Open 0khasawneh opened 1 year ago
same problem :'(
Same problem also
the same problem
Same problem
did anyone get this after clearing there settings.json?
yeah i did clear but still i getting this error, can anyone please help me clear the error
yeah i did clear but still i getting this error, can anyone please help me clear the error
We are also unable fix the error
when you login in gave your id and pass than try the same id as target. then it will work. so we need the password of the target to use this tool. if you found any other way to solve this let me know please.
when you login in gave your id and pass than try the same id as target. then it will work. so we need the password of the target to use this tool. if you found any other way to solve this let me know please.
yse me too
what is the problem
Whay
I found the solution,
First, the error: the http response is returning an html output. We are expecting a json format, thus the parsing error in the first block.
Why is this returning an http response? Looking into the instagram-private-api library, we see that there is an authentication issue.
Solution Flow:
FOLLOW ME ON YT!
If we know the password of the target than why we need to use this one.
thanks it worked!
Attempt to login... Error parsing error response: Expecting value: line 1 column 1 (char 0) Traceback (most recent call last): File "/usr/local/lib/python3.11/dist-packages/instagram_private_api/client.py", line 523, in _call_api response = self.opener.open(req, timeout=self.timeout) ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ File "/usr/lib/python3.11/urllib/request.py", line 525, in open response = meth(req, response) ^^^^^^^^^^^^^^^^^^^ File "/usr/lib/python3.11/urllib/request.py", line 634, in http_response response = self.parent.error( ^^^^^^^^^^^^^^^^^^ File "/usr/lib/python3.11/urllib/request.py", line 563, in error return self._call_chain(args) ^^^^^^^^^^^^^^^^^^^^^^^ File "/usr/lib/python3.11/urllib/request.py", line 496, in _call_chain result = func(args) ^^^^^^^^^^^ File "/usr/lib/python3.11/urllib/request.py", line 643, in http_error_default raise HTTPError(req.full_url, code, msg, hdrs, fp) urllib.error.HTTPError: HTTP Error 404: Not Found
During handling of the above exception, another exception occurred:
Traceback (most recent call last):
File "/home/monu9012/Desktop/Osintgram/main.py", line 121, in
when you login in gave your id and pass than try the same id as target. then it will work. so we need the password of the target to use this tool. if you found any other way to solve this let me know please.
yes to osint a acc you need to have hes password
I found the solution,
First, the error: the http response is returning an html output. We are expecting a json format, thus the parsing error in the first block.
Why is this returning an http response? Looking into the instagram-private-api library, we see that there is an authentication issue.
Solution Flow:
1. Use a username and password and then use that username as the victim, (this worked) 2. Since the above worked, the program itself is not the issue. 3. The issue lies in the authentication of your account with instagram 4. One thought, since I just made this (BOT) account, it could be that there is an account life time that needs to true in order for the apis to work. 5. I used an older bot account and was able to get it working! 6. Use an older IG account or get a developer account.
FOLLOW ME ON YT! does it really work
when you login in gave your id and pass than try the same id as target. then it will work. so we need the password of the target to use this tool. if you found any other way to solve this let me know please.
same here plz help
i use my offical account but it didnt work. it only work on same accont as in settings but when i try to implement it in other account it gives me this error
Error parsing error response: Expecting value: line 1 column 1 (char 0) Traceback (most recent call last): File "/usr/local/lib/python3.11/dist-packages/instagram_private_api/client.py", line 523, in _call_api response = self.opener.open(req, timeout=self.timeout) ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ File "/usr/lib/python3.11/urllib/request.py", line 525, in open response = meth(req, response) ^^^^^^^^^^^^^^^^^^^ File "/usr/lib/python3.11/urllib/request.py", line 634, in http_response response = self.parent.error( ^^^^^^^^^^^^^^^^^^ File "/usr/lib/python3.11/urllib/request.py", line 563, in error return self._call_chain(args) ^^^^^^^^^^^^^^^^^^^^^^^ File "/usr/lib/python3.11/urllib/request.py", line 496, in _call_chain result = func(args) ^^^^^^^^^^^ File "/usr/lib/python3.11/urllib/request.py", line 643, in http_error_default raise HTTPError(req.full_url, code, msg, hdrs, fp) urllib.error.HTTPError: HTTP Error 404: Not Found
During handling of the above exception, another exception occurred:
Traceback (most recent call last):
File "/home/fluffy/Osintgram/main.py", line 186, in
I found the solution,
First, the error: the http response is returning an html output. We are expecting a json format, thus the parsing error in the first block.
Why is this returning an http response? Looking into the instagram-private-api library, we see that there is an authentication issue.
Solution Flow:
- Use a username and password and then use that username as the victim, (this worked)
- Since the above worked, the program itself is not the issue.
- The issue lies in the authentication of your account with instagram
- One thought, since I just made this (BOT) account, it could be that there is an account life time that needs to true in order for the apis to work.
- I used an older bot account and was able to get it working!
- Use an older IG account or get a developer account.
FOLLOW ME ON YT!
it didn't work with me
i use my offical account but it didnt work. it only work on same accont as in settings but when i try to implement it in other account it gives me this error
Error parsing error response: Expecting value: line 1 column 1 (char 0) Traceback (most recent call last): File "/usr/local/lib/python3.11/dist-packages/instagram_private_api/client.py", line 523, in _call_api response = self.opener.open(req, timeout=self.timeout) ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ File "/usr/lib/python3.11/urllib/request.py", line 525, in open response = meth(req, response) ^^^^^^^^^^^^^^^^^^^ File "/usr/lib/python3.11/urllib/request.py", line 634, in http_response response = self.parent.error( ^^^^^^^^^^^^^^^^^^ File "/usr/lib/python3.11/urllib/request.py", line 563, in error return self._call_chain(args) ^^^^^^^^^^^^^^^^^^^^^^^ File "/usr/lib/python3.11/urllib/request.py", line 496, in _call_chain result = func(args) ^^^^^^^^^^^ File "/usr/lib/python3.11/urllib/request.py", line 643, in http_error_default raise HTTPError(req.full_url, code, msg, hdrs, fp) urllib.error.HTTPError: HTTP Error 404: Not Found
During handling of the above exception, another exception occurred:
Traceback (most recent call last): File "/home/fluffy/Osintgram/main.py", line 186, in _cmd() File "/home/fluffy/Osintgram/src/Osintgram.py", line 110, in change_target self.setTarget(line) File "/home/fluffy/Osintgram/src/Osintgram.py", line 61, in setTarget self.following = self.check_following() ^^^^^^^^^^^^^^^^^^^^^^ File "/home/fluffy/Osintgram/src/Osintgram.py", line 1159, in check_following return self.api._call_api(endpoint)['user_detail']['user']['friendship_status']['following'] ^^^^^^^^^^^^^^^^^^^^^^^^^^^^ File "/usr/local/lib/python3.11/dist-packages/instagram_private_api/client.py", line 527, in _call_api ErrorHandler.process(e, error_response) File "/usr/local/lib/python3.11/dist-packages/instagram_private_api/errors.py", line 135, in process raise ClientError(error_msg, http_error.code, error_response) instagram_private_api.errors.ClientError: Not Found
same
my brothers the problem of osintgram i can resolve the problem
my brothers the problem of osintgram i can resolve the problem
How ?
CAN ANYONE SOLVE THIS PLEASE
A porcaria do instagram mudou a API.... O retorno da chamada de API específico que, o Instagram o descontinuou, fazendo com que o programa não funcionasse totalmente.
O que você realmente precisa fazer é:
1 - vai até o diretorio: a src 2 - procure o arquivo Osintgram.py 3 - na linha onde está escrito self.following = self.check_following() na função setTarget(self, target) 4 - comente essa linha, assim: #self.following = self.check_following(). 5 - Prontinho sejam felizes.... Isso deve funcionar.
A porcaria do instagram mudou a API.... O retorno da chamada de API específico que, o Instagram o descontinuou, fazendo com que o programa não funcionasse totalmente.
O que você realmente precisa fazer é:
1 - vai até o diretorio: a src 2 - procure o arquivo Osintgram.py 3 - na linha onde está escrito self.following = self.check_following() na função setTarget(self, target) 4 - comente essa linha, assim: #self.following = self.check_following(). 5 - Prontinho sejam felizes.... Isso deve funcionar.
Thanks man it worked
Logged in, but still dodnt work ...
_ \ ___|| _/ | ____ ____
/ | \ / _/ |/ \ _\/ _ _ \ / \
/ | \ | | | \ | / // > | \// | Y Y \
____ /__ >|_| /_| \ /|| (__ /|_| /
\/ \/ \/ /___/ \/ \/
Version 1.1 - Developed by Giuseppe Criscione
Type 'list' to show all allowed commands
Type 'FILE=y' to save results to files like '
Type 'FILE=n' to disable saving to files'
Type 'JSON=y' to export results to a JSON files like '
Type 'JSON=n' to disable exporting to files'
Run a command: likes
Impossible to execute command: user has private profile
Do you want send a follow request? [Y/N]:
Instagram has modified its API, leading to the discontinuation of a specific API callback. As a result, the program is experiencing functionality issues.
To address this, follow these steps:
self.following = self.check_following()
.#self.following = self.check_following()
.Save the changes.
This adjustment should resolve the issue.
Omong kosong Instagram mengubah API.... Callback API spesifik itu, Instagram menghentikannya, menyebabkan program tidak bekerja sepenuhnya.
Yang benar-benar perlu Anda lakukan adalah:
1 - buka direktori: src 2 - cari file Osintgram.py 3 - pada baris yang bertuliskan self.following = self.check_following() di fungsi setTarget(self, target) 4 - beri komentar pada baris ini, seperti ini: #self.following = self.check_following(). 5 - Siap, berbahagialah.... Ini seharusnya berhasil.
this work
Logged in, but still dodnt work ...
\ **|| / | ____ __ / | \ / / |/ \ / __ \ / \ / | \ | | | \ | / // > | // | Y Y \ _ /__ >|**_| /| _ /|**| (__ /**|| / / / / /____/ / /
Version 1.1 - Developed by Giuseppe Criscione
Type 'list' to show all allowed commands Type 'FILE=y' to save results to files like '.txt (default is disabled)' Type 'FILE=n' to disable saving to files' Type 'JSON=y' to export results to a JSON files like '.json (default is disabled)' Type 'JSON=n' to disable exporting to files' Run a command: likes Impossible to execute command: user has private profile Do you want send a follow request? [Y/N]:
FAQ
Can I access the contents of a private profile? No, you cannot get information on private profiles. You can only get information from a public profile or a profile you follow. The tools that claim to be successful are scams!
What is and how I can bypass the challenge_required error? The challenge_required error means that Instagram notice a suspicious behavior on your profile, so needs to check if you are a real person or a bot. To avoid this you should follow the suggested link and complete the required operation (insert a code, confirm email, etc)
In v2 repo, this self.following = self.check_following()
is not existing anymore. But still same error.
Attempt to login... Error parsing error response: Expecting value: line 1 column 1 (char 0) Traceback (most recent call last): File "/usr/local/lib/python3.11/dist-packages/instagram_private_api/client.py", line 523, in _call_api response = self.opener.open(req, timeout=self.timeout) ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ File "/usr/lib/python3.11/urllib/request.py", line 525, in open response = meth(req, response) ^^^^^^^^^^^^^^^^^^^ File "/usr/lib/python3.11/urllib/request.py", line 634, in http_response response = self.parent.error( ^^^^^^^^^^^^^^^^^^ File "/usr/lib/python3.11/urllib/request.py", line 563, in error return self._call_chain(args) ^^^^^^^^^^^^^^^^^^^^^^^ File "/usr/lib/python3.11/urllib/request.py", line 496, in _call_chain result = func(args) ^^^^^^^^^^^ File "/usr/lib/python3.11/urllib/request.py", line 643, in http_error_default raise HTTPError(req.full_url, code, msg, hdrs, fp) urllib.error.HTTPError: HTTP Error 404: Not Found
During handling of the above exception, another exception occurred:
Traceback (most recent call last): File "/home/kali/inst/Osintgram/main.py", line 121, in
api = Osintgram(args.id, args.file, args.json, args.command, args.output, args.cookies)
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
File "/home/kali/inst/Osintgram/src/Osintgram.py", line 48, in init
self.setTarget(target)
File "/home/kali/inst/Osintgram/src/Osintgram.py", line 61, in setTarget
self.following = self.check_following()
^^^^^^^^^^^^^^^^^^^^^^
File "/home/kali/inst/Osintgram/src/Osintgram.py", line 1159, in check_following
return self.api._call_api(endpoint)['user_detail']['user']['friendship_status']['following']
^^^^^^^^^^^^^^^^^^^^^^^^^^^^
File "/usr/local/lib/python3.11/dist-packages/instagram_private_api/client.py", line 527, in _call_api
ErrorHandler.process(e, error_response)
File "/usr/local/lib/python3.11/dist-packages/instagram_private_api/errors.py", line 135, in process
raise ClientError(error_msg, http_error.code, error_response)
instagram_private_api.errors.ClientError: Not Found