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How to make a program that replaces newlines in python file with a string [duplicate]

This question already has answers here:
Why doesn't calling a string method (such as .replace or .strip) modify (mutate) the string?
(3 answers)
Closed 3 years ago.
I am trying to display my python file in html and therefore I would like to replace every time the file jumps to a newline with < br> but the program I've written is not working.
I've looked on here and tried changing the code around a bit I have gotten different results but not the ones I need.
with open(path, "r+") as file:
contents = file.read()
contents.replace("\n", "<br>")
print(contents)
file.close()
I want to have the file display < br> every time I have a new line but instead the code dosen't change anything to the file.

Here is an example program that works:
path = "example"
contents = ""
with open(path, "r") as file:
contents = file.read()
new_contents = contents.replace("\n", "<br>")
with open(path, "w") as file:
file.write(new_contents)
Your program doesn't work because the replace method does not modify the original string; it returns a new string.
Also, you need to write the new string to the file; python won't do it automatically.
Hope this helps :)
P.S. a with statement automatically closes the file stream.

Your code reads from the file, saves the contents to a variable and replaces the newlines. But the result is not saved anywhere. And to write the result into a file you must open the file for writing.
with open(path, "r+") as file:
contents = file.read()
contents = contents.replace("\n", "<br>")
with open(path, "w+") as file:
contents = file.write(contents)

there are some issues in this code snippet.
contents.replace("\n", "<br>") will return a new object which replaced \n with <br>, so you can use html_contents = contents.replace("\n", "<br>") and print(html_contents)
when you use with the file descriptor will close after leave the indented block.

Try this:
import re
with open(path, "r") as f:
contents = f.read()
contents = re.sub("\n", "<br>", contents)
print(contents)

Borrowed from this post:
import tempfile
def modify_file(filename):
#Create temporary file read/write
t = tempfile.NamedTemporaryFile(mode="r+")
#Open input file read-only
i = open(filename, 'r')
#Copy input file to temporary file, modifying as we go
for line in i:
t.write(line.rstrip()+"\n")
i.close() #Close input file
t.seek(0) #Rewind temporary file to beginning
o = open(filename, "w") #Reopen input file writable
#Overwriting original file with temporary file contents
for line in t:
o.write(line)
t.close() #Close temporary file, will cause it to be deleted

How to amend an existing python file

I am trying to ammend a group of files in a folder, by adding F to the 4th line (which is number 3 in python, if I'm correct). With the following code below, the code is just continuously running and not making the amendments, anyone got any ideas?
import os
from glob import glob
list_of_files = glob('*.gjf') # get list of all .gjf files
for file in list_of_files:
# read file:
with open(file, 'r+') as f:
lines=f.readlines()
for line in lines:
lines.insert(3,'F')

for file in files:
# read your file's lines
with open(file, 'r') as f:
lines = f.readlines()
# add the value to insert in the list of lines at index 3
# don't forget line-break (\n)
lines.insert(3, 'F\n')
# join lines to create a string
text = ''.join(lines)
# don't forget to write the string back in your file
with open(file, 'w') as f:
f.write(text)

You are not able to edit a file with Python in this way. You would need to create a temporary file and then do some cleanup at the end by renaming the temporary file and removing the temporary file. The core logic is to read through the original files and add an F to the 4th line, otherwise just add the line. This can be done with a function like this:
import os
def add_f_4th_line(filename):
with open(filename, 'r') as f_in:
with opne('temp', 'w') as f_out:
for line_number, line_contents in enumerate(f_in):
if line_number == 3:
f_out.write(line_contents.replace('\n', 'F\n'))
else:
f_out.write(line_contents)
os.rename('temp', filename)
os.remove('temp')
enumerate will keep make an iterator of the line number and the contents of the line, so just iterate over each line, and you can find the 4th line with line_number == 3. Then, it will take that line and replace \n with F\n. \n is a new line character, so I'm assuming the end of the line is this character. So after adding this function, you'll just need to call this function for each file you get with your glob call.
import os
from glob import glob
list_of_files = glob('*.gjf') # get list of all .gjf files
for file in list_of_files:
add_f_4th_line(file)

search replace the string from number of .txt files in python

there are multiple files in directory with extension .txt, .dox, .qcr etc.
i need to list out txt files, search & replace the text from each txt files only.
need to search the $$\d ...where \d stands for the digit 1,2,3.....100.
need to replace with xxx.
please let me know the python script for this .
thanks in advance .
-Shrinivas
#created following script, it works for single txt files, but it is not working for txt files more than one lies in directory.
-----
def replaceAll(file,searchExp,replaceExp):
for line in fileinput.input(file, inplace=1):
if searchExp in line:
line = line.replace(searchExp,replaceExp)
sys.stdout.write(line)
#following code is not working, i expect to list out the files start #with "um_*.txt", open the file & replace the "$$\d" with replaceAll function.
for um_file in glob.glob('*.txt'):
t = open(um_file, 'r')
replaceAll("t.read","$$\d","xxx")
t.close()

fileinput.input(...) is supposed to process a bunch of files, and must be ended with a corresponding fileinput.close(). So you can either process all in one single call:
def replaceAll(file,searchExp,replaceExp):
for line in fileinput.input(file, inplace=True):
if searchExp in line:
line = line.replace(searchExp,replaceExp)
dummy = sys.stdout.write(line) # to avoid a possible output of the size
fileinput.close() # to orderly close everythin
replaceAll(glob.glob('*.txt'), "$$\d","xxx")
or consistently close fileinput after processing each file, but it rather ignores the main fileinput feature.

Try out this.
import re
def replaceAll(file,searchExp,replaceExp):
for line in file.readlines():
try:
line = line.replace(re.findall(searchExp,line)[0],replaceExp)
except:
pass
sys.stdout.write(line)
#following code is not working, i expect to list out the files start #with "um_*.txt", open the file & replace the "$$\d" with replaceAll function.
for um_file in glob.glob('*.txt'):
t = open(um_file, 'r')
replaceAll(t,"\d+","xxx")
t.close()
Here we are sending file handler to the replaceAll function rather than a string.

You can try this:
import os
import re
the_files = [i for i in os.listdir("foldername") if i.endswith("txt")]
for file in the_files:
new_data = re.sub("\d+", "xxx", open(file).read())
final_file = open(file, 'w')
final_file.write(new_data)
final_file.close()

Odd behavior of os

I'm trying to write a Python script that will take any playlist and recreate it on another file structure. I have it written now so that all the filenames are stripped off the original playlist and put into a file. That works. Then the function findsong() is supposed to walk thru the new directory and find the same songs and make a new playlist based on the new directory structure.
Here's where it gets weird. If I use 'line' as my argument in the line 'If line in files' I get an empty new playlist. If I use ANY file that I know is there as the argument the entire playlist is recreated, not just the file I used as the argument. That's how I have it set up in this code. I cannot figure out this weird behavior. As long as the file exists, the whole playlist is recreated with the new paths. Wut??
Here is the code:
import os
def check():
datafile = open('testlist.m3u')
nopath = open('nopath.txt', 'w')
nopath.truncate()
for line in datafile:
if 'mp3' in line:
nopath.write(os.path.basename(line))
if 'wma' in line:
nopath.write(os.path.basename(line))
nopath.close()
def findsong():
nopath = open('nopath.txt')
squeezelist = open('squeezelist.m3u' ,'w')
squeezelist.truncate()
for line in nopath:
print line
for root, dirs, files in os.walk("c:\\Documents and Settings\\"):
print files
if ' Tuesday\'s Gone.mp3' in files:
squeezelist.write(os.path.join(root, line))
squeezelist.close()
check()
findsong()

When you iterate over the lines in a file Python retains the trailing newlines \n. You'll want to strip those off:
for line in nopath:
line = line.rstrip()

python printing a blank line on the first line when writing to a file

I'm stuck on why my code is printing a blank line before writing text to a file. What I am doing is reading two files from a zipped folder and writing the text to a new text file. I am getting the expected results in the file, except for the fact that there is a blank line on the first line of the file.
def test():
if zipfile.is_zipfile(r'C:\Users\test\Desktop\Zip_file.zip'):
zf = zipfile.ZipFile(r'C:\Users\test\Desktop\Zip_file.zip')
for filename in zf.namelist():
with zf.open(filename, 'r') as f:
words = io.TextIOWrapper(f)
new_file = io.open(r'C:\Users\test\Desktop\new_file.txt', 'a')
for line in words:
new_file.write(line)
new_file.write('\n')
else:
pass
zf.close()
words.close()
f.close()
new_file.close()
Output in new_file (there is a blank line before the first "This is a test line...")
This is a test line...
This is a test line...
this is test #2
this is test #2
Any ideas?
Thanks!

My guess is that the first file in zf.namelist() doesn't contain anything, so you skip the for line in words loop for that file and just do new_file.write('\n'). It's difficult to tell without seeing the files that you're looping over; perhaps add some debug statements that print out the files' names and some info, e.g. their size.