I recieve some files with .ini file with them. I have to recieve file names from [FILES] section.
Sometimes there is an extra witespace in another section of .ini-file which raises exception in ConfigParser module
The example of "bad" ini-file:
[LETTER]
SUBJECT=some text
some text
and text with whitespace in the beggining
[FILES]
0=file1.txt
1=file2.doc
My code(Python 3.7):
import configparser
def get_files_from_ini_file(info_file):
ini = configparser.ConfigParser(allow_no_value=True)
ini.read(info_file) # ERROR is here
if ini.has_section("FILES"):
pocket_files = [ini.get("FILES", i) for i in ini.options("FILES")]
return pocket_files
print(get_files_from_ini_file("D:\\bad.ini"))
Traceback (most recent call last):
File "D:/test.py", line 10, in <module>
print(get_files_from_ini_file("D:\\bad.ini"))
File "D:/test.py", line 5, in get_files_from_ini_file
ini.read(info_file) # ERROR
File "C:\Users\ap\AppData\Local\Programs\Python\Python37-32\lib\configparser.py", line 696, in read
self._read(fp, filename)
File "C:\Users\ap\AppData\Local\Programs\Python\Python37-32\lib\configparser.py", line 1054, in _read
cursect[optname].append(value)
AttributeError: 'NoneType' object has no attribute 'append'
I can't influence on files I recieve so that is there any way to ignore this error? In fact I need only [FILES] section to parse.
Have tried empty_lines_in_values=False with no result
May be that's invalid ini file and I should write my own parser?
If you only need the "FILES" part, a simple way is to:
open the file and read into a string
get the part after "[FILES]" using .split() method
add "[FILES]" before the string
use the configparser read_string method on the string
This is a hacky solution but it should work:
import configparser
def get_files_from_ini_file(info_file):
with open(info_file, 'r') as file:
ini_string = file.read()
useful_part = "[FILES]" + ini_string.split("[FILES]")[-1]
ini = configparser.ConfigParser(allow_no_value=True)
ini.read_string(useful_part) # ERROR is here
if ini.has_section("FILES"):
pocket_files = [ini.get("FILES", i) for i in ini.options("FILES")]
return pocket_files
print(get_files_from_ini_file("D:\\bad.ini"))
Related
I would appreciate a hand with this.
It has previously popped corrupt file errors when opening the word file, but if I change .doc to .docx and remove some hyperlinks (I understand from another post somewhere that hyperlinks, footnotes, comments all cause errors), this time IDLE pops out the following error:
Traceback (most recent call last):
File "C:\Users\User\AppData\Local\Programs\Python\Python38-32\Files_tempfiller_\tempfiller.py", line 43, in
document.write('TBCO.docx')
File "C:\Users\User\AppData\Local\Packages\PythonSoftwareFoundation.Python.3.8_qbz5n2kfra8p0\LocalCache\local-packages\Python38\site-packages\mailmerge.py", line 129, in write
output.writestr(zi.filename, self.zip.read(zi))
File "C:\Program Files\WindowsApps\PythonSoftwareFoundation.Python.3.8_3.8.1776.0_x64__qbz5n2kfra8p0\lib\zipfile.py", line 1475, in read
with self.open(name, "r", pwd) as fp:
File "C:\Program Files\WindowsApps\PythonSoftwareFoundation.Python.3.8_3.8.1776.0_x64__qbz5n2kfra8p0\lib\zipfile.py", line 1532, in open
raise BadZipFile("Truncated file header")
zipfile.BadZipFile: Truncated file header
Because this is in the mailmerge.py file I can't really understand what this is.
My code is as follows:
from __future__ import print_function
from mailmerge import MailMerge
from datetime import date
template = "TBCO.docx" #add .docx suffix if failing
print('1')
document = MailMerge(template)
print(document.get_merge_fields())
print('2')
document.merge(
date = '1.1.1',
name = 'Bob',
nhs = '2223')
print('3')
document.write('TBCO.docx')
print('4')
The prints were for me to see what was happening when it was giving set() repeatedly, but that's fixed. The sense I get from the error message is that it is struggling with the file type for some reason, but I can't make head nor tail of the error. Any help would be appreciated.
Thank you
When I try to save the file lines in a list I with the following code I get the below error:
Traceback (most recent call last):
File "C:\Users\emiel\AppData\Roaming\Sublime Text 3\Packages\User\Making_a_list_from_file.py", line 7, in <module>
with filename as file_object:
AttributeError: __enter__
This is the code:
with filename as file_object:
lines=file_object.readlines()
for line in lines:
print(line.strip())
The error is telling you that the object you're using in the with statement is not of the right type. __enter__ is one of the methods called as part of the context manager protocol, and the type you're using doesn't have that method.
Based on your variable name, it looks like you may be using a file name where you want to be using a file object. The error message will then make sense, as strings are not context managers the way file objects are. Try changing your with statement to:
with open(filename) as file_object:
How do I turn this format of TXT file into a CSV file?
Date,Open,high,low,close
1/1/2017,1,2,1,2
1/2/2017,2,3,2,3
1/3/2017,3,4,3,4
I am sure you can understand? It already has the comma -eparated values.
I tried using numpy.
>>> import numpy as np
>>> table = np.genfromtxt("171028 A.txt", comments="%")
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "C:\Users\Smith\AppData\Local\Continuum\anaconda3\lib\site-packages\numpy\lib\npyio.py", line 1551, in genfromtxt
fhd = iter(np.lib._datasource.open(fname, 'rb'))
File "C:\Users\Smith\AppData\Local\Continuum\anaconda3\lib\site-packages\numpy\lib\_datasource.py", line 151, in open
return ds.open(path, mode)
File "C:\Users\Smith\AppData\Local\Continuum\anaconda3\lib\site-packages\numpy\lib\_datasource.py", line 501, in open
raise IOError("%s not found." % path)
OSError: 171028 A.txt not found.
I have (S&P) 500 txt files to do this with.
You can use csv module. You can find more information here.
import csv
txt_file = 'mytext.txt'
csv_file = 'mycsv.csv'
in_txt = csv.reader(open(txt_file, "r"), delimiter=',')
out_csv = csv.writer(open(csv_file, 'w+'))
out_csv.writerows(in_txt)
Per #dclarke's comment, check the directory from which you run the code. As you coded the call, the file must be in that directory. When I have it there, the code runs without error (although the resulting table is a single line with four nan values). When I move the file elsewhere, I reproduce your error quite nicely.
Either move the file to be local, add a local link to the file, or change the file name in your program to use the proper path to the file (either relative or absolute).
I wrote this python script in order to create Xml content and i would like to write this "prettified" xml to a file (50% done):
My script so far:
data = ET.Element("data")
project = ET.SubElement(data, "project")
project.text = "This project text"
rawString = ET.tostring(data, "utf-8")
reparsed = xml.dom.minidom.parseString(rawString)
cleanXml = reparsed.toprettyxml(indent=" ")
# This prints the prettified xml i would like to save to a file
print cleanXml
# This part does not work, the only parameter i can pass is "data"
# But when i pass "data" as a parameter, a xml-string is written to the file
tree = ET.ElementTree(cleanXml)
tree.write("config.xml")
The error i get when i pass cleanXml as parameter:
Traceback (most recent call last):
File "app.py", line 45, in <module>
tree.write("config.xml")
File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/xml/etree/ElementTree.py", line 817, in write
self._root, encoding, default_namespace
File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/xml/etree/ElementTree.py", line 876, in _namespaces
iterate = elem.getiterator # cET compatibility
AttributeError: 'unicode' object has no attribute 'getiterator'
Anybody knows how i can get my prettified xml to a file ?
Thanks and Greetings!
The ElementTree constructor can be passed a root element and a file. To create an ElementTree from a string, use ElementTree.fromstring.
However, that isn't what you want. Just open a file and write the string directly:
with open("config.xml", "w") as config_file:
config_file.write(cleanXml)
I am creating a program that loads and runs python scripts from a compressed file. Along with those python scripts, I have a config file that I previously used configparser to load info from in an uncompressed version of the program.
Is it possible to directly read config files in zip files directly with configparser? or do I have to unzip it into a temp folder and load it from there?
I have tried directly giving the path:
>>> sysconf = configparser.ConfigParser()
>>> sysconf.read_file("compressed.zip/config_data.conf")
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/usr/local/lib/python3.4/configparser.py", line 691, in read_file
self._read(f, source)
File "/usr/local/lib/python3.4/configparser.py", line 1058, in _read
raise MissingSectionHeaderError(fpname, lineno, line)
configparser.MissingSectionHeaderError: File contains no section headers.
file: '<???>', line: 1
Didn't work. no surprises there.
Then I tried using zipfile
>>> zf = zipfile.ZipFile("compressed.zip")
>>> data = zf.read("config_data.conf")
>>> sysconf = configparser.ConfigParser()
>>> sysconf.read_file(data)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/usr/local/lib/python3.4/configparser.py", line 691, in read_file
self._read(f, source)
File "/usr/local/lib/python3.4/configparser.py", line 1009, in _read
if line.strip().startswith(prefix):
AttributeError: 'int' object has no attribute 'strip'
and found that it didn't work either.
so I've resorted to creating a temp folder, uncompressing to it, and reading the conf file there. I would really like to avoid this if possible as the conf files are the only limiting factor. I can (and am) loading the python modules from the zip file just fine at this point.
I can get the raw text of the file if there's a way to pass that directly to configparser, but searching the docs I came up empty handed.
Update:
I tried using stringIO as a file object, and it seems to work somewhat.
configparser doesn't reject it, but it doesn't like it either.
>>> zf = zipfile.ZipFile("compressed.zip")
>>> data = zf.read(config_data.conf)
>>> confdata = io.StringIO(str(data))
>>> sysconf = configparser.ConfigParser()
>>> sysconf.readfp(confdata)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/usr/local/lib/python3.4/configparser.py", line 736, in readfp
self.read_file(fp, source=filename)
File "/usr/local/lib/python3.4/configparser.py", line 691, in read_file
self._read(f, source)
File "/usr/local/lib/python3.4/configparser.py", line 1058, in _read
raise MissingSectionHeaderError(fpname, lineno, line)
configparser.MissingSectionHeaderError: File contains no section headers.
file: '<???>', line: 1
(continues to spit out the entire contents of the file)
If I use read_file instead, it doesn't error out, but doesn't load anything either.
>>> zf = zipfile.ZipFile("compressed.zip")
>>> data = zf.read(config_data.conf)
>>> confdata = io.StringIO(str(data))
>>> sysconf = configparser.ConfigParser()
>>> sysconf.read_file(confdata)
>>> sysconf.items("General") #(this is the main section in the file)
Traceback (most recent call last):
File "/usr/local/lib/python3.4/configparser.py", line 824, in items
d.update(self._sections[section])
KeyError: 'General'
During handling of the above exception, another exception occurred:
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/usr/local/lib/python3.4/configparser.py", line 827, in items
raise NoSectionError(section)
configparser.NoSectionError: No section: 'General'
can get the raw text of the file if there's a way to pass that directly to configparser
Try configparser.ConfigParser.read_string
When coupled with an appropriate ZIP file, this code works for me:
import zipfile
import configparser
zf = zipfile.ZipFile("compressed.zip")
zf_config = zf.open("config_data.conf", "rU")
zf_config_data = zf_config.read().decode('ascii')
config = configparser.ConfigParser()
config.read_string(zf_config_data)
assert config['today']['lunch']=='cheeseburger'
Upon reflection, the following might be more appropriate:
import zipfile
import configparser
import io
zf = zipfile.ZipFile("compressed.zip")
zf_config = zf.open("config_data.conf", "rU")
zf_config = io.TextIOWrapper(zf_config)
config = configparser.ConfigParser()
config.read_file(zf_config)
assert config['today']['lunch']=='cheeseburger'
As written in comments, #matthewatabet answer won't work with Python 3.4 (and newer vesions). It's because ZipFile.open now returns a "bytes-like" object and not a "file-like" object anymore. You can use:
codecs.getreader("utf-8")(config_file)
To convert the config_file bytes-like object into a file-like object using the UTF-8 encoding. The code is now:
import zipfile, configparser, codecs
# Python >= 3.4
with zipfile.ZipFile("compressed.zip") as zf:
config_file = zf.open("config_data.conf") # binary mode
sysconfig = configparser.ConfigParser()
sysconfig.read_file(codecs.getreader("utf-8")(config_file))
That seems more satisfactory than creating a string, but I don't know if it's more efficient...
EDIT Since Python 3.9, the zipfile module provides a zipfile.Path.open method that can handle text and binary modes. Default is text mode. The following code works fine:
# Python >= 3.9
with zipfile.ZipFile("compressed.zip") as zf:
zip_path = zipfile.Path(zf)
config_path = zip_path / "config_data.conf"
config_file = config_path.open() # text mode
sysconfig = configparser.ConfigParser()
sysconfig.read_file(config_file)
ZipFile not only supports read but also open, which returns a file-like object. So, you could do something like this:
zf = zipfile.ZipFile("compressed.zip")
config_file = zf.open("config_data.conf")
sysconfig = configparser.ConfigParser()
sysconfig.readfp(config_file)