I have a package which contains static files I want to reuse among applications. Based on https://webassets.readthedocs.io/en/latest/environment.html#webassets.env.Environment.load_path I came up with the following code snippet, to be used in each application's __init__.py (the shared package is loutilities):
with app.app_context():
# js/css files
asset_env.append_path(app.static_folder)
# os.path.split to get package directory
asset_env.append_path(os.path.join(os.path.split(loutilities.__file__)[0], 'tables-assets', 'static'))
but when ASSETS_DEBUG = False, this causes a ValueError exception for one of the files found in the package. (See https://github.com/louking/rrwebapp/issues/366 for detailed traceback -- this is possibly related to https://github.com/miracle2k/webassets/issues/387).
ValueError: Cannot determine url for /var/www/sandbox.scoretility.com/rrwebapp/lib/python2.7/site-packages/loutilities/tables-assets/static/branding.css
Changed code to use a url parameter which now works fine for ASSETS_DEBUG = False
asset_env.append_path(os.path.join(os.path.split(loutilities.__file__)[0], 'tables-assets', 'static'), '/loutilities')
however now when ASSETS_DEBUG = True, I see that the file failed to load in the javascript console
Failed to load resource: the server responded with a status of 404 (NOT FOUND) branding.css
Have worked around the Catch-22 using the inelegant code as follows, but wondering how to choose the append_path() url parameter which will work for both ASSETS_DEBUG = True or False.
with app.app_context():
# js/css files
asset_env.append_path(app.static_folder)
# os.path.split to get package directory
loutilitiespath = os.path.split(loutilities.__file__)[0]
# kludge: seems like assets debug doesn't like url and no debug insists on it
if app.config['ASSETS_DEBUG']:
url = None
else:
url = '/loutilities'
asset_env.append_path(os.path.join(loutilitiespath, 'tables-assets', 'static'), url)
One solution is to create a route for /loutilities/static, thus
# add loutilities tables-assets for js/css/template loading
# see https://adambard.com/blog/fresh-flask-setup/
# and https://webassets.readthedocs.io/en/latest/environment.html#webassets.env.Environment.load_path
# loutilities.__file__ is __init__.py file inside loutilities; os.path.split gets package directory
loutilitiespath = os.path.join(os.path.split(loutilities.__file__)[0], 'tables-assets', 'static')
#app.route('/loutilities/static/<path:filename>')
def loutilities_static(filename):
return send_from_directory(loutilitiespath, filename)
with app.app_context():
# js/css files
asset_env.append_path(app.static_folder)
asset_env.append_path(loutilitiespath, '/loutilities/static')
Related
I have a web.py server hosted on pythonanywhere.com doing some handy things with python.
Now I'd like to just serve a straightforward html file from the same server i.e. just return the contents of a static html file to the client
The comments/answers below state that it should be possible, out of the box, to serve static files in the static directory, located in the same directory as the main python file which contains the following :
import web
urls = (
'/', 'hello'
)
app = web.application(urls, globals())
class hello:
def GET(self):
return 'Hello, Joe'
if __name__ == "__main__":
app.run()
The server above works fine, when I go to http://myhost/ it displays "Hello , Joe".
The directory static exists and contains a file small.jpg but when I try the url http://myhost/static/small.jpg it gives me "not found"
Previous text of question up to Nov 9th 2022 is below :
original question title : Trying to return a html file in web.py but getting "No template named ....." error message
So I've looked at the web.py documentation on serving static files and templating and I think the following code should work :
import web
render = web.template.render('static/')
# have also tried render = web.template.render('/full/path/to/static/')
urls = (
'/getlatlongEIRCODE', 'getlatlongEIRCODE', #other stuff
'/getlatlongGOOGLE', 'getlatlongGOOGLE', #other stuff
'/getmonthlyPV', 'getmonthlyPV', #other stuff
'/Tomas', 'Tomas',
)
class Tomas:
def GET(self):
return render.Tomas()
I have created a folder static at the same level as my file above (which works fine for the other scripts) and i have created a file Tomas.html in the static folder containing
<h1>Help me</h1>
However I get an error message when I go to https://example.com/Tomas
<class 'AttributeError'> at /Tomas
No template named Tomas
P.S. From the static files page it seems to say I should just be able to put the Tomas.html file in a folder called "static" and then access is via https://example.com/static/Tomas.html but that is not working (it returns "not found")
You're using a relative path to your template directory without paying attention to the working directory. See https://help.pythonanywhere.com/pages/NoSuchFileOrDirectory/
You're working too hard. 'static' is built in.
As the documentation says, http://localhost/static/logo.png will return the file logo.png from the existing directory static, which is relative to your webserver root.
Do not use render() for this (not needed). Also, do not list your desired file ('/Tomas') in the urls list (not needed).
Anything under the static directory can be accessed with the url https://localhost/static/...
"static" is hardcoded in the web.py server, so you cannot (easily) change this to some other folder. The suggestion in the web.py documents is to have nginx or apache host your application and use an Alias there to go to web.py static. (I think you can also add StaticMiddleware to your web.py application, but you'd need to investigate that yourself -- look at web.application.run()
The case of the disappearing /static/ directory was related to the fact that I'm hosting on pythonanywhere.com
Even though the web.py documentation says that the /static/ folder is plugged in by default, that's not the case in pythonanywhere and you need to expressly make the link between the url http://yourhost/static/ and /path/to/static in the Web part of the dashboard.
By using CodeIgniter 4 framework, I've developed RESTful api and there I need to access file (.json and .txt) to get content. But not able to access php inbuilt function file_get_contents().
For more details, pls check attached screenshot API_curl_file_get_content_error.PNG
And test.txt file is also accessible with same file path. For more details pls check screenshot Input-txt-file-content.png
NOTE : 1) test.txt file and respective directories have full permission.
2) Development environment :
<---->Apache/2.4.47 (Win64) OpenSSL/1.1.1k PHP/8.1.2
<---->Database client version: libmysql - mysqlnd 8.1.2
<---->PHP version: 8.1.2
<---->CodeIgniter 4
Product.php (index method)
<?php
namespace App\Controllers;
use CodeIgniter\RESTful\ResourceController;
class Product extends ResourceController
{
use \CodeIgniter\API\ResponseTrait;
public function index()
{
helper("filesystem");
ini_set('max_execution_time', 300);
$data['msg'] = "Product Index wizard";
$data['status'] = 200;
$file_path = base_url() . '/assets/data/test.txt'; // Read JSON
$json = file_get_contents($file_path, true);
$json_data = json_decode($json, true);
return $this->respond($data);
}
}
Explanation:
Remember that "http://localhost:8085/" most likely points to the document root of the project, which is usually the /public path. So, unless the "assets" folder resides in the /public path, file_get_contents("http://localhost:8085/assets/data/test.txt"); will fail to find the requested server resource.
Solution:
Since your file resource (test.txt) is on the local filesystem,
Instead of:
file_get_contents("http://localhost:8085/assets/data/test.txt");
Use this:
constant ROOTPATH
The path to the project root directory. Just above APPPATH.
file_get_contents(ROOTPATH . "assets/data/test.txt");
Addendum:
I believe you also forgot to add the $json_data output to the returned $data variable in the Product::index resource controller method.
I've been struggling to deploy my shiny app using the tesseractpackage. It seems that it can't 'reach' the downloaded languages. In my case: English and Dutch.
When setting up the language, the resulting object should 'point' to a path. That's the part where shiny can't open the connection.
Any help would be much appriciated!
Kind regards, R
Below I've copied the error message and the relevant code.
This is the error message I get after deployment:
Warning in file(con, "wb") :
cannot open file '/usr/share/tesseract-ocr/tessdata/nld.traineddata': Permission denied
Error in value[3L] : cannot open the connection
Calls: local ... tryCatch -> tryCatchList -> tryCatchOne ->
Execution halted
This is my code
#loading software requirement
library(tesseract)
#download language (dutch)
tesseract_download('nld')
tesseract_download('eng')
#set language parameters for later use.
dutch <- tesseract('nld')
english <- tesseract('eng')
I've managed to get it working myself. The key was taking the following steps:
Creating a subdirectory (of folder) which is called 'tessdata'. This subdirecty is the directory you can download the languages in and 'set' the languages.
When deploying your app, you have to deploy this tessdata-subdirectory as well. So in the deployement prompt, you 'tick' the boxes of this folder as well.
Then make sure the tesseract engine points at the following path:
Screenshot of how to upload the tessdata-folder along with the app
enter image description here
Please see the code below
#loading software requirementlibrary(tesseract)
#Make sure the tesseract package is 'pointing' at the right 'parent directory'
#which is in this case the path your shiny app is working from.
#That's why you need the dot ("."). Which is in essence the workdir.
Sys.setenv(TESSDAT_PREFIX = ".")
#so combining the workdir and the pre-installed folder 'tessdata'
path <- paste0(getwd(), '/tessdata')
#use this path for downloading
#download languages (dutch and english)
tesseract_download('nld', datapath = path)
tesseract_download('eng', datapath = path)
#set language parameters for later use, using the same path
dutch <- tesseract('nld', datapath = path)
english <- tesseract('eng', datapath = path)
I just can't figure out what's going on with my RSync. I'm running RSync on RHEL5, ip = xx.xx.xx.97. It's getting files from RHEL5, ip = xx.xx.xx.96.
Here's what the log (which I specified on the RSync command line) shows on xx.97 (the one requesting the files):
(local time)
2015/08/30 13:40:01 [17353] #ERROR: auth failed on module tomcat_backup
2015/08/30 13:40:01 [17353] rsync error: error starting client-server protocol (code 5) at main.c(1530) [receiver=3.0.6]
Here's what the log(which is specified in the rsyncd.conf file) shows on xx.96 (the one supplying the files):
(UTC time)
2015/08/30 07:40:01 [8836] name lookup failed for xx.xx.xx.97: Name or service not known
2015/08/30 07:40:01 [8836] connect from UNKNOWN (xx.xx.xx.97)
2015/08/30 07:40:01 [8836] auth failed on module tomcat_backup from unknown (xx.xx.xx.97): password mismatch
Here's the actual rsync.sh command called from xx.xx.xx.97 (the requester):
export RSYNC_PASSWORD=rsyncclient
rsync -havz --log-file=/usr/local/bin/RSync/test.log rsync://rsyncclient#xx.xx.xx.96/tomcat_backup/ProcessSniffer/ /usr/local/bin/ProcessSniffer
Here's the rsyncd.conf on xx.xx.xx.97:
lock file = /var/run/rsync.lock
log file = /var/log/rsyncd.log
pid file = /var/run/rsyncd.pid
[files]
name = tomcat_backup
path = /usr/local/bin/
comment = The copy/backup of tomcat from .96
uid = tomcat
gid = tomcat
read only = no
list = yes
auth users = rsyncclient
secrets file = /etc/rsyncd.secrets
hosts allow = xx.xx.xx.96/255.255.255.0
Here's the rsyncd.secrets on xx.xx.xx.97:
files:files
Here's the rsyncd.conf on xx.xx.xx.96 (the supplier of files):
Note: there is a 'cwrsync' (Windows version of rsync) successfully calling for files also (xx.xx.xx.100)
Note: yes, there is the possibility of xx.96 requesting files from xx.97. However, this is NOT actually happening.
It's commented out of the init.d mechanism.
lock file = /var/run/rsync.lock
log file = /var/log/rsync.log
pid file = /var/run/rsync.pid
strict modes = false
[files]
name = tomcat_backup
path = /usr/local/bin
comment = The copy/backup of tomcat from xx.97
uid = tomcat
gid = tomcat
read only = no
list = yes
auth users = rsyncclient
secrets file = /etc/rsyncd.secrets
hosts allow = xx.xx.xx.97/255.255.255.0, xx.xx.xx.100/255.255.255.0
Here's the rsyncd.secrets on xx.xx.xx.97:
files:files
It was something else. I had a script calling the rsync command, and that was causing the problem. The actual rsync command line was ok.
Apologies.
This is what I have been through when I got this error. My first thinking was to check rsync server log. and it is not in the place configured in rsync.conf. Then I checked the log printed in systemctl status rsyncd
rsyncd[23391]: auth failed on module signaling from unknown (172.28.15.10): missing secret for user "rsync_backup"
rsyncd[23394]: Badly formed boolean in configuration file: "no # rsync daemon before transmission, change to the root directory and limited within.".
rsyncd[23394]: params.c:Parameter() - Ignoring badly formed line in configuration file: ignore errors # ignore some io error informations.
rsyncd[23394]: Badly formed boolean in configuration file: "false # if true, cannot upload file to this server.".
rsyncd[23394]: Badly formed boolean in configuration file: "false # if true, cannot download file from this server.".
rsyncd[23394]: Badly formed boolean in configuration file: "false # if true, can only list files here.".
Combining the fact that log configuration does not come into play. It seems that the comment after each line of configuration in rsync.conf makes configurations invalid. So I deleted those # ... and restart rsyncd.
While trying to compile a proto file named UserOptions.proto which has an import named Account.proto using the below command
protoc --proto_path=/home/project_new1/account --java_out=/home/project_new1/source /home/project_new1/settings/Useroptions.proto
I get the following error :
/home/project_new1/settings/UserOptions.proto: File does not reside within any path specified using --proto_path (or -I). You must specify a --proto_path which encompasses this file.
PS: UserOptions.proto present in the directory /home/project_new1/settings
imports Account.proto present in the directory
/home/project_new1/account
Proto descriptor files:
UserOptions.proto
package settings;
import "Account.proto";
option java_outer_classname = "UserOptionsVOProto";
Account.proto
package account;
option java_outer_classname = "AccountVOProto";
message Object
{
optional string userId = 1;
optional string service = 2;
}
As the error message states, the file you pass on the command line needs to be in one of the --proto_paths. In your case, you have only specified one --proto_path of:
/home/project_new1/
But the file you're passing is:
/home/project_new1/settings/UserOptions.proto
Notice that the file is not in the account subdirectory; it's in settings instead.
You have two options:
(Not recommended) Pass a second --proto_path argument to add .../settings to the path.
(Recommended) Use the root of your source tree as the proto path. E.g.:
protoc --proto_path=/home/project_new1/ --java_out=/home/project_new1 /home/project_new1/settings/UserOptions.proto
In this case, to import Account.proto, you'll need to write:
import "account/Account.proto";
For those of us who want this really spelled out, here is an example where I have installed the protoc beta for gRPC using NuGet Packages Google.Protobuf, Grpc.Core and Grpc.Tools. My solution packages are one level above my Grpc directory (i.e. at BruTrader\packages). My .proto files are at BruTrader\Grpc\protos.
1. My .proto file:
syntax = "proto3";
import "timestamp.proto";
import "enums.proto";
package BruTrader.Grpc;
message DividendMessage {
double amount = 1;
google.protobuf.Timestamp dateUnix = 2;
}
2. my GenerateProto.bat file:
..\packages\Google.Protobuf.3.0.0-beta2\tools\protoc.exe -I..\Grpc\protos -I..\packages\Google.Protobuf.3.0.0-beta2\tools\google\protobuf --csharp_out=..\Grpc\Generated --grpc_out=..\Grpc\Generated --plugin=protoc-gen-grpc=..\packages\Grpc.Tools.0.13.0\tools\grpc_csharp_plugin.exe %1
3. my BuildProtos.bat
call GenerateProto ..\Grpc\protos\masterinstrument.proto
call GenerateProto .\protos\instrument.proto
etc.
4. BuildProtos.bat is executed as a Pre-build event on my Grpc project like this:
CD $(ProjectDir)
CALL "$(ProjectDir)BuildProtos.bat"
For my environment, Windows 10 Pro operating system and C++ programming languaje, I used the protoc-3.12.2-win64.zip that you can downloat it from here. You should open a Windows PowerShell inside the protoc-3.12.2-win64\bin path and then you must execute one of the next commands:
.\protoc.exe -I=C:\Users\UserName\Desktop\SRC --cpp_out=C:\Users\UserName\Desktop\DST C:\Users\UserName\Desktop\SRC\addressbook.proto
Or
.\protoc.exe --proto_path=C:\Users\UserName\Desktop\SRC --cpp_out=C:\Users\UserName\Desktop\DST C:\Users\UserName\Desktop\SRC\addressbook.proto
Note:
1- My source folder is in: C:\Users\UserName\Desktop\SRC
2- My destination folder is in: C:\Users\UserName\Desktop\DST
3- My .proto file is in: C:\Users\UserName\Desktop\SRC\addressbook.proto