Question

我有一个表格

的字符串

**var beforeDate = new Date('2015-08-21')**

这里我不知道paranthesis（）之间的值。我想将此日期替换为任何其他日期。我怎么能在Python中做到这一点？我想打开文件，然后使用该语言的标准替换函数，但由于未知的值beween（），这是不可能的。

此代码段以及此代码段之后会有很多代码，因此用新行替换整行将无法正常工作，因为它会覆盖此代码段周围的代码。

Answer 1

使用正则表达式怎么样？例如：

TEMP.TXT

print "I've got a lovely bunch of coconuts"
var beforeDate = new Date('2015-08-21') #date determined by fair die roll
print "Here they are, standing in a row"

main.py

import re

new_value = "'1999-12-31'"
with open("temp.txt") as infile:
    data = infile.read()
    data = re.sub(r"(var beforeDate = new Date\().*?(\))", "\\1"+new_value+"\\2", data)
with open("output.txt", "w") as outfile:
    outfile.write(data)

运行后

output.txt：

print "I've got a lovely bunch of coconuts"
var beforeDate = new Date('1999-12-31') #date determined by fair die roll
print "Here they are, standing in a row"

Answer 2

就个人而言，我通常认为re.split（）比re.sub（）更易于使用。这将重用Kevin的代码，并将捕获它所做的一切（加上中间组），然后替换中间组：

import re

new_value = "'1999-12-31'"
with open("temp.txt") as infile:
    data = infile.read()

data = re.split(r"(var beforeDate = new Date\()(.*?)(\))", data)
# data[0] is everything before the first capture
# data[1] is the first capture
# data[2] is the second capture -- the one we want to replace
data[2] = new_value

with open("output.txt", "w") as outfile:
    outfile.write(''.join(stuff))

您可以放弃捕获中间组，但随后您将在列表中插入内容。只做替换更容易。

OTOH，这个特殊的问题可能足够小，不需要锤子。这是相同的代码，没有re：

new_value = "'1999-12-31'"
with open("temp.txt") as infile:
    data = infile.read()

data = list(data.partition('var beforeDate = new Date('))
data += data.pop().partition(')')
data[2] = new_value

with open("output.txt", "w") as outfile:
    for stuff in data:
        outfile.write(stuff)

替换其周围环境已知但字符串值不为Python的字符串

2 个答案: