I am trying to open Notepad using popen and write something into it. I can't get my head around it. I can open Notepad using command:
notepadprocess=subprocess.Popen('notepad.exe')
I am trying to identify how can I write anything in the text file using python. Any help is appreciated.
You can at first write something into txt file (ex. foo.txt) and then open it with notepad:
import os
f = open('foo.txt','w')
f.write('Hello world!')
f.close()
os.system("notepad.exe foo.txt")
You may be confusing the concept of (text) file with the processes that manipulate them.
Notepad is a program, of which you can create a process. A file, on the other hand, is just a structure on your hard drive.
From a programming standpoint, Notepad doesn't edit files. It:
reads a file into computer memory
modifies the content of that memory
writes that memory back into a file (which could be similarly named, or otherwise - which is known as the "Save as" operation).
Your program, just as any other program, can manipulate files, just as notepad does. In particular, you can perform exactly the same sequence as Notepad:
my_file= "myfile.txt" #the name/path of the file
with open(file, "rb") as f: #open the file for reading
content= f.read() #read the file into memory
content+= "mytext" #change the memory
with open(file, "wb") as f: #open the file for writing
f.write( content ) #write the memory into the file
Found the exact solution from Alex K's comment. I used pywinauto to perform this task.
Related
Started Python a week ago and I have some questions to ask about reading and writing to the same files. I've gone through some tutorials online but I am still confused about it. I can understand simple read and write files.
openFile = open("filepath", "r")
readFile = openFile.read()
print readFile
openFile = open("filepath", "a")
appendFile = openFile.write("\nTest 123")
openFile.close()
But, if I try the following I get a bunch of unknown text in the text file I am writing to. Can anyone explain why I am getting such errors and why I cannot use the same openFile object the way shown below.
# I get an error when I use the codes below:
openFile = open("filepath", "r+")
writeFile = openFile.write("Test abc")
readFile = openFile.read()
print readFile
openFile.close()
I will try to clarify my problems. In the example above, openFile is the object used to open file. I have no problems if I want write to it the first time. If I want to use the same openFile to read files or append something to it. It doesn't happen or an error is given. I have to declare the same/different open file object before I can perform another read/write action to the same file.
#I have no problems if I do this:
openFile = open("filepath", "r+")
writeFile = openFile.write("Test abc")
openFile2 = open("filepath", "r+")
readFile = openFile2.read()
print readFile
openFile.close()
I will be grateful if anyone can tell me what I did wrong here or is it just a Pythong thing. I am using Python 2.7. Thanks!
Updated Response:
This seems like a bug specific to Windows - http://bugs.python.org/issue1521491.
Quoting from the workaround explained at http://mail.python.org/pipermail/python-bugs-list/2005-August/029886.html
the effect of mixing reads with writes on a file open for update is
entirely undefined unless a file-positioning operation occurs between
them (for example, a seek()). I can't guess what
you expect to happen, but seems most likely that what you
intend could be obtained reliably by inserting
fp.seek(fp.tell())
between read() and your write().
My original response demonstrates how reading/writing on the same file opened for appending works. It is apparently not true if you are using Windows.
Original Response:
In 'r+' mode, using write method will write the string object to the file based on where the pointer is. In your case, it will append the string "Test abc" to the start of the file. See an example below:
>>> f=open("a","r+")
>>> f.read()
'Test abc\nfasdfafasdfa\nsdfgsd\n'
>>> f.write("foooooooooooooo")
>>> f.close()
>>> f=open("a","r+")
>>> f.read()
'Test abc\nfasdfafasdfa\nsdfgsd\nfoooooooooooooo'
The string "foooooooooooooo" got appended at the end of the file since the pointer was already at the end of the file.
Are you on a system that differentiates between binary and text files? You might want to use 'rb+' as a mode in that case.
Append 'b' to the mode to open the file in binary mode, on systems
that differentiate between binary and text files; on systems that
don’t have this distinction, adding the 'b' has no effect.
http://docs.python.org/2/library/functions.html#open
Every open file has an implicit pointer which indicates where data will be read and written. Normally this defaults to the start of the file, but if you use a mode of a (append) then it defaults to the end of the file. It's also worth noting that the w mode will truncate your file (i.e. delete all the contents) even if you add + to the mode.
Whenever you read or write N characters, the read/write pointer will move forward that amount within the file. I find it helps to think of this like an old cassette tape, if you remember those. So, if you executed the following code:
fd = open("testfile.txt", "w+")
fd.write("This is a test file.\n")
fd.close()
fd = open("testfile.txt", "r+")
print fd.read(4)
fd.write(" IS")
fd.close()
... It should end up printing This and then leaving the file content as This IS a test file.. This is because the initial read(4) returns the first 4 characters of the file, because the pointer is at the start of the file. It leaves the pointer at the space character just after This, so the following write(" IS") overwrites the next three characters with a space (the same as is already there) followed by IS, replacing the existing is.
You can use the seek() method of the file to jump to a specific point. After the example above, if you executed the following:
fd = open("testfile.txt", "r+")
fd.seek(10)
fd.write("TEST")
fd.close()
... Then you'll find that the file now contains This IS a TEST file..
All this applies on Unix systems, and you can test those examples to make sure. However, I've had problems mixing read() and write() on Windows systems. For example, when I execute that first example on my Windows machine then it correctly prints This, but when I check the file afterwards the write() has been completely ignored. However, the second example (using seek()) seems to work fine on Windows.
In summary, if you want to read/write from the middle of a file in Windows I'd suggest always using an explicit seek() instead of relying on the position of the read/write pointer. If you're doing only reads or only writes then it's pretty safe.
One final point - if you're specifying paths on Windows as literal strings, remember to escape your backslashes:
fd = open("C:\\Users\\johndoe\\Desktop\\testfile.txt", "r+")
Or you can use raw strings by putting an r at the start:
fd = open(r"C:\Users\johndoe\Desktop\testfile.txt", "r+")
Or the most portable option is to use os.path.join():
fd = open(os.path.join("C:\\", "Users", "johndoe", "Desktop", "testfile.txt"), "r+")
You can find more information about file IO in the official Python docs.
Reading and Writing happens where the current file pointer is and it advances with each read/write.
In your particular case, writing to the openFile, causes the file-pointer to point to the end of file. Trying to read from the end would result EOF.
You need to reset the file pointer, to point to the beginning of the file before through seek(0) before reading from it
You can read, modify and save to the same file in python but you have actually to replace the whole content in file, and to call before updating file content:
# set the pointer to the beginning of the file in order to rewrite the content
edit_file.seek(0)
I needed a function to go through all subdirectories of folder and edit content of the files based on some criteria, if it helps:
new_file_content = ""
for directories, subdirectories, files in os.walk(folder_path):
for file_name in files:
file_path = os.path.join(directories, file_name)
# open file for reading and writing
with io.open(file_path, "r+", encoding="utf-8") as edit_file:
for current_line in edit_file:
if condition in current_line:
# update current line
current_line = current_line.replace('john', 'jack')
new_file_content += current_line
# set the pointer to the beginning of the file in order to rewrite the content
edit_file.seek(0)
# delete actual file content
edit_file.truncate()
# rewrite updated file content
edit_file.write(new_file_content)
# empties new content in order to set for next iteration
new_file_content = ""
edit_file.close()
I have code that looks something like this:
import csv
import os
import tempfile
from azure.storage import CloudStorageAccount
account = CloudStorageAccount(
account_name=os.environ['AZURE_ACCOUNT'],
account_key=os.environ['AZURE_ACCOUNT_KEY'],
)
service = account.create_block_blob_service()
with tempfile.NamedTemporaryFile(mode='w') as f:
writer = csv.DictWriter(f, fieldnames=['foo', 'bar'])
writer.writerow({'foo': 'just an example', 'bar': 'of what I do'})
with open(f.name, 'rb') as stream:
service.create_blob_from_stream(
container_name='test',
blob_name='nothing_secret.txt',
stream=stream,
)
Now, this is ugly. I don't like having to open the file twice. I know that the Azure API provides a way to upload text and binary, but my file has the potential to be several hundred MB large so I'm not too interested in sticking the whole thing in memory at a time (not that it would be the end of the world, but still).
Azure doesn't support uploading a file in text mode (that I can see), and csv doesn't seem to support writing to a binary file (at least not text data).
Is there a way that I can have two handles to the same file, one in binary and one in text mode? Of course I could write my own file wrapper, but I'd prefer to use something I don't have to maintain. Is there a better way to do this than what I've got?
Files opened in text mode have a buffer attribute. This object is the same one you would get by opening the file in binary mode, the text mode is just a wrapper on top of it.
Open your file in text mode, use it for read it, then seek the buffer back to the start and use it for uploading. Make sure you use + mode for reading and writing from the same handle.
with tempfile.NamedTemporaryFile(mode='w+') as f:
...
f.seek(0)
service.create_blob_from_stream(
...
stream=f.buffer,
)
You can go the other way too, by opening in binary mode then wrapping with io.TextIOWRapper(f).
I am new to python.
I wanted to know if I could create a text file in the script itself before writing into.
I do not want to create a text file using the command prompt.
I have written this script to write the result into the file
with open('1.txt', 'r') as flp:
data = flp.readlines()
however I know that 1.txt has to be created before writing into it.
Any help would be highly appreciated.
Open can be used in a several modes, in your case you have opened in read mode ('r'). To write to a file you use the write mode ('w').
So you can get a file object with:
open('1.txt', 'w')
If 1.txt doesn't exist it will create it. If it does exist it will truncate it.
You can use open() to create files.
Example:
open("log.txt", "a")
This will create the file if it doesn't exist yet, and will append to it if the file already exists.
Using open(filename, 'w') creates the file if it's not there. Be careful though, if the file exists it will be overritten!
You can read more details here:
Started Python a week ago and I have some questions to ask about reading and writing to the same files. I've gone through some tutorials online but I am still confused about it. I can understand simple read and write files.
openFile = open("filepath", "r")
readFile = openFile.read()
print readFile
openFile = open("filepath", "a")
appendFile = openFile.write("\nTest 123")
openFile.close()
But, if I try the following I get a bunch of unknown text in the text file I am writing to. Can anyone explain why I am getting such errors and why I cannot use the same openFile object the way shown below.
# I get an error when I use the codes below:
openFile = open("filepath", "r+")
writeFile = openFile.write("Test abc")
readFile = openFile.read()
print readFile
openFile.close()
I will try to clarify my problems. In the example above, openFile is the object used to open file. I have no problems if I want write to it the first time. If I want to use the same openFile to read files or append something to it. It doesn't happen or an error is given. I have to declare the same/different open file object before I can perform another read/write action to the same file.
#I have no problems if I do this:
openFile = open("filepath", "r+")
writeFile = openFile.write("Test abc")
openFile2 = open("filepath", "r+")
readFile = openFile2.read()
print readFile
openFile.close()
I will be grateful if anyone can tell me what I did wrong here or is it just a Pythong thing. I am using Python 2.7. Thanks!
Updated Response:
This seems like a bug specific to Windows - http://bugs.python.org/issue1521491.
Quoting from the workaround explained at http://mail.python.org/pipermail/python-bugs-list/2005-August/029886.html
the effect of mixing reads with writes on a file open for update is
entirely undefined unless a file-positioning operation occurs between
them (for example, a seek()). I can't guess what
you expect to happen, but seems most likely that what you
intend could be obtained reliably by inserting
fp.seek(fp.tell())
between read() and your write().
My original response demonstrates how reading/writing on the same file opened for appending works. It is apparently not true if you are using Windows.
Original Response:
In 'r+' mode, using write method will write the string object to the file based on where the pointer is. In your case, it will append the string "Test abc" to the start of the file. See an example below:
>>> f=open("a","r+")
>>> f.read()
'Test abc\nfasdfafasdfa\nsdfgsd\n'
>>> f.write("foooooooooooooo")
>>> f.close()
>>> f=open("a","r+")
>>> f.read()
'Test abc\nfasdfafasdfa\nsdfgsd\nfoooooooooooooo'
The string "foooooooooooooo" got appended at the end of the file since the pointer was already at the end of the file.
Are you on a system that differentiates between binary and text files? You might want to use 'rb+' as a mode in that case.
Append 'b' to the mode to open the file in binary mode, on systems
that differentiate between binary and text files; on systems that
don’t have this distinction, adding the 'b' has no effect.
http://docs.python.org/2/library/functions.html#open
Every open file has an implicit pointer which indicates where data will be read and written. Normally this defaults to the start of the file, but if you use a mode of a (append) then it defaults to the end of the file. It's also worth noting that the w mode will truncate your file (i.e. delete all the contents) even if you add + to the mode.
Whenever you read or write N characters, the read/write pointer will move forward that amount within the file. I find it helps to think of this like an old cassette tape, if you remember those. So, if you executed the following code:
fd = open("testfile.txt", "w+")
fd.write("This is a test file.\n")
fd.close()
fd = open("testfile.txt", "r+")
print fd.read(4)
fd.write(" IS")
fd.close()
... It should end up printing This and then leaving the file content as This IS a test file.. This is because the initial read(4) returns the first 4 characters of the file, because the pointer is at the start of the file. It leaves the pointer at the space character just after This, so the following write(" IS") overwrites the next three characters with a space (the same as is already there) followed by IS, replacing the existing is.
You can use the seek() method of the file to jump to a specific point. After the example above, if you executed the following:
fd = open("testfile.txt", "r+")
fd.seek(10)
fd.write("TEST")
fd.close()
... Then you'll find that the file now contains This IS a TEST file..
All this applies on Unix systems, and you can test those examples to make sure. However, I've had problems mixing read() and write() on Windows systems. For example, when I execute that first example on my Windows machine then it correctly prints This, but when I check the file afterwards the write() has been completely ignored. However, the second example (using seek()) seems to work fine on Windows.
In summary, if you want to read/write from the middle of a file in Windows I'd suggest always using an explicit seek() instead of relying on the position of the read/write pointer. If you're doing only reads or only writes then it's pretty safe.
One final point - if you're specifying paths on Windows as literal strings, remember to escape your backslashes:
fd = open("C:\\Users\\johndoe\\Desktop\\testfile.txt", "r+")
Or you can use raw strings by putting an r at the start:
fd = open(r"C:\Users\johndoe\Desktop\testfile.txt", "r+")
Or the most portable option is to use os.path.join():
fd = open(os.path.join("C:\\", "Users", "johndoe", "Desktop", "testfile.txt"), "r+")
You can find more information about file IO in the official Python docs.
Reading and Writing happens where the current file pointer is and it advances with each read/write.
In your particular case, writing to the openFile, causes the file-pointer to point to the end of file. Trying to read from the end would result EOF.
You need to reset the file pointer, to point to the beginning of the file before through seek(0) before reading from it
You can read, modify and save to the same file in python but you have actually to replace the whole content in file, and to call before updating file content:
# set the pointer to the beginning of the file in order to rewrite the content
edit_file.seek(0)
I needed a function to go through all subdirectories of folder and edit content of the files based on some criteria, if it helps:
new_file_content = ""
for directories, subdirectories, files in os.walk(folder_path):
for file_name in files:
file_path = os.path.join(directories, file_name)
# open file for reading and writing
with io.open(file_path, "r+", encoding="utf-8") as edit_file:
for current_line in edit_file:
if condition in current_line:
# update current line
current_line = current_line.replace('john', 'jack')
new_file_content += current_line
# set the pointer to the beginning of the file in order to rewrite the content
edit_file.seek(0)
# delete actual file content
edit_file.truncate()
# rewrite updated file content
edit_file.write(new_file_content)
# empties new content in order to set for next iteration
new_file_content = ""
edit_file.close()
I'm learning Python, and have run into a bit of a problem. On my OSX install of Python 3.1, this happens in the console:
>>> filename = "test"
>>> reader = open(filename, 'r')
>>> writer = open(filename, 'w')
>>> reader.read()
''
>>> writer.write("hello world\n")
12
>>> reader.read()
''
And calling more test in BASH confirms that there is nothing in test. What's going on?
Thanks.
There are two potential reasons why you are seeing this behaviour.
When you open a file for writing (with the "w" open mode in Python), the OS removes the original file and creates a totally new one. So by opening the file for reading first and then writing, the original reading handle refers to a file that no longer has a name (the file still exists until you close it). At that point you're reading from a different file than you're writing to.
After you swap the order of opening so you open for writing and then reading, you won't necessarily be able to read the data from the file until you flush it:
>>> writer.flush()
>>> reader.read()
'hello world\n'
Flushing the file writes any data that might be in Python's file buffers to the OS, so that when you read from the file from the other handle, the OS will return the data. Note that Python itself doesn't know these two handles refer to the same file, but the OS does.
You're probably trashing your file. It's not usually a good idea to open a file for reading and writing at the same time.
Buffering. If you really want to read and write to the same file open one handle using "w+".
And with the buttering, you will need to force the buffer to be emptied before reading. Closing the file is a good way to do this.