I have multiple input telegraf input plugins for every application and they are separated by the app.toml files. output plugin for all of them is configured in the global telegraf.conf file to write to InfluxDB
Now I am trying to introduce new app but write it's metrics to local filesystem and everything else continue to write to influxDB.
Now when I have my new app with the below configuration, the output file /tmp/metrics.out is collecting all the metrics from other apps as well.
[[inputs.app]]
context = "/jolokia/read"
servers = [":7090"]
metrics = ["/jvm_heap_usage"]
[[outputs.file]]
## Files to write to, "stdout" is a specially handled file.
files = ["/tmp/metrics.out"]
How can I send all the metrics to InfluxDB and only one app where I added outputs.file to send to local filesystem?
This can be achieved by using the namepass option on the output plugin.
[[outputs.file]]
files = ["stdout", "./metrics.out"]
data_format = "influx"
namepass = ["mem"]
[[inputs.mem]]
# No configuration options
The name of the input plugin can also be modified using name_override, name_prefix, or name_suffix.
[[outputs.file]]
files = ["stdout", "./metrics.out"]
data_format = "influx"
namepass = ["app1"]
[[inputs.mem]]
name_override = "app1"
Related
I am trying to create a database for my Flask application in the main directory of my project. This is my code for initializing a database:
app.config["SQLALCHEMY_DATABASE_URI"] = 'sqlite:///users.db'
app.config["SQLALCHEMY_TRACK_MODIFICATIONS"] = False
db = SQLAlchemy(app)
Flask requires application context, so this is how I create the database:
$ flask shell
>>> db.create_all()
I also tried doing it with:
$ python
>>> from app import app, db
>>> app.app_context().push()
>>> db.create_all()
Both of these options create the database in the /instance directory. Is there any way to get around this and create it in the main directory of the project?
The instance path is the preferred and default location for the database. I recommend you to use this one for security reasons. However, it is also possible to choose an alternative solution in which the full length of the path is specified in the configuration.
The following configuration corresponds to an outdated variant, where the database is created in the current working directory. Please don't use this anymore.
app.config['SQLALCHEMY_DATABASE_URI'] ='sqlite:///' + os.path.join(os.getcwd(), 'users.db')
This corresponds to the current solution.
app.config['SQLALCHEMY_DATABASE_URI'] ='sqlite:///' + os.path.join(app.instance_path, 'users.db')
I am trying to create wordpress web application on Azure with Terraform. Each web app has own database. I manage to create resource groups, database server and databases but i cannot create wordpress web app. I can create a web app and all works fine but not wordpress. When i make wordpress web app manually and import data to see what is different i see that wordpress has repo_url and branch pointing to wordpress-azure repo on github. When i try to incorporate this in code i get error message.
resource "azurerm_mysql_database" "testtt" {
name = "testtt"
resource_group_name = azurerm_resource_group.RG_mok_2024.name
server_name = azurerm_mysql_server.wp-db-mok-2024.name
charset = "utf8"
coll`enter code here`ation = "utf8_unicode_ci"
}
resource "azurerm_app_service" "testtt" {
name = "testtt"
location = azurerm_resource_group.RG_mok_2024.location
resource_group_name = azurerm_resource_group.RG_mok_2024.name
app_service_plan_id = azurerm_app_service_plan.appserviceplan-wordpress-mok-6.id
site_config {
dotnet_framework_version = "v4.0"
scm_type = "GitHub"
default_documents = ["Default.htm","Default.html","Default.asp","index.htm","index.html","iistart.htm","default.aspx","index.php","hostingstart.html"]
}
source_control {
repo_url = "https://github.com/azureappserviceoss/wordpress-azure"
branch = "master"
}
connection_string {
name = "defaultConnection"
type = "MySQL"
value = "Database=testtt;Data Source=wp-db-mok-2024.mysql.database.azure.com;User Id=mysqladminun#wp-db-mok-2024;Password=password"
}
}
The error message i get when i am using source_control part of a code is
Error: "source_control": this field cannot be set
The source_control field is only exported. It cannot be used to connect a deploymentsource.
To execute an automated deployment directly, there is currently only the workaround via a local-exec null_resource.
The sourcecontrol integration can be created using an azure cli / powershell script which is then executed by the local-exec provisioner.
This works as follows:
resource "null_resource" "scm_integration" {
provisioner "local-exec" {
command = "${path.module}/enable_scm.ps1 -webAppName ${azurerm_app_service.testtt.name} -appResourceGroupName ${azurerm_resource_group.RG_mok_2024.name} -scmBranch master -repoUrl https://github.com/azureappserviceoss/wordpress-azure"
interpreter = ["pwsh", "-Command"]
}
}
In addition you need the powershell script enable_scm.ps1.
In this GitHub Issue the workaround incl. script is described completely
According to terraform doc about azurerm app service, the field source_control is only exported. And it is ONLY exported when the scm_type is set to LocalGit. You have set it to GitHub, and it is an output value, so according to the documentation, you dont need that.
Furthermore in line 6 there is enter code here but I guess that this was pasted here by accident.
Finally, I hope that in the connection string value, your database password is not "password".
Can you try to set the source_control section before the site_config? There is an open issue for the Terraform azurerm_app_service provider that suggests this might be a work-around.
https://github.com/terraform-providers/terraform-provider-azurerm/issues/3696
I have created Cosmos db with Mongo db api. Its working fine for my dev env. But now i want to copy it from dev to stage and prod env.
So i didn't find any option there on azure portal.
Any help would be greatly appreciated.
Using Azure Resource Manager templates, you can easily automate the deployment and configuration of your Azure resources. This tutorial shows how to deploy a web application and automatically configure Azure Cosmos DB account connection information.
You can find a step by step guide using the deployment template on here
Let me know if this helps.
Software dependencies
Download Python from https://www.python.org/downloads/windows/
Download the script folder and run the following command
py -m pip install azure.cosmos
py .\deploy_db.py "Endpoint of Cosmosdb" "Primary Key" "DatabaseName"
Create seed data python script file as seed_data.py
import json
import os
def get_employees():
employees = []
for file in os.listdir("Data\Employees"):
if file.endswith(".json"):
employeesFile = os.path.join("Data\Employees", file)
with open(employeesFile, encoding="utf8") as f:
employees.append(json.load(f))
return employees
def get_employee_details():
employee_details = {};
with open('SeedData/EmployeesDetails.json', encoding="utf8") as f:
employee_details = json.load(f)
return employee_details
Create python script file names as deploy_db.py
from azure.cosmos import exceptions, CosmosClient, PartitionKey
import seed_data
import sys
# Initialize the Cosmos client
endpoint = sys.argv[1]
key = sys.argv[2]
database = sys.argv[3]
# Create cosmosClient
client = CosmosClient(endpoint, key)
# Create a database
database_name = database
database = client.create_database_if_not_exists(id=database_name)
# Create Employees Container
container_name = 'Employees'
container = database.create_container_if_not_exists(
id=container_name,
partition_key=PartitionKey(path="/id")
)
# Add see data to the Employees
employee_details_to_create = seed_data.get_employee_details()
container.upsert_item(body=employee_details_to_create)
Add this script in your DevOps pipeline and add to the pipline as
automation, that runs on each build
I know how to copy a file in sftp server. But, how to put files in a sftp folder to a suitescript array? Also, is it possible to move or delete a file in sftp server using suite script 2.0? If yes, how to do it?
The following coding is for copying a file in sftp:
var myPwdGuid = "B34672495064525E5D65032D63B52301";
var myHostKey = "AAA1234567890Q=";
var connection = sftp.createConnection({
username: 'myuser',
passwordGuid: myPwdGuid,
url: 'host.somewhere.com',
directory: 'myuser/wheres/my/file',
hostKey: myHostKey
});
var downloadedFile = connection.download({
directory: 'relative/path/to/file',
filename: 'downloadMe.js'
});
connection.upload({
directory: 'relative/path/to/remote/dir',
filename: 'copy_of_downloadme.js',
file: downloadedFile,
replaceExisting: true
});
Indeed, the SFTP module in its current version does not include a way to list the contents of a remote directory. In some scenarios the file name is known or predictable. For the other scenarios one can use a file with a known name to store the names of the other files or middleware to retrieve the list of files. Ask me if you want to know about an existing middleware solution.
This is solved; thanks to #vmontco's solution: I was missing MEDIA_URL, now it works perfectly.
----------original question below-----------
I welcome suggestions from every angle; I am fairly new to Django and Python. I'm sure I am missing something simple.
Using a Model Form, with a FileField, I upload and save an Excel file to a folder structure under MEDIA_ROOT. This works.
I want to read that same file later to perform operations using Pyexcel. This is where I am stuck. I am attempting to upload the file using the FileField stored in the DB.
This is where I have problems, and I am not sure if am misunderstanding MEDIA_ROOT, or some other aspect of Django.
When I pass the pk to the 2nd view, I then instantiate an object based on the Model. It has the FileField 'docfile', which I am trying to use to access the file to do some operations using Pyexcel,
here is the FileField declaration from models.py:
docfile = models.FileField(
verbose_name="Choose file to upload:",
upload_to='Excel_CSV_Assets/%Y/%m/%d')
EDIT: If I hard-code the pth to the file like this, everything works, including operations afterwards:
thedocfile='site_static/site/original_assets/Excel_CSV_Assets/2016/04/23/Animals_oglc4DV.xlsx'
book=pyexcel.get_book(file_name=thedocfile)
:END OF EDIT
Here is the code from the 2nd view, where I attempt to read the file into memory, and make a 'book' class object using Pyexcel. I am stuck here:
asset = Excel_CSV_Asset.objects.get(id=assetid)
book=pyexcel.get_book(file_name=asset.docfile)
Here is my error description:
Here is the info right at where my code breaks:
Although it says "Wrong filename", I can see the file is in the folder:
I'm able to open the file by double-clicking; the file is not corrupted.
EDIT:
If I cast the 'asset.docfile' to str, like so:
asset = Excel_CSV_Asset.objects.get(id=assetid)
book=pyexcel.get_book(file_name=str(asset.docfile))
I get a different error:
[Errno 2] No such file or directory: 'Excel_CSV_Assets/2016/04/23/Animals_oglc4DV.xlsx'
...but this is the correct directory, located beneath the MEDIA_ROOT file structure.
Here is settings.py MEDIA_ROOT:
MEDIA_ROOT = 'site_static/site/original_assets/'
Here is urls.py:
urlpatterns = [
url(r'^admin/', include(admin.site.urls)),
url(r'^e/', include('excel_to_mongo.urls')),
] + static(settings.STATIC_URL, document_root=settings.STATIC_ROOT)
Here is the url.py of that app:
url(r'^efactory/(?P<assetid>\d+)/$', 'display_sheet_column_choices', {}),
I think your problem is that you don't fully understand the media files management with Django.
What are media files?
Media files are all the files that are user-uploaded (at running time).
You must not mistake them with Static files that are assets needed by your project to work and that you add at development time (CSS, background picture and JS files for instance).
You shouldn't mix them because they are managed differently by the server and that it could lead to security problems (cf. the warning here):
Static files management :
You put your static files as a part of the code either in one static subdirectory from the installed django applications, either in one of the locations you added to STATICFILES_DIRS.
Static files have to be gathered before starting the server by calling ./manage.py collectstatic, this command will collect (copy) the static files into the a directory (STATIC_ROOT's value).
You then have to set STATIC_URL to choose with wich url you should serve your static files. An usual choice would be /static/. To access the static file you should then try to reach /static/path/to/static/file/in/static_root/dir.
Media files management :
Your media files are added at running time. They are stored in the MEDIA_ROOT location that has to be an absolute path. Hence the fact I suggested you to join the BASE_DIR value (an absolute path) and the subdir you would choose with something like :
MEDIA_ROOT = os.path.join(BASE_DIR, "/media/subdir")
You then have to set an URL for your media files, by using the MEDIA_URL variable. To access your media files, the urls will start with the value you choose :
MEDIA_URL = '/media/'
Then, add this to your urls.py file :
if settings.DEBUG:
urlpatterns = urlpatterns + static(settings.MEDIA_URL, document_root=settings.MEDIA_ROOT)
With the current example, your mymediafile.txt will be located at /path/to/your/project/media/subdir/path/in/media/root/mymediafile.txt and served at http://127.0.0.1:8000/media/path/in/media/root/mymediafile.txt.
But this is suitable only for a development use as told here. And this would work only for DEBUG == TRUE
For a production use, you should consider deploying your media files with your http server (apache for instance).
Conclusion :
Take the time to understand this. Because I suspect you don't really understood what you did and this lack of understanding could lead to future bugs and errors.