python-requests

问题 I am using Python requests get method to query the MediaWiki API, but it takes a lot of time to receive the response. The same requests receive the response very fast through a web browser. I have the same issue requesting google.com. Here are the sample codes that I am trying in Python 3.5 on Windows 10: response = requests.get("https://www.google.com") response = requests.get("https://en.wikipedia.org/wiki/Main_Page") response = requests.get("http://en.wikipedia.org/w/api.php?", params={

问题 I have a very simple problem. I'm trying to get the job description from the html of a linkedIn page, but instead of getting the html of the page I'm getting few lines that look like a javascript code instead. I'm very new to this so any help will be greatly appreciated! Thanks Here's my code: import requests url = "https://www.linkedin.com/jobs/view/inside-sales-manager-at-stericycle-1089095836/" page_html = requests.get(url).text print(page_html) When I run this I don't get the html that I

问题 I have a very simple problem. I'm trying to get the job description from the html of a linkedIn page, but instead of getting the html of the page I'm getting few lines that look like a javascript code instead. I'm very new to this so any help will be greatly appreciated! Thanks Here's my code: import requests url = "https://www.linkedin.com/jobs/view/inside-sales-manager-at-stericycle-1089095836/" page_html = requests.get(url).text print(page_html) When I run this I don't get the html that I

问题 music = pygame.mixer.music.load('not.mp3') pygame.mixer.music.play(loops=-1) when executing this i got this error Traceback (most recent call last): File "C:\Users\Admin\AppData\Local\Programs\Python\Python38-32\Escape it!.py", line 15, in <module> music = pygame.mixer.music.load('not.mp3') pygame.error: Failed loading libmpg123.dll: Attempt to access invalid address. i has tried everything from giving full path to only the name but everytime this shows uo 回答1: Restart your IDE if that doesn

问题 I'm currently working on my Python Keitaro Admin API request-library which I called keitaropy. Keitaro Admin API docs Keitaro install on domain and have path to api, I'll use https://example.com/admin_api/v1 in this question. To get a response from the API I need to send 2 types of requests: GET, POST I want to work with Affiliate Networks, Campaigns, Streams, Offers and Traffic Sources. I'll call them targets. Every target has different path in url like: to get all campaigns I need to send

问题 I'm currently working on my Python Keitaro Admin API request-library which I called keitaropy. Keitaro Admin API docs Keitaro install on domain and have path to api, I'll use https://example.com/admin_api/v1 in this question. To get a response from the API I need to send 2 types of requests: GET, POST I want to work with Affiliate Networks, Campaigns, Streams, Offers and Traffic Sources. I'll call them targets. Every target has different path in url like: to get all campaigns I need to send

问题 curl has an option connect-to --connect-to <HOST1:PORT1:HOST2:PORT2> For a request to the given HOST:PORT pair, connect to CONNECT-TO-HOST:CONNECT-TO-PORT instead. This option is suitable to direct requests at a specific server, e.g. at a specific cluster node in a cluster of servers. This option is only used to establish the network connection. It does NOT affect the host- name/port that is used for TLS/SSL (e.g. SNI, certificate verification) or for the application protocols. "host" and

问题 Some case study here. I am trying to play with PIL library in Google Colab, and can't get ImageFont to read my originally zipped file. The code: import requests, zipfile, io r3 = requests.get('https://sources.archlinux.org/other/community/ttf-roboto/ttf-roboto-hinted-2.138.zip') z3 = zipfile.ZipFile(io.BytesIO(r3.content)) z3.extractall() So far so good, and if I browse my directory with ls, it shows me the elements: ls Shows: LICENSE RobotoCondensed-Regular.ttf __MACOSX/ Roboto-Italic.ttf

问题 Some case study here. I am trying to play with PIL library in Google Colab, and can't get ImageFont to read my originally zipped file. The code: import requests, zipfile, io r3 = requests.get('https://sources.archlinux.org/other/community/ttf-roboto/ttf-roboto-hinted-2.138.zip') z3 = zipfile.ZipFile(io.BytesIO(r3.content)) z3.extractall() So far so good, and if I browse my directory with ls, it shows me the elements: ls Shows: LICENSE RobotoCondensed-Regular.ttf __MACOSX/ Roboto-Italic.ttf

问题 Some case study here. I am trying to play with PIL library in Google Colab, and can't get ImageFont to read my originally zipped file. The code: import requests, zipfile, io r3 = requests.get('https://sources.archlinux.org/other/community/ttf-roboto/ttf-roboto-hinted-2.138.zip') z3 = zipfile.ZipFile(io.BytesIO(r3.content)) z3.extractall() So far so good, and if I browse my directory with ls, it shows me the elements: ls Shows: LICENSE RobotoCondensed-Regular.ttf __MACOSX/ Roboto-Italic.ttf