I'm new to pyramid. When trying to use chameleon as the templating engine, it fails to find the template when specified with a relative path - It is looking for it at env35/lib/python3.5/site-packages/pyramid/ where env35 is the virtual environment I created. It will work however if the full path is specified. It will also work using a relative path using jinja2 as the templating engine.
Why can I not use a chameleon template using relative path?
From the manual
add_view(...., renderer,...)
This is either a single string term (e.g. json) or a string implying a path or asset specification (e.g. templates/views.pt)
naming a renderer implementation. If the renderer value does not
contain a dot ., the specified string will be used to look up a
renderer implementation, and that renderer implementation will be used
to construct a response from the view return value. If the renderer
value contains a dot (.), the specified term will be treated as a
path, and the filename extension of the last element in the path will
be used to look up the renderer implementation, which will be passed
the full path. The renderer implementation will be used to construct a
response from the view return value.
Note that if the view itself returns a response (see View Callable Responses), the specified renderer implementation is never called.
When the renderer is a path, although a path is usually just a simple relative pathname (e.g. templates/foo.pt, implying that a
template named "foo.pt" is in the "templates" directory relative to
the directory of the current package of the Configurator), a path can
be absolute, starting with a slash on UNIX or a drive letter prefix on
Windows. The path can alternately be a asset specification in the form
some.dotted.package_name:relative/path, making it possible to address
template assets which live in a separate package.
The renderer attribute is optional. If it is not defined, the "null" renderer is assumed (no rendering is performed and the value is
passed back to the upstream Pyramid machinery unmodified).
Here are the steps I undertook to set up my environment...
export VENV=~/Documents/app_projects/pyramid_tutorial/env35
python3 -m venv $VENV
source $VENV/bin/activate #activate the virtual environment
pip install --upgrade pip
pip install pyramid
pip install wheel
pip install pyramid_chameleon
pip install pyramid_jinja2
My file structure:
pyramid_tutorial
env35
bin
...
templates
hello.jinja2
hello.pt
test_app.py
test_app.py:
from wsgiref.simple_server import make_server
from pyramid.config import Configurator
from pyramid.response import Response
from pyramid.view import view_config
def hello(request):
return dict(name='Bugs Bunny')
if __name__ == '__main__':
config = Configurator()
config.include('pyramid_chameleon')
config.include('pyramid_jinja2')
#This does not work... http://localhost:6543/chameleon
config.add_route('hello_world_1', '/chameleon')
config.add_view(hello, route_name='hello_world_1', renderer='templates/hello.pt')
# ValueError: Missing template asset: templates/hello.pt (/home/david/Documents/app_projects/pyramid_tutorial/env35/lib/python3.5/site-packages/pyramid/templates/hello.pt)
#This works... http://localhost:6543/chameleon2
config.add_route('hello_world_2', '/chameleon2')
config.add_view(hello, route_name='hello_world_2', renderer='/home/david/Documents/app_projects/pyramid_tutorial/templates/hello.pt')
#This works... http://localhost:6543/jinja
config.add_route('hello_world_3', '/jinja')
config.add_view(hello, route_name='hello_world_3', renderer='templates/hello.jinja2')
app = config.make_wsgi_app()
server = make_server('0.0.0.0', 6543, app)
print ('Serving at http://127.0.0.1:6543')
server.serve_forever()
hello.pt:
<p>Hello <strong>${name}</strong>! (Chameleon renderer)</p>
hello.jinja2:
<p>Hello <strong>{{name}}</strong>! (jinja2 renderer)</p>
It will work if you specify renderer=__name__ + ':templates/hello.pt'. The resolution logic doesn't work in this case because the file is not being executed as a python package and thus some weird stuff can occur. pyramid_chameleon could likely be updated with better support here but by far the common case for real apps is to write your code as a package which will work as expected.
It might also work if you tweak things slighty run your script as a module via python -m test_app.
Related
As far as I'm aware I'm using best practices to define paths (using raw strings) and how I go about joining them (using os.path.join()), e.g.
import os
fdir = r'C:\Code\...\samples'
fpath = os.path.join(fdir, 'fname.ext')
and doing so has not caused me any problems when running my code within a Python or command shell. If I print fpath to the console I get consistent use of \s in the path:
C:\Code...\samples\fname.ext
But when I run a Docker containerized version of the code and run the image I get the error:
FileNotFoundError: [Errno 2] No such file or directory:
'C:\Code\...\samples/fname.ext'
I don't understand why os.path.join() has used a / to join fdir and fname.ext when the rest of the path included \\. It doesn't do this when I run the code outside of the container.
I have tried using os.path.normpath():
fpath = os.path.join(fdir, 'fname.ext')
fpath = os.path.normpath(fpath)
as discussed here, and os.sep.join():
fpath = os.sep.join([fdir, 'fname.ext'])
as covered here, and Path().joinpath():
from pathlib import Path
fpath = Path(fdir).joinpath('fname.ext')
as well as Path() / 'path_to_add':
fpath = Path(fdir) / 'fname.ext'
as discussed here, but in every case I end up with the same result using os.path.join().
Can someone please help me to understand what is going on and how to create consistent paths that will work whether I run the code in Python in a Windows environment, or in a Docker container?
Update Nov. 16:
In trying to keep my question brief I think I've left out details that are crucial. Apologies to those who have kindly taken the time to offer suggestions based on my incomplete description of the problem.
My code needs to import/export files from/to directories that are defined within a user-specified configuration file.
So the configuration file has a section of code where the user defines variables and paths, e.g.
samplesDir = r"path-to-samples-directory"
The variables are stored in a dictionary of dictionaris and stored as a .json.
At the start of the code the user defines the key that selects the dictionary of interest so that at various parts in my code when a file needs to be imported/exported, the paths are at hand.
So back to my example, samplesDir is stored in the configuration dictionary, cfgDict, so all I need to do is append the file name:
sampleFpath = os.path.join(sampleDir, sampleFname)
and sampleFname is determined based on other variables.
Because of the dynamic nature of the variables (including directory paths and file paths), I think it rules out the use of static path defined in a .yml with Docker Compose.
Update Nov. 18:
It may help to include a few more details and some screenshots.
The above screenshot shows the file and folder structure of the src directory containing the source code, the main app.py script for command-line use, the Dockerfile, etc.
The configs folder contains JSON files that includes variables, paths to directories and files. The user can create configuration files either by copying an existing one and modifying the entries, or configuration files can be generated by calling config.py.
Within config.py I have pre-set variables and paths, so that the directory path to the configuration files (configs), sample files (sample_DROs) and others (e.g. fiducials) are all within src.
I don't anticipate any reason why the user would want to store the config files anywhere else, nor do I expect them to want to use different sample files (or move them elsewhere). However, they will undoubtedly create their own fiducials and may decide not to store them in the fiducials directory (i.e. somewhere not within the src directory).
Likewise I have pre-set the download directory (based on the parameters stored within the configuration files, files are fetched from a server and downloaded) to be the default Downloads directory:
rootDownloadDir = os.path.join(Path.home(), "Downloads", "xnat_downloads")
Those files are later imported, processed, and the outputs are (by default) exported into sub-directories within rootDownloadDir.
Within Dockerfile I set the working directory of the container to be that of the source code and copy all of the contents of src (with the exception of some directories defined in .dockerignore):
WORKDIR C:/Code/WP1.3_multiple_modalities/src
...
COPY . .
so that the structure of the container mimics that of WORKDIR:
Hence I have allowed for flexibility in import/export directories, and they are by default a combination of paths within and outside of the src directory. And so, the code executed within the container will need to access files both within and outside of src.
That said, I don't know what rootDownloadDir will look like when os.path.join(Path.home(), "Downloads", "xnat_downloads") is run within the container.
This has got me thinking - Is it bad practice to set the download directory outside of src?
Returning to the original error:
the sample file is in the container:
From the actual behavior I can suppose that the container is based on Unix-like image. Path separator is / in such systems.
To build an environment-independent path which works inside and outside of the container you need the following steps:
Mounting of host folder to container directory.
Environment variable inside and outside the container.
I can show an example of how this is achievable via docker-compose tool and its configuration file docker-compose.yml:
# docker-compose.yml file
version: '3'
services:
<service_name>: # your service name here
image: <image_name> # name of image your container is built on
environment:
- SAMPLES_PATH=/samples
volumes:
- C:\Code\somepath\samples:/samples
In your python code you can use the following structure:
import os
fdir = os.getenv('SAMPLES_PATH', r'C:\Code\...\samples')
fpath = os.path.join(fdir, 'fname.ext')
I created a website which generates PDF using PDFKIT and I know how to install and setup environment variable path on Window. I managed to deploy my first website on Heroku but now I'm getting error "No wkhtmltopdf executable found: "b''" When trying to generate the PDF.
I have no idea, How to install and setup WKHTMLTOPDF on Heroku because this is first time I'm dealing with Linux.
I really tried everything before asking this but even following this not working for me.
Python 3 flask install wkhtmltopdf on heroku
If possible, please guide me with step by step on how to install and setup this.
I followed all the resource and everything but couldn't make it work. Every time I get the same error.
I'm using Django version 2. Python version 3.7.
This is what I get if I do heroku stack
Available Stacks
cedar-14
container
heroku-16
* heroku-18
Error, I'm getting when generating the PDF.
No wkhtmltopdf executable found: "b''"
If this file exists please check that this process can read it. Otherwise please install wkhtmltopdf - https://github.com/JazzCore/python-pdfkit/wiki/Installing-wkhtmltopdf
My website works very well on localhost without any problem and as far as I know, I'm sure that I have done something wrong in installing wkhtmltopdf.
Thank you
It's non-trivial. If you want to avoid all of the below's headache, you can just use my service, api2pdf: https://github.com/api2pdf/api2pdf.python. Otherwise, if you want to try and work through it, see below.
1) Add this to your requirements.txt to install a special wkhtmltopdf pack for heroku as well as pdfkit.
git+git://github.com/johnfraney/wkhtmltopdf-pack.git
pdfkit==0.6.1
2) I created a pdf_manager.py in my flask app. In pdf_manager.py I have a method:
def _get_pdfkit_config():
"""wkhtmltopdf lives and functions differently depending on Windows or Linux. We
need to support both since we develop on windows but deploy on Heroku.
Returns:
A pdfkit configuration
"""
if platform.system() == 'Windows':
return pdfkit.configuration(wkhtmltopdf=os.environ.get('WKHTMLTOPDF_BINARY', 'C:\\Program Files\\wkhtmltopdf\\bin\\wkhtmltopdf.exe'))
else:
WKHTMLTOPDF_CMD = subprocess.Popen(['which', os.environ.get('WKHTMLTOPDF_BINARY', 'wkhtmltopdf')], stdout=subprocess.PIPE).communicate()[0].strip()
return pdfkit.configuration(wkhtmltopdf=WKHTMLTOPDF_CMD)
The reason I have the platform statement in there is that I develop on a windows machine and I have the local wkhtmltopdf binary on my PC. But when I deploy to Heroku, it runs in their linux containers so I need to detect first which platform we're on before running the binary.
3) Then I created two more methods - one to convert a url to pdf and another to convert raw html to pdf.
def make_pdf_from_url(url, options=None):
"""Produces a pdf from a website's url.
Args:
url (str): A valid url
options (dict, optional): for specifying pdf parameters like landscape
mode and margins
Returns:
pdf of the website
"""
return pdfkit.from_url(url, False, configuration=_get_pdfkit_config(), options=options)
def make_pdf_from_raw_html(html, options=None):
"""Produces a pdf from raw html.
Args:
html (str): Valid html
options (dict, optional): for specifying pdf parameters like landscape
mode and margins
Returns:
pdf of the supplied html
"""
return pdfkit.from_string(html, False, configuration=_get_pdfkit_config(), options=options)
I use these methods to convert to PDF.
Just follow these steps to Deploy Django app(pdfkit) on Heroku:
Step 1:: Add following packages in requirements.txt file
wkhtmltopdf-pack==0.12.3.0
pdfkit==0.6.0
Step 2: Add below lines in the views.py to add path of binary file
import os, sys, subprocess, platform
if platform.system() == "Windows":
pdfkit_config = pdfkit.configuration(wkhtmltopdf=os.environ.get('WKHTMLTOPDF_BINARY', 'C:\\Program Files\\wkhtmltopdf\\bin\\wkhtmltopdf.exe'))
else:
os.environ['PATH'] += os.pathsep + os.path.dirname(sys.executable)
WKHTMLTOPDF_CMD = subprocess.Popen(['which', os.environ.get('WKHTMLTOPDF_BINARY', 'wkhtmltopdf')],
stdout=subprocess.PIPE).communicate()[0].strip()
pdfkit_config = pdfkit.configuration(wkhtmltopdf=WKHTMLTOPDF_CMD)
Step 3: And then pass pdfkit_config as argument as below
pdf = pdfkit.from_string(html,False,options, configuration=pdfkit_config)
How do I load a python module, that is not built in. I'm trying to create a plugin system for a small project im working on. How do I load those "plugins" into python? And, instaed of calling "import module", use a string to reference the module.
Have a look at importlib
Option 1: Import an arbitrary file in an arbiatrary path
Assume there's a module at /path/to/my/custom/module.py containing the following contents:
# /path/to/my/custom/module.py
test_var = 'hello'
def test_func():
print(test_var)
We can import this module using the following code:
import importlib.machinery
myfile = '/path/to/my/custom/module.py'
sfl = importlib.machinery.SourceFileLoader('mymod', myfile)
mymod = sfl.load_module()
The module is imported and assigned to the variable mymod. We can then access the module's contents as:
mymod.test_var
# prints 'hello' to the console
mymod.test_func()
# also prints 'hello' to the console
Option 2: Import a module from a package
Use importlib.import_module
For example, if you want to import settings from a settings.py file in your application root folder, you could use
_settings = importlib.import_module('settings')
The popular task queue package Celery uses this a lot, rather than giving you code examples here, please check out their git repository
I'm using Tornado working with Python 3 and Linux server, when I edit and save some text or XML files I want Tornado to restart itself. I checked the document and found the autoreload module and the watch function here.
It seems it only worked for pyo files. What can I do if I want it to reload when a certain URI is modified?
Setting the debug flag to True in settings forces Tornado to reload whenever a file is modified or whenever a URI is changed in app.py (or where ever you have defined your handlers). Tornado also automatically reloads template files so any changes in there will be seen instantly.
settings = {
'debug':True,
# other stuff
}
tornado.web.Application.__init__(self, handlers, **settings)
the file added must be an absolute path.
def addwatchfiles(*paths):
for p in paths:
autoreload.watch(os.path.abspath(p))
addwatchfiles('config.xml')
config.xml is at the same directory in where the server's python file start at.
You need to turn autoreload on:
tornado.autoreload.start()
tornado.autoreload.watch('myfile')
Complete example at https://gist.github.com/renaud/10356841
I'm writing a groovy script that I want to be controlled via a properties file stored in the same folder. However, I want to be able to call this script from anywhere. When I run the script it always looks for the properties file based on where it is run from, not where the script is.
How can I access the path of the script file from within the script?
You are correct that new File(".").getCanonicalPath() does not work. That returns the working directory.
To get the script directory
scriptDir = new File(getClass().protectionDomain.codeSource.location.path).parent
To get the script file path
scriptFile = getClass().protectionDomain.codeSource.location.path
As of Groovy 2.3.0 the #SourceURI annotation can be used to populate a variable with the URI of the script's location. This URI can then be used to get the path to the script:
import groovy.transform.SourceURI
import java.nio.file.Path
import java.nio.file.Paths
#SourceURI
URI sourceUri
Path scriptLocation = Paths.get(sourceUri)
Note that this will only work if the URI is a file: URI (or another URI scheme type with an installed FileSystemProvider), otherwise a FileSystemNotFoundException will be thrown by the Paths.get(URI) call. In particular, certain Groovy runtimes such as groovyshell and nextflow return a data: URI, which will not typically match an installed FileSystemProvider.
This makes sense if you are running the Groovy code as a script, otherwise the whole idea gets a little confusing, IMO. The workaround is here: https://issues.apache.org/jira/browse/GROOVY-1642
Basically this involves changing startGroovy.sh to pass in the location of the Groovy script as an environment variable.
As long as this information is not provided directly by Groovy, it's possible to modify the groovy.(sh|bat) starter script to make this property available as system property:
For unix boxes just change $GROOVY_HOME/bin/groovy (the sh script) to do
export JAVA_OPTS="$JAVA_OPTS -Dscript.name=$0"
before calling startGroovy
For Windows:
In startGroovy.bat add the following 2 lines right after the line with
the :init label (just before the parameter slurping starts):
#rem get name of script to launch with full path
set GROOVY_SCRIPT_NAME=%~f1
A bit further down in the batch file after the line that says "set
JAVA_OPTS=%JAVA_OPTS% -Dgroovy.starter.conf="%STARTER_CONF%" add the
line
set JAVA_OPTS=%JAVA_OPTS% -Dscript.name="%GROOVY_SCRIPT_NAME%"
For gradle user
I have same issue when I'm starting to work with gradle. I want to compile my thrift by remote thrift compiler (custom by my company).
Below is how I solved my issue:
task compileThrift {
doLast {
def projectLocation = projectDir.getAbsolutePath(); // HERE is what you've been looking for.
ssh.run {
session(remotes.compilerServer) {
// Delete existing thrift file.
cleanGeneratedFiles()
new File("$projectLocation/thrift/").eachFile() { f ->
def fileName=f.getName()
if(f.absolutePath.endsWith(".thrift")){
put from: f, into: "$compilerLocation/$fileName"
}
}
execute "mkdir -p $compilerLocation/gen-java"
def compileResult = execute "bash $compilerLocation/genjar $serviceName", logging: 'stdout', pty: true
assert compileResult.contains('SUCCESSFUL')
get from: "$compilerLocation/$serviceName" + '.jar', into: "$projectLocation/libs/"
}
}
}
}
One more solution. It works perfect even you run the script using GrovyConsole
File getScriptFile(){
new File(this.class.classLoader.getResourceLoader().loadGroovySource(this.class.name).toURI())
}
println getScriptFile()
workaround: for us it was running in an ANT environment and storing some location parent (knowing the subpath) in the Java environment properties (System.setProperty( "dirAncestor", "/foo" )) we could access the dir ancestor via Groovy's properties.get('dirAncestor').
maybe this will help for some scenarios mentioned here.