IOError: [Errno 22] Invalid argument with url
0
I am newbie in python and i am trying to open an localhost with write mode
but i got following errors
with io.open('http:\localhost:3000\assets\i18n\locale-ru.json', 'w') as outfile:
IOError: [Errno 22] Invalid argument: 'http:\localhost:3000\assets\i18n\locale-ru.json'
Here is a piece of code:
with io.open('http://localhost:3000//assets//i18n//locale-ru.json', 'w') as outfile:
str_ = json.dumps(data_ru,
indent=4, sort_keys=True,
separators=(',', ': '), ensure_ascii=False)
outfile.write(to_unicode(str_))
python flask
edited Nov 12 at 20:04
vencaslac
1,002217
asked Nov 12 at 19:40
user987
12
add a comment |
0
I am newbie in python and i am trying to open an localhost with write mode
but i got following errors
with io.open('http:\localhost:3000\assets\i18n\locale-ru.json', 'w') as outfile:
IOError: [Errno 22] Invalid argument: 'http:\localhost:3000\assets\i18n\locale-ru.json'
Here is a piece of code:
with io.open('http://localhost:3000//assets//i18n//locale-ru.json', 'w') as outfile:
str_ = json.dumps(data_ru,
indent=4, sort_keys=True,
separators=(',', ': '), ensure_ascii=False)
outfile.write(to_unicode(str_))
python flask
edited Nov 12 at 20:04
vencaslac
1,002217
asked Nov 12 at 19:40
user987
12
1
Just a note, you have only one slash after `http:`.
– vishes_shell
Nov 12 at 19:42
with io.open('http:/localhost:3000//assets//i18n//locale-ru.json', 'w') as outfile: IOError: [Errno 22] Invalid argument: 'http:/localhost:3000//assets//i18n//locale-ru.json'
– user987
Nov 12 at 19:46
still got error
– user987
Nov 12 at 19:47
add a comment |
0
0
0
I am newbie in python and i am trying to open an localhost with write mode
but i got following errors
with io.open('http:\localhost:3000\assets\i18n\locale-ru.json', 'w') as outfile:
IOError: [Errno 22] Invalid argument: 'http:\localhost:3000\assets\i18n\locale-ru.json'
Here is a piece of code:
with io.open('http://localhost:3000//assets//i18n//locale-ru.json', 'w') as outfile:
str_ = json.dumps(data_ru,
indent=4, sort_keys=True,
separators=(',', ': '), ensure_ascii=False)
outfile.write(to_unicode(str_))
python flask
edited Nov 12 at 20:04
vencaslac
1,002217
asked Nov 12 at 19:40
user987
12
I am newbie in python and i am trying to open an localhost with write mode
but i got following errors
with io.open('http:\localhost:3000\assets\i18n\locale-ru.json', 'w') as outfile:
IOError: [Errno 22] Invalid argument: 'http:\localhost:3000\assets\i18n\locale-ru.json'
Here is a piece of code:
with io.open('http://localhost:3000//assets//i18n//locale-ru.json', 'w') as outfile:
str_ = json.dumps(data_ru,
indent=4, sort_keys=True,
separators=(',', ': '), ensure_ascii=False)
outfile.write(to_unicode(str_))
python flask
python flask
edited Nov 12 at 20:04
vencaslac
1,002217
asked Nov 12 at 19:40
user987
12
edited Nov 12 at 20:04
vencaslac
1,002217
asked Nov 12 at 19:40
user987
12
edited Nov 12 at 20:04
vencaslac
1,002217
edited Nov 12 at 20:04
vencaslac
1,002217
edited Nov 12 at 20:04
vencaslac
1,002217
1,002217
asked Nov 12 at 19:40
user987
12
asked Nov 12 at 19:40
user987
12
asked Nov 12 at 19:40
user987
12
12
1
Just a note, you have only one slash after `http:`.
– vishes_shell
Nov 12 at 19:42
with io.open('http:/localhost:3000//assets//i18n//locale-ru.json', 'w') as outfile: IOError: [Errno 22] Invalid argument: 'http:/localhost:3000//assets//i18n//locale-ru.json'
– user987
Nov 12 at 19:46
still got error
– user987
Nov 12 at 19:47
add a comment |
1
Just a note, you have only one slash after `http:`.
– vishes_shell
Nov 12 at 19:42
with io.open('http:/localhost:3000//assets//i18n//locale-ru.json', 'w') as outfile: IOError: [Errno 22] Invalid argument: 'http:/localhost:3000//assets//i18n//locale-ru.json'
– user987
Nov 12 at 19:46
still got error
– user987
Nov 12 at 19:47
1
1
Just a note, you have only one slash after `http:`.
– vishes_shell
Nov 12 at 19:42
Just a note, you have only one slash after `http:`.
– vishes_shell
Nov 12 at 19:42
with io.open('http:/localhost:3000//assets//i18n//locale-ru.json', 'w') as outfile: IOError: [Errno 22] Invalid argument: 'http:/localhost:3000//assets//i18n//locale-ru.json'
– user987
Nov 12 at 19:46
with io.open('http:/localhost:3000//assets//i18n//locale-ru.json', 'w') as outfile: IOError: [Errno 22] Invalid argument: 'http:/localhost:3000//assets//i18n//locale-ru.json'
– user987
Nov 12 at 19:46
still got error
– user987
Nov 12 at 19:47
still got error
– user987
Nov 12 at 19:47
add a comment |
1 Answer
1
active
oldest
votes
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According to the doc, the io.open function only reads a local file given a file path. Considering you're trying to read from a http url, I guess a better tools might be requests
answered Nov 12 at 19:44
Old Panda
574621
getting struggle with reguests documentation, i would really appreciate if you send me with source code
– user987
Nov 12 at 20:10
what about urllib.urlopen in Python 2.7
– user987
Nov 13 at 11:24
` with urllib.urlopen('localhost:3000//assets/i18n//locale-kg.json', 'w') as outfile: str_ = json.dumps(data_kg, indent=4, sort_keys=True, separators=(',', ': '), ensure_ascii=False) outfile.write(to_unicode(str_))`
– user987
Nov 13 at 11:25
As far as I know, to update a file via HTTP you have to send a POST or a PUT request with a file on your local, just like you upload a file to some website. So if I wanted to update a file via an URL, I would firstly download it to my local, then do changes on it, and finally upload the file back. As for file uploading, you could refer stackoverflow.com/a/26791188/2191173 for details.
– Old Panda
Nov 13 at 18:26
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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oldest
votes
active
oldest
votes
0
According to the doc, the io.open function only reads a local file given a file path. Considering you're trying to read from a http url, I guess a better tools might be requests
answered Nov 12 at 19:44
Old Panda
574621
getting struggle with reguests documentation, i would really appreciate if you send me with source code
– user987
Nov 12 at 20:10
what about urllib.urlopen in Python 2.7
– user987
Nov 13 at 11:24
` with urllib.urlopen('localhost:3000//assets/i18n//locale-kg.json', 'w') as outfile: str_ = json.dumps(data_kg, indent=4, sort_keys=True, separators=(',', ': '), ensure_ascii=False) outfile.write(to_unicode(str_))`
– user987
Nov 13 at 11:25
As far as I know, to update a file via HTTP you have to send a POST or a PUT request with a file on your local, just like you upload a file to some website. So if I wanted to update a file via an URL, I would firstly download it to my local, then do changes on it, and finally upload the file back. As for file uploading, you could refer stackoverflow.com/a/26791188/2191173 for details.
– Old Panda
Nov 13 at 18:26
add a comment |
0
According to the doc, the io.open function only reads a local file given a file path. Considering you're trying to read from a http url, I guess a better tools might be requests
answered Nov 12 at 19:44
Old Panda
574621
getting struggle with reguests documentation, i would really appreciate if you send me with source code
– user987
Nov 12 at 20:10
what about urllib.urlopen in Python 2.7
– user987
Nov 13 at 11:24
` with urllib.urlopen('localhost:3000//assets/i18n//locale-kg.json', 'w') as outfile: str_ = json.dumps(data_kg, indent=4, sort_keys=True, separators=(',', ': '), ensure_ascii=False) outfile.write(to_unicode(str_))`
– user987
Nov 13 at 11:25
As far as I know, to update a file via HTTP you have to send a POST or a PUT request with a file on your local, just like you upload a file to some website. So if I wanted to update a file via an URL, I would firstly download it to my local, then do changes on it, and finally upload the file back. As for file uploading, you could refer stackoverflow.com/a/26791188/2191173 for details.
– Old Panda
Nov 13 at 18:26
add a comment |
0
0
0
According to the doc, the io.open function only reads a local file given a file path. Considering you're trying to read from a http url, I guess a better tools might be requests
answered Nov 12 at 19:44
Old Panda
574621
According to the doc, the io.open function only reads a local file given a file path. Considering you're trying to read from a http url, I guess a better tools might be requests
answered Nov 12 at 19:44
Old Panda
574621
answered Nov 12 at 19:44
Old Panda
574621
answered Nov 12 at 19:44
Old Panda
574621
answered Nov 12 at 19:44
Old Panda
574621
574621
getting struggle with reguests documentation, i would really appreciate if you send me with source code
– user987
Nov 12 at 20:10
what about urllib.urlopen in Python 2.7
– user987
Nov 13 at 11:24
` with urllib.urlopen('localhost:3000//assets/i18n//locale-kg.json', 'w') as outfile: str_ = json.dumps(data_kg, indent=4, sort_keys=True, separators=(',', ': '), ensure_ascii=False) outfile.write(to_unicode(str_))`
– user987
Nov 13 at 11:25
As far as I know, to update a file via HTTP you have to send a POST or a PUT request with a file on your local, just like you upload a file to some website. So if I wanted to update a file via an URL, I would firstly download it to my local, then do changes on it, and finally upload the file back. As for file uploading, you could refer stackoverflow.com/a/26791188/2191173 for details.
– Old Panda
Nov 13 at 18:26
add a comment |
getting struggle with reguests documentation, i would really appreciate if you send me with source code
– user987
Nov 12 at 20:10
what about urllib.urlopen in Python 2.7
– user987
Nov 13 at 11:24
` with urllib.urlopen('localhost:3000//assets/i18n//locale-kg.json', 'w') as outfile: str_ = json.dumps(data_kg, indent=4, sort_keys=True, separators=(',', ': '), ensure_ascii=False) outfile.write(to_unicode(str_))`
– user987
Nov 13 at 11:25
As far as I know, to update a file via HTTP you have to send a POST or a PUT request with a file on your local, just like you upload a file to some website. So if I wanted to update a file via an URL, I would firstly download it to my local, then do changes on it, and finally upload the file back. As for file uploading, you could refer stackoverflow.com/a/26791188/2191173 for details.
– Old Panda
Nov 13 at 18:26
getting struggle with reguests documentation, i would really appreciate if you send me with source code
– user987
Nov 12 at 20:10
getting struggle with reguests documentation, i would really appreciate if you send me with source code
– user987
Nov 12 at 20:10
what about urllib.urlopen in Python 2.7
– user987
Nov 13 at 11:24
what about urllib.urlopen in Python 2.7
– user987
Nov 13 at 11:24
` with urllib.urlopen('localhost:3000//assets/i18n//locale-kg.json', 'w') as outfile: str_ = json.dumps(data_kg, indent=4, sort_keys=True, separators=(',', ': '), ensure_ascii=False) outfile.write(to_unicode(str_))`
– user987
Nov 13 at 11:25
` with urllib.urlopen('localhost:3000//assets/i18n//locale-kg.json', 'w') as outfile: str_ = json.dumps(data_kg, indent=4, sort_keys=True, separators=(',', ': '), ensure_ascii=False) outfile.write(to_unicode(str_))`
– user987
Nov 13 at 11:25
As far as I know, to update a file via HTTP you have to send a POST or a PUT request with a file on your local, just like you upload a file to some website. So if I wanted to update a file via an URL, I would firstly download it to my local, then do changes on it, and finally upload the file back. As for file uploading, you could refer stackoverflow.com/a/26791188/2191173 for details.
– Old Panda
Nov 13 at 18:26
As far as I know, to update a file via HTTP you have to send a POST or a PUT request with a file on your local, just like you upload a file to some website. So if I wanted to update a file via an URL, I would firstly download it to my local, then do changes on it, and finally upload the file back. As for file uploading, you could refer stackoverflow.com/a/26791188/2191173 for details.
– Old Panda
Nov 13 at 18:26
add a comment |
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1
Just a note, you have only one slash after `http:`.
– vishes_shell
Nov 12 at 19:42
with io.open('http:/localhost:3000//assets//i18n//locale-ru.json', 'w') as outfile: IOError: [Errno 22] Invalid argument: 'http:/localhost:3000//assets//i18n//locale-ru.json'
– user987
Nov 12 at 19:46
still got error
– user987
Nov 12 at 19:47