Develop Reference

How to open and print the contents of a file line by line using subprocess?

I am trying to write a python script which SSHes into a specific address and dumps a text file. I am currently having some issues. Right now, I am doing this:
temp = "cat file.txt"
need = subprocess.Popen("ssh {host} {cmd}".format(host='155.0.1.1', cmd=temp),shell=True,stdout=subprocess.PIPE, stderr=subprocess.PIPE).communicate()
print(need)
This is the naive approach where I am basically opening the file, saving its output to a variable and printing it. However, this really messes up the format when I print "need". Is there any way to simply use subprocess and read the file line by line? I have to be SSHed into the address in order to dump the file otherwise the file will not be detected, that is why I am not simply doing
f = open(temp, "r")
file_contents = f.read()
print (file_contents)
f.close()
Any help would be appreciated :)

You don't need to use the subprocess module to print the entire file line by line. You can use pure python.
f = open(temp, "r")
file_contents = f.read()
f.close()
# turn the file contents into a list
file_lines = file_contents.split("\n")
# print all the items in the list
for file_line in file_lines:
print(file_line)

Attempt to use the open() function failing

I'm trying to learn to manipulate files on python, but I can't get the open function to work. I have made a .txt file called foo that holds the content "hello world!" in my user directory (/home/yonatan) and typed this line into the shell:
open('/home/yonatan/foo.txt')
What i get in return is:
<_io.TextIOWrapper name='/home/yonatan/foo.txt' mode='r' encoding='UTF-8'>
I get what that means, but why don't I get the content?

open() returns a file object.
You then need to use read() to read the whole file
f = open('/home/yonatan/foo.txt', 'r')
contents = f.read()
Or you can use readline() to read just one line
line = f.readline()
and don't forget to close the file at the end
f.close()

An example iterating through the lines of the file (using with which ensures file.close() gets called on the end of it's lexical scope):
file_path = '/home/yonatan/foo.txt'
with open(file_path) as file:
for line in file:
print line
A great resource on I/O and file handling operations.

You haven't specified the mode you want to open it in.
Try:
f = open("home/yonatan/foo.txt", "r")
print(f.read())

string.replace method in python

I am a newbie with python, so kindly excuse for asking basic question.
I am trying to use the string.replace method in python and getting a weird behavior. here is what I am doing:
# passing through command line a file name
with open(sys.argv[2], 'r+') as source:
content = source.readlines()
for line in content:
line = line.replace(placeholerPattern1Replace,placeholerPattern1)
#if I am printing the line here, I am getting the correct value
source.write(line.replace(placeholerPattern1Replace,placeholerPattern1))
try:
target = open('baf_boot_flash_range_test_'+subStr +'.gpj', 'w')
for line in content:
if placeholerPattern3 in line:
print line
target.write(line.replace(placeholerPattern1, <variable>))
target.close()
When I am checking the values in the new file, then these are not replaced. I could see that the value of the source is also not changed, but the content had changed, what am I doing wrong here?

Rather do something like this -
contentList = []
with open('somefile.txt', 'r') as source:
for line in source:
contentList.append(line)
with open('somefile.txt','w') as w:
for line in contentList:
line = line.replace(stringToReplace,stringToReplaceWith)
w.write(line)

Because with will close your file after runing all the statements wrapped within it, which means the content local variable will be nil in the second loop.

You are reading from the file source and also writing to it. Don't do that. Instead, you should write to a NamedTemporaryFile and then rename it over the original file after you finish writing and close it.

Try this:
# Read the file into memory
with open(sys.argv[2], 'r') as source:
content = source.readlines()
# Fix each line
new_content = list()
for line in content:
new_content.append(line.replace(placeholerPattern1Replace, placeholerPattern1))
# Write the data to a temporary file name
with open(sys.argv[2] + '.tmp', 'w') as dest:
for line in new_content:
dest.write(line)
# Rename the temporary file to the input file name
os.rename(sys.argv[2] + '.tmp', sys.argv[2])

How can I tell python to edit another python file?

Right now, I have file.py and it prints the word "Hello" into text.txt.
f = open("text.txt")
f.write("Hello")
f.close()
I want to do the same thing, but I want to print the word "Hello" into a Python file. Say I wanted to do something like this:
f = open("list.py")
f.write("a = 1")
f.close
When I opened the file list.py, would it have a variable a with a value 1? How would I go about doing this?

If you want to append a new line to the end of a file
with open("file.py", "a") as f:
f.write("\na = 1")
If you want to write a line to the beginning of a file try creating a new one
with open("file.py") as f:
lines = f.readlines()
with open("file.py", "w") as f:
lines.insert(0, "a = 1")
f.write("\n".join(lines))

with open("list.py","a") as f:
f.write("a=1")
This is simple as you see. You have to open that file in write and read mode (a). Also with open() method is safier and more clear.
Example:
with open("list.py","a") as f:
f.write("a=1")
f.write("\nprint(a+1)")
list.py
a=1
print(a+1)
Output from list.py:
>>>
2
>>>
As you see, there is a variable in list.py called a equal to 1.

I would recommend you specify opening mode, when you are opening a file for reading, writing, etc. For example:
for reading:
with open('afile.txt', 'r') as f: # 'r' is a reading mode
text = f.read()
for writing:
with open('afile.txt', 'w') as f: # 'w' is a writing mode
f.write("Some text")
If you are opening a file with 'w' (writing) mode, old file content will be removed. To avoid that appending mode exists:
with open('afile.txt', 'a') as f: # 'a' as an appending mode
f.write("additional text")
For more information, please, read documentation.

os.path.join does not yield the contents of the file

print "Which category would you like to view? Savory, Dessert, Cake, Soup or Drink? "
category = raw_input()
for x in os.listdir(category): print x
name = raw_input("Which recipe would wou like to view? ")
fullname = os.path.join(category, name)
f = open(fullname, "r");
print f
I am writing a program that will allow users to view the contents of .txt files saved in specific directories. When I run this code I don't get the contents and instead get a message that says this:
open file 'savory/b.txt', mode 'r' at 0x1004bd140
any ideas. I am new to python so i dont have much of an idea as to what is causing the error but i assume it is due to some missing code.
Thank you.

The return value of open is a file object (not the file contents!) .You need to call a method on your file object to actually read the file:
f = open(fullname, "r")
print f.read()
f.close()
If it's a big file you may want to iterate over the file line-by-line
f = open(fullname, "r")
for line in f:
print line
f.close()
On a side note, here's alternate syntax to you don't have to remember to call the close method:
with open(fullname, "r") as f:
for line in f:
print line