Develop Reference

How do you get Python to open a text file using print and input? I have the code I am using below

The code I have so far which is only producing an error is below
print(input("What is the text file's name?"))
file = input
openfile = open(file)
Any help would be appreciated! If you need more code/background please ask :)

Python 3.x
filename = input("What is the text file's name?")
with open(filename) as f:
for line in f:
print(line)
Python 2.x
filename = raw_input("What is the text file's name?")
with open(filename) as f:
for line in f:
print line
Note that you may either use:
A file name if the file is in the current working directory, e.g.
data.txt
A relative path from your working directory
..\folder\data.txt
An absolute path
C:\Folder\Subfolder\data.txt

Python: Unable to open and read a file

I am totally new to python.
I was trying to read a file which I already created but getting the below error
File "C:/Python25/Test scripts/Readfile.py", line 1, in <module>
filename = open('C:\Python25\Test scripts\newfile','r')
IOError: [Errno 2] No such file or directory: 'C:\\Python25\\Test scripts\newfile
My code:
filename = open('C:\Python25\Test scripts\newfile','r')
print filename.read()
Also I tried
filename = open('C:\\Python25\\Test scripts\\newfile','r')
print filename.read()
But same errors I am getting.

Try:
fpath = r'C:\Python25\Test scripts\newfile'
if not os.path.exists(fpath):
print 'File does not exist'
return
with open(fpath, 'r') as src:
src.read()
First you validate that file, that it exists.
Then you open it. With wrapper is more usefull, it closes your file, after you finish reading. So you will not stuck with many open descriptors.

I think you're probably having this issue because you didn't include the full filename.
You should try:
filename = open('C:\Python25\Test scripts\newfile.txt','r')
print filename.read()
*Also if you're running this python file in the same location as the target file your are opening, you don't need to give the full directory, you can just call:
filename = open(newfile.txt

I had the same problem. Here's how I got it right.
your code:
filename = open('C:\\Python25\\Test scripts\\newfile','r')
print filename.read()
Try this:
with open('C:\\Python25\\Test scripts\\newfile') as myfile:
print(myfile.read())
Hope it helps.

I am using VS code. If I am not using dent it would not work for the print line. So try to have the format right then you will see the magic.
with open("mytest.txt") as myfile:
print(myfile.read())
or without format like this:
hellofile=open('mytest.txt', 'r')
print(hellofile.read())

Python not printing beautifulsoup data to .txt file (or I can't find it)

I'm trying to put all the anchor text on a page in a txt file
print(anchor.get('href'))
with open('file.txt', 'a') as fd:
fd.write(anchor.get('href') + '\n')
and the script executes with no errors but I cannot find file.txt anywhere on my computer. Am I missing out on something really obvious?

With file open mode a (append) the file will be created if it doesn't already exist, otherwise writes will be appended to the end of the file. As written in the question, the file will be created in the current directory of the running process... look there.
from bs4 import BeautifulSoup
import requests
response = requests.get('http://httpbin.org/')
soup = BeautifulSoup(response.text)
with open('file.txt', 'a') as fd:
links = (link.get('href') for link in soup.find_all('a'))
fd.write('\n'.join(links) + '\n')

Try to set a full path to filename, so you can find in this path.
Example:
print(anchor.get('href'))
with open('/home/wbk/file.txt', 'a') as fd:
fd.write(anchor.get('href') + '\n')
And yes, you can use 'a' to create a new file, although this is not a good practice. The 'a' is to append data to the end of an existing file as describes doc

The following code works for me:
with open("example.txt","a") as a:
a.write("hello")
are you sure that the file isn't in your computer? In which ways you have demonstrated it? In my case, i work with eclipse IDE, so I have to refresh the Eclipse file explorer to see it..
Try this:
with open("prova.txt","a") as a:
a.write("ciao")
try: open("prova.txt","r")
except: print "file doesn't exist"

Error when trying to read and write multiple files

I modified the code based on the comments from experts in this thread. Now the script reads and writes all the individual files. The script reiterates, highlight and write the output. The current issue is, after highlighting the last instance of the search item, the script removes all the remaining contents after the last search instance in the output of each file.
Here is the modified code:
import os
import sys
import re
source = raw_input("Enter the source files path:")
listfiles = os.listdir(source)
for f in listfiles:
filepath = source+'\\'+f
infile = open(filepath, 'r+')
source_content = infile.read()
color = ('red')
regex = re.compile(r"(\b be \b)|(\b by \b)|(\b user \b)|(\bmay\b)|(\bmight\b)|(\bwill\b)|(\b's\b)|(\bdon't\b)|(\bdoesn't\b)|(\bwon't\b)|(\bsupport\b)|(\bcan't\b)|(\bkill\b)|(\betc\b)|(\b NA \b)|(\bfollow\b)|(\bhang\b)|(\bbelow\b)", re.I)
i = 0; output = ""
for m in regex.finditer(source_content):
output += "".join([source_content[i:m.start()],
"<strong><span style='color:%s'>" % color[0:],
source_content[m.start():m.end()],
"</span></strong>"])
i = m.end()
outfile = open(filepath, 'w+')
outfile.seek(0)
outfile.write(output)
print "\nProcess Completed!\n"
infile.close()
outfile.close()
raw_input()

The error message tells you what the error is:
No such file or directory: 'sample1.html'
Make sure the file exists. Or do a try statement to give it a default behavior.

The reason why you get that error is because the python script doesn't have any knowledge about where the files are located that you want to open.
You have to provide the file path to open it as I have done below. I have simply concatenated the source file path+'\\'+filename and saved the result in a variable named as filepath. Now simply use this variable to open a file in open().
import os
import sys
source = raw_input("Enter the source files path:")
listfiles = os.listdir(source)
for f in listfiles:
filepath = source+'\\'+f # This is the file path
infile = open(filepath, 'r')
Also there are couple of other problems with your code, if you want to open the file for both reading and writing then you have to use r+ mode. More over in case of Windows if you open a file using r+ mode then you may have to use file.seek() before file.write() to avoid an other issue. You can read the reason for using the file.seek() here.

Confusing Error when Reading from a File in Python

I'm having a problem opening the names.txt file. I have checked that I am in the correct directory. Below is my code:
import os
print(os.getcwd())
def alpha_sort():
infile = open('names', 'r')
string = infile.read()
string = string.replace('"','')
name_list = string.split(',')
name_list.sort()
infile.close()
return 0
alpha_sort()
And the error I got:
FileNotFoundError: [Errno 2] No such file or directory: 'names'
Any ideas on what I'm doing wrong?

You mention in your question body that the file is "names.txt", however your code shows you trying to open a file called "names" (without the ".txt" extension). (Extensions are part of filenames.)
Try this instead:
infile = open('names.txt', 'r')

As a side note, make sure that when you open files you use universal mode, as windows and mac/unix have different representations of carriage returns (/r/n vs /n etc.). Universal mode gets python to handle this, so it's generally a good idea to use it whenever you need to read a file. (EDIT - should read: a text file, thanks cameron)
So the code would just look like this
infile = open( 'names.txt', 'rU' ) #capital U indicated to open the file in universal mode

This doesn't solve that issue, but you might consider using with when opening files:
with open('names', 'r') as infile:
string = infile.read()
string = string.replace('"','')
name_list = string.split(',')
name_list.sort()
return 0
This closes the file for you and handles exceptions as well.