bill'],
['Mike']]
def print_names(name):
cnt = 0
while cnt < 1:
try:
print(next(name))
except StopIteration:
print("You have reached the end of the list!")
break
cnt += 1
first_name = iter(listofnames)
for i in range (3):
print_names(first_name)

Answer 2

我根据你的评论和你想要完成的任务编辑了我的答案。

这个解决方案利用了闭包。闭包是嵌套函数，其中外部函数def包含内部函数def，外部函数返回内部函数。在内部函数中声明为非局部的变量在调用者命名空间中调用内部函数时保持其值。在本例中，内部函数有2个非局部函数。一个用于跟踪处理的行总数，另一个是迭代器的副本。

对于你的例子来说，闭包可能需要做很多工作，但我认为从长远来看，它使你的代码更易于移植和可用。处理N行的所有智能都在内部函数中。在那里更改它，闭包行为的所有"实例"也会更新。

另外，我使用的不是'try'，而是unittest.mock.sentinel，因为它比try要少得多:…而且，它看起来更干净。

import unittest.mock
from unittest.mock import sentinel
sheetData = [
['George'],
['Bill'],
['Mike']
]
def set_iter(sheet_data):
total_rows = 0
sheet_data_iter = iter(sheet_data)
# Create a closure that will track the total number of times we have called
# the iterator. A closure may seem like a lot of extra work in this example,
# but it shows how usefule they can be when you variables in a function
# to persist after they have been called.
def get_next(rows_to_print):
# Declaring total_rows as nonlocal means this function uses the total_rows from
# the enclosing fucntion, and that this variable will keep its value in subsequent
# calls to the this function after set_iter is called.
nonlocal total_rows
nonlocal sheet_data_iter
rows_printed = 0
while rows_printed < rows_to_print:
rows_printed += 1
print(f"Printing Row: {rows_printed} of {rows_to_print} Rows to Print this time through.")
row_data = next(sheet_data_iter, sentinel.END_OF_ITERATION)
if row_data == sentinel.END_OF_ITERATION:
print("There are no more items to iterate over")
return None
else:
# row_data = next(sheet_data_iter)
print(f"0th Element of Row With Index: {total_rows} is: '{row_data[0]}'")
# total_rows maintains it's value through subsequent calls to the returned
# get_next function.  It's keeps track of the total number of times the
# iterator has been called.
total_rows += 1
return True
# get_next(row_count)
return get_next

# print_next_n_rows is a function, as returned by set_iter()
print_next_n_rows = set_iter(sheetData)
# another_print_n_rows is also a function, and it encapsulate a fresh iterator,
# it tracks separate from print_next_n_rows
another_print_n_rows = set_iter(sheetData)
# This line should print George
if print_next_n_rows(1):
print("There were still rows.n")
else:
print("Caught out of rows...n")

# This should print the next 2 lines
if print_next_n_rows(2):
print("There were still rows.n")
else:
print("Caught out of rows...n")

# This should print the first 2 lines...
if another_print_n_rows(2):
print("There were still rows.n")
else:
print("Caught out of rows...n")

# This should catch the end of the list
if print_next_n_rows(1):
print("There were still rows.n")
else:
print("Caught out of rows...n")

以下是上述脚本的输出:

Printing Row: 1 of 1 Rows to Print this time through.
0th Element of Row With Index: 0 is: 'George'
There were still rows.
Printing Row: 1 of 2 Rows to Print this time through.
0th Element of Row With Index: 1 is: 'Bill'
Printing Row: 2 of 2 Rows to Print this time through.
0th Element of Row With Index: 2 is: 'Mike'
There were still rows.
Printing Row: 1 of 2 Rows to Print this time through.
0th Element of Row With Index: 0 is: 'George'
Printing Row: 2 of 2 Rows to Print this time through.
0th Element of Row With Index: 1 is: 'Bill'
There were still rows.
Printing Row: 1 of 1 Rows to Print this time through.
There are no more items to iterate over
Caught out of rows...

hth,

大卫