I want to write some data, e.g. "hello" on a file that is located on a remote server, not a local server. This is the code that I used to read from server:
import urllib2
var = "hello"
url = "http://url:port/log/log.txt"
response = urllib2.urlopen(url)
txt = response.read();
print txt
As an output I was able to get the data from the log file.
Now I want to write some data, e.g. "hello" onto the same file. How to achieve that?
What your code is doing is actually not "reading a file" but sending an HTTP get request to a given url and print the HTTP response's body. What you get as content for this url is up to the HTTP server serving this url, and the fact it actually comes from a file stored on this server is totally irrelevant - it might as well come from a database, from another web server, or be typed in real time by a monkey FWIW.
If you want to use the HTTP protocol to modify the content of a given file on this server, the server must provide this service (as a given url where you're supposed to send a POST or PUT http request with the new content).
Related
guys I want to create a python API and request msg and get a response with JSON and send it to knime workflow and work with it
this is my script in python the fact is that he give me json.decoder.jsondecodeError
from importlib.metadata import files
import requests
url ='https://api.edination.com/v2/edifact/read'
headers = {'Ocp-Apim-Subscription-Key': '3ecf6b1c5cf34bd797a5f4c57951a1cf'}
files = {'file':open('C:\\Users\\hcharafeddine\\Desktop\\EDI\\Interchange_1654767219416.edi','rb')}
r = requests.post(url,files=files)
r.json()
We'll need more info to help further. I understand if you don't want to share the content of the EDI message, so here are a few things to try:
The EDINation website allows you to paste an EDI message in and it'll show you the JSON output that the API will return
It also has a sample EDIFACT document you can select and then save locally to run through your python script - you can then share the results here
You can also use HTTPToolkit to inspect the API request and response to troubleshoot further.
There's a website that has a button which downloads an Excel file. After I click, it takes around 20 seconds for the server API to generate the file and send it back to my browser for download.
If I monitor the communication after I click the button, I can see how the browser sends a POST request to a server with a series of headers and form values.
Is there a way that I can simulate a similar POST request programmatically using Python, and retrieve the Excel file after the server sends it over?
Thank you in advance
The requests module is used for sending all kinds of request types.
requests.post sends the post requests synchronously.
The payload data can be set using data=
The response can be accessed using .content.
Be sure to check the .status_code and only save on a successful response code
Also note the use of "wb" inside open, because we want to save the file as a binary instead of text.
Example:
import requests
payload = {"dao":"SampleDAO",
"condigId": 1,
...}
r = requests.post("http://url.com/api", data=payload)
if r.status_code == 200:
with open("file.save","wb") as f:
f.write(r.content)
Requests Documentation
I guess You could similarly do this:
file_info = request.get(url)
with open('file_name.extension', 'wb') as file:
file.write(file_info.content)
I honestly do not know how to explain this tho since I have little understanding how it works
I am trying to update an already saved form on a system using HTTP requests. Due to the server configuration for the third party app we use, updating by POST requires sending a fully filled out payload every single time.
I want to get round this by recovering the form data already present on the server and converting it into a dictionary. Then changing any values I need and reposting to make changes sever side.
The application we use sends a POST request when the save button is clicked for a particular form.
Here I send a post request with no payload.
[This simulates pressing the save button and is also the point where dev tools shows me a the payload I want to capture]
post_test = self.session.post(url_to_retrieve_from)
I thought that now I should be able to print the output, which should resemble what Google Dev tools Form data captures.
print(post_test.text)
This just gives me html found on the webpage.
If Dev Tools can get this from the server then I should also be able to?
Example of Data I am trying to get via requests:
Form Data
If Dev Tools can get this from the server then I should also be able to?
Yes, of course. In requests you pass form data in data keyword:
import requests
url = 'http://www.example.com'
data = {
'name': 'value',
}
response = requests.post(url, data=data)
You can get the data you sent with a request from the response in this way:
import requests
response = requests.post('http://your_url', data=data) # send request
body = response.request.body
parsed_data = dict(data.split('=') for data in body.split('&')) # parse request body
Here you can find more information about data argument
In the documentation, in the class requests.Response we can find the attribute:
request = None
The PreparedRequest object to which this is a response.
In requests.PreparedRequest class we can read:
body = None
request body to send to the server.
I have a public folder in Google Drive, in which I store pictures.
In Python, I am trying to detect if a picture with a particular name exist or not. I am using this code:
import urllib2
url = "http://googledrive.com/host/0B7K23HtYjKyBfnhYbkVyUld3YUVqSWgzWm1uMXdrMzQ0NlEwOXVUd3o0MWVYQ1ZVMlFSNms/0000.png"
resp = urllib2.urlopen(url)
print resp.getcode()
And even though there is no file with this name in this folder, this code is not throwing an exception and is printing "200" as the return code. I have checked in my browser and this URL (http://googledrive.com/host/0B7K23HtYjKyBfnhYbkVyUld3YUVqSWgzWm1uMXdrMzQ0NlEwOXVUd3o0MWVYQ1ZVMlFSNms/0000.png) does return a 404, after a few redirects.
Why doesn't urllib2 detect that this file actually doesn't exist?
When you make the request, your request goes to google's web servers and is processed there. If and only if google's servers were to return a 404, would you see a 404 on your end; urllub2 simply encapsulates the underlying handshaking and data transfer logic.
In this particular case, google's server side code requires the request to be authenticated, and your request url is simply unauthenticated. As such, the request is redirected to the login page, and since this is a valid existing page/response, urllib2 shows the correct code 200. You can get the same page if you open the link in a private window.
However, if you are authenticated and then open the url (basically logged into your gmail/googgle docs account), you would get the 404 error.
I am new to Python and it is first post. Want to upload a TXT file to server, (as of now, it is local host).Every time, I run the script, the local file uploaded and updated on server. I am using Requests module
import requests, time
url ='http://192.168.49.205/test/database/data.txt' # where i want to write
files = {'file':('data.txt','C:\Python27\data.txt','rb')}
#r = requests.post(url,files=files) # this works too
r= requests.post('http://192.168.49.205/test/database/data.txt',
data={'upload_type': 'standard', 'upload_to': '0'},files=files)
print r.status_code
print r.text
data.txt is not updated. Just seeing old data (I put some values when file created).I am not using any Forms in PHP or HTML. Is it possible to upload with method?
Believe,I got some clarity after posting the question. Now, there is PHP file on server side listening to client. Here is the "post.php". So, this will replace the text file on Client side. PHP file get the Name, Task, Value from client and post to "a.txt" on server (local)
<?php
if(isset($_GET["Name"])){
$name=$_GET["Name"];
}
if(isset($_GET["Task"])){
$task=$_GET["Task"];
}
if(isset($_GET["Value"])){
$value=$_GET["Value"];
}
$f=fopen("a.txt","w") or exit("Unable to open file!");
fwrite($f,$name);
fwrite($f," ");
fwrite($f,$task);
fwrite($f," ");
fwrite($f,$value);
fclose($f);
?>`
So now Requests look like this
import requests, time
url = 'http://192.168.49.205/test/test.php'
post_data = {'Name':'job','Task':'008','Value':'8'}
r= requests.post('http://192.168.49.205/test/post.php', data= post_data)
print r.status_code
print r.text
Stil the values is not reaching "a.txt". What i am missing? Please advise!
You're not reading the files content in your code. The requests documentation states:
files – (optional) Dictionary of ‘name’: file-like-objects (or
{‘name’: (‘filename’, fileobj)}) for multipart encoding upload.
Your code suggests you want to use the second option. However:
files = {'file':('data.txt','C:\Python27\data.txt','rb')}
You're not creating the structure as the documentation indicates - you're passing a tuple of 3 strings instead of a tuple of (string, file_obj). You probably wanted to do this:
files = {'file':('data.txt',open('C:\Python27\data.txt','rb'))}