I want to set the URL in the schema.prisma setting by concatenating the strings as follows, but an error occurs.
My desire is to set it without adding the DATABASE_URL environment variable.
Is such a configuration possible?
I would be happy if I could simply use schema.prisma to concatenate the strings, but...
datasource db {
provider = "postgresql"
url = "postgresql://" + env("PG_USER") + ":" + env("PG_PASS") + "#" + env("PG_HOST") + ":" + env("PG_PORT") + "/" + env("PG_DATABASE") + "?schema=public"
}
schema.prisma file does not understand how to concatenate the environment variables as it is not a javascript or typescript file.
Prisma expects that you specify your environment variable in .env file
// .env file
DATABASE_URL_WITH_SCHEMA=${DATABASE_URL}?schema=foo
and in your schema.prisma file, you load it as shown
// schema.prisma
datasource db {
provider = "postgresql"
url = env("DATABASE_URL")
}
To learn more about using environment variables with Prisma, check out the docs
Related
I have created a data transformation which picks the 2 set of files from Azure Data Lake directory and then do the join and sink the resultant file in Azure Data Lake directory.
For the above scenario I created a Dataset which is pointing to the desired location where I have the 2 set of file which are .csv format. My dataset connection is failing and giving me below error
Error code: DFExecutorUserError
Details: abfss://<data lake path>#<XXXXXX.dfs.core.windows.net>/ has invalid authority.
When I change the directory location to something else, the dataset connection works fine. I did the research on this error, but not getting any guidance on this.
Error suggests like a permissions issue with SAS authentication.
You can construct or modify SAS URI and assign SignedPermission
To construct the string-to-sign for an account SAS, use the following format:
StringToSign = accountname + "\n" +
signedpermissions + "\n" +
signedservice + "\n" +
signedresourcetype + "\n" +
signedstart + "\n" +
signedexpiry + "\n" +
signedIP + "\n" +
signedProtocol + "\n" +
signedversion + "\n"
Use this article for more information
Just in case someone finds this. In my case I was getting the
has invalid authority
because my path was missing an extra "/" forward slash
abfss:/blah#XXXXXX.dfs.core.windows.net/some/path
instead of
abfss://blah#XXXXXX.dfs.core.windows.net/some/path
because my use of pathlib mangled it
from pathlib import Path
path = Path("abfss://blah#XXXXXX.dfs.core.windows.net") / "some/other/path"
print(str(path))
abfss:/blah#XXXXXX.dfs.core.windows.net/some/other/path
I solved this by only specifying the file path in the wildcard options
I have application.conf file which contains secrets(password for DB etc..) And this secret will be mounted as a file(the file content will contain the actual secrets) in the running pod. How can scala config library be tweaked to handle this. i.e
instead of normal application.conf
db {
user = "username"
password = "xxx"
}
I would have something like this...
db {
user = "username"
password = "${file_location}"
}
As the file is parsed, it should identify that the value of key password, needs to be resolved by looking up the file and loading its contents.
A simple function can be written to load the content of this file, how can this is be integrated with seamlessly with scala config. ie. The rest of the code will continue to use
config.getString(db.password)
I assume you are using configuration of HOCON format and Typesafe configuration library for it.
I don't think it has such feature out of the box, but as an possible alternative you can take a look at include feature - you can include content of another file into your application.conf:
db {
user = "username"
}
include /path/to/pod.conf //include env specific configuration file
and put inside /path/to/pod.conf:
db {
password = "pod_db_pass"
}
So eventually contents of both files will be merged inside application during loading, and your final config will contain password at path db.password
UPDATE
Another possible option load password from file and merge into config file with withFallback method. Example:
import com.typesafe.config._
val passord = "password_from_file"
val passwordConfig = ConfigFactory.parseString(s"db.password=$passord")
val applicationConfig = ConfigFactory.parseString(s"db.user=db_user")// Replace this with `ConfigFactory.load()`
val config = applicationConfig.withFallback(passwordConfig)
println(config)
Printout result:
Config(SimpleConfigObject({"db":{"password":"password_from_file","user":"db_user"}}))
Scatie: https://scastie.scala-lang.org/WW3weuqiT9WRUKfdrZgwcw
I am trying to encrypt a column in my prostrgres DB. The column name is "test" of type "bytea".
My enity code is below,
#ColumnTransformer(read = "pgp_sym_decrypt(" + " test, "
+ " current_setting('encrypt.key')"
+ ")", write = "pgp_sym_encrypt( " + " ?, "
+ " current_setting('encrypt.key')" + ") ")
#Column(columnDefinition = "bytea")
private String test;
postgresql.conf configuration file:
encrypt.key = 'Wow! So much security.
Placed the postgresql.conf configuration file in src/main/resources of spring boot appln. But the encryption.key value is not being picked up. And is there a way to pass the key using application.properties?
postgresql.conf is the configuration file for the Postgres server. It's stored inside the data directory (aka "cluster") on the server.
You can't put it on the client side (where your application runs). It has no meaning there.
To change values in there, you need to edit the file (on the server) or use ALTER SYSTEM.
If you want to change a configuration setting for the current session, use SET or set_config()
The latter two are probably the ones you are looking for to initialize the custom property for your encryption/decryption functions.
The way to use encrypt.key, not only for current session, it's store it in postgresql.conf.
The correct place is at the end of this file, in the "Customized Options" section:
#------------------------------------------------------------------------------
# CUSTOMIZED OPTIONS
#------------------------------------------------------------------------------
# Add settings for extensions here
encrypt.key=123456
Reload the configuration of the database server:
systemctl reload postgresql.service
To testing if it's working correctly. Open a pgsql session and type:
mydb=# show encrypt.key;
encrypt.key
-------------
123456
(1 row)
Example of encrypt:
mydb=# select pgp_sym_encrypt('Hola mundo',current_setting('encrypt.key'));
pgp_sym_encrypt
------------------------------------------------------------------------------------------------------------------------------------------------------------
\xc30d04070302255230e388dfe25e7dd23b01c5b8e62d148088a3417d3c27ed2cc11655d863b271672b9f076fffb82f1a7f074f2ecbe973df04642cd7a4f76ca5cff4a13b9a71e7cc6e693827
(1 row)
Example of decrypt:
mydb=# select pgp_sym_decrypt('\xc30d04070302255230e388dfe25e7dd23b01c5b8e62d148088a3417d3c27ed2cc11655d863b271672b9f076fffb82f1a7f074f2ecbe973df04642cd7a4f76ca5cff4a13b9a71e7cc6e693827',current_setting('encrypt.key'));
pgp_sym_decrypt
-----------------
Hola mundo
(1 row)
I'm trying to develop an Word-AddIn where I can upload a file, or where I could make changes to the file itself. I can do that with any File on the Local NTFS Windows Filesystem. But when a File is located to any "OneDrive"-Folder it won't work, as the file-Document path starts with:
https://companyname-my.sharepoint.com/personal/name_whatever/Documents
according to the return values of:
var doc = Globals.ThisAddIn.Application.ActiveDocument;
MessageBox.Show("Document Name : " + doc.Name);
MessageBox.Show("Document Full Name : " + doc.FullName);
MessageBox.Show("Document Path : " + doc.Path);
rather than expected to be:
C:\Users\myname\OneDrive-My_Company_Name\Documents
So basically the local folder is not callable. So does anybody has already faced this kind of issue?
You can use the Save or SaveAs methods of the Document to save the document on the local folder.
I have nearly a hundred data sources in a WebSphere Application Server (WAS) and due to office relocation, the IP of the database servers have changed and I need to update the datasource IP addresses in my WAS too.
Considering it error-prone to update hundred IPs through admin console.
Is there any way that I can make the change by updating config files or running a script? My version of WAS is 7.0.
You should be able to use the WAS Admin Console's built-in "command assistance" to capture simple code snippets for listing datasources and changing them by just completing those operations in the UI once.
Take those snippets and create a new jython script to list and update all of them.
More info on command assistance:
http://www.ibm.com/developerworks/websphere/library/techarticles/0812_rhodes/0812_rhodes.html
wsadmin scripting library:
https://github.com/wsadminlib/wsadminlib
You can achieve this using wsadmin scripting. Covener has the right idea with using the admin console's command assistance to do the update once manually (to get the code) and then dump that into a script that you can automate.
The basic idea is that a datasource has a set of properties nested under it, one of which is the ip address. So you want to write a script that will query for the datasource, find its nested property set, and iterate over the property set looking for the 'ipAddress' property to update.
Here is a function that will update the value of the "ipAddress" property.
import sys
def updateDataSourceIP(dsName, newIP):
ds = AdminConfig.getid('/Server:myServer/JDBCProvider:myProvider/DataSource:' + dsName + '/')
propertySet = AdminConfig.showAttribute(ds, 'propertySet')
propertyList = AdminConfig.list('J2EEResourceProperty', propertySet).splitlines()
for prop in propertyList:
print AdminConfig.showAttribute(prop, 'name')
if (AdminConfig.showAttribute(prop, 'name') == 'ipAddress'):
AdminConfig.modify(prop, '[[value '" + newIP + "']]')
AdminConfig.save();
# Call the function using command line args
updateDataSourceIP(sys.argv[0], sys.argv[1])
To run this script you would invoke the following from the command line:
$WAS_HOME/bin/wsadmin.sh -lang jython -f /path/to/script.py myDataSource 127.0.0.1
**Disclaimer: untested script. I don't know the name of the "ipAddress" property off the top of my head, but if you run this once it will print out all the properties on your ds, so you can get it there
Some useful links:
Basics about jython scripting
Modifying config objects using wsadmin
As an improvement to aguibert's script, to avoid having to provide all 100 datasource names and to update it to correct the containment path of the configuration id, consider this script which will update all datasources, regardless of the scope at which they're defined. As always, backup your configuration before beginning and once you're satisified the script is working as expected, replace the AdminConfig.reset() with save(). Note, these scripts will likely not work properly if you're using connection URLs in your configuration.
import sys
def updateDataSourceIP(newIP):
datasources = AdminConfig.getid('/DataSource:/').splitlines()
for datasource in datasources:
propertySet = AdminConfig.showAttribute(datasource, 'propertySet')
propertyList = AdminConfig.list('J2EEResourceProperty', propertySet).splitlines()
for prop in propertyList:
if (AdminConfig.showAttribute(prop, 'name') == 'serverName'):
oldip = AdminConfig.showAttribute(prop, 'value')
print "Updating serverName attribute of datasource '" + datasource + "' from " + oldip + " to " + sys.argv[0]
AdminConfig.modify(prop, '[[value ' + newIP + ']]')
AdminConfig.reset();
# Call the function using command line arg
updateDataSourceIP(sys.argv[0])
The script should be invoked similarly to the above, but without the datasource parameter, the only parameter is the new hostname or ip address:
$WAS_HOME/bin/wsadmin.sh -lang jython -f /path/to/script.py 127.0.0.1