harbor | An open source trusted cloud native registry project that stores, signs, and scans content | Continuous Deployment library
kandi X-RAY | harbor Summary
kandi X-RAY | harbor Summary
Note: The master branch may be in an unstable or even broken state during development. Please use releases instead of the master branch in order to get a stable set of binaries. Harbor is an open source trusted cloud native registry project that stores, signs, and scans content. Harbor extends the open source Docker Distribution by adding the functionalities usually required by users such as security, identity and management. Having a registry closer to the build and run environment can improve the image transfer efficiency. Harbor supports replication of images between registries, and also offers advanced security features such as user management, access control and activity auditing. Harbor is hosted by the Cloud Native Computing Foundation (CNCF). If you are an organization that wants to help shape the evolution of cloud native technologies, consider joining the CNCF. For details about who's involved and how Harbor plays a role, read the CNCF announcement.
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QUESTION
ANSWER
Answered 2022-Apr-16 at 13:15You can wrap your
- in a and center it like this:
QUESTION
I am trying to create k8s secrets with the storing username and secret in results module of ansible but i created one loop for the name and namespace section and i am trying to create one more loop from the json results output. but it was taking one secret for all the projects
variables:
...ANSWER
Answered 2022-Mar-22 at 07:58You have to link secret with projectName:
QUESTION
ITNOA
I try to creating redis enterprise cluster with redis operator.
For declaration of my cluster I write something like below
...ANSWER
Answered 2022-Feb-08 at 10:59Posting comment as the community wiki answer for better visibility
Is it possible that you had previously created a Redis Enterprise Cluster with the same name before? I am thinking the PVC could be from a previous run. Can you check if the PVC is older than the REC by comparing their creation timestamp?
QUESTION
Skaffold should pull the image from insecure Harbor registry running on HTTP. I have tried everything from these docs:
https://skaffold.dev/docs/environment/image-registries/#insecure-image-registries
but without success.
Jib is pushing image to the insecure Harbor registry without a problem, but error is thrown when trying to pull the image and deploy microservice to Kubernetes:
192.168.2.24:30002/trm/redis-spring:latest@sha256:0f8d21819d845bd55aa699afa8b21e141d41f10d9d9fb1a2c6dbb2d468d89e81 can't be pulled.
Specified image can be pulled using docker:
docker pull 192.168.2.24:30002/trm/redis-spring:latest@sha256:0f8d21819d845bd55aa699afa8b21e141d41f10d9d9fb1a2c6dbb2d468d89e81
- Skaffold version: v1.35.1
- Operating system: Windows 10 Home
- Installed via: skaffold.dev
- Contents of skaffold.yaml:
ANSWER
Answered 2022-Feb-02 at 20:15You need to configure a registry pull secret for your cluster, and then either annotate your pod-specs or your service account to use this registry pull secret.
QUESTION
I'm trying to use Harbor
registry with SCDF 2.9.1
in microk8s 1.18.20
.
I successfully configure SCDF to retrieve the Docker Labels of my apps by adding this in SCDF server Config Map :
ANSWER
Answered 2022-Jan-12 at 09:20The solution is quite simple : I only had to copy the Harbor ca.crt
file (downloaded from Harbor UI) in the /ets/ssl/certs
directory.
You can also create the secret directly without configuring Docker daemon with something like :
QUESTION
I have this script and I would like to print a single title before executing the conditional if
My code
...ANSWER
Answered 2021-Dec-27 at 16:05Since there was no input example, I used your "Output I have" as input.
I also checked if the whole line contains the word terror or bird, but you can change it if you need the column where it is.
QUESTION
Here is small portion of my data in dictionary format.
...ANSWER
Answered 2021-Dec-25 at 16:34The day that has the highest number of complaints can be find like this.
QUESTION
Here is how my data looks like.
...ANSWER
Answered 2021-Dec-24 at 17:51First make that dictionary a Pandas Dataframe:
QUESTION
As you see, there is a map and I want to pathfinding to identify which cities are liked each other.
The yellow tiles in the map are the Land, and blue ones are the ocean. The red font means there is a waterway, and the green font means there is a road. The correct path should be linked as road-road, waterway-waterway, road-harbor-waterway or waterway-harbor-road. Therefore,
2,6City can link to 2,4City via (2,6City)-(1,6)-(0,6)-(1,5)-(2,5)-(3,4Harbor)-(2,4City),
2,6City can link to 0,0City via (2,6City)-(1,6)-(0,6)-(1,5)-(2,5)-(3,4Harbor)-(2,4City)– (1,4)-(0,3City)-(0,2)-(0,1)-(0,0City),
2,6City can link to 3,0City via (2,6City)-(1,6)-(0,6)-(1,5)-(2,5)-(3,4Harbor)-(3,3)– (3,2)-(4,1Harbor)-(3,0City).
However, when I use GKGridGraph to create a map for Pathfinding, I don’t know how to tell the situation that waterway is not accessible to road. You can see, I DON‘T want:
2,6City can link to 2,4City via (2,6City)-(2,5)-(2.4City) or
2,4City is linked to 2,2City because (2,4City)-(3,4Harbor)-(3,3)-(3,2)-(2,2City)
So, any suggestion? Thanks a lot.
...ANSWER
Answered 2021-Dec-18 at 08:33If you are using GKGridGraph(fromStartingGridAt:width:height:diagonalsAllowed:)
to create your graph then every node has connections to each of its neighbors. So a better picture of your map would be this:
Since every node is connected it will find the shortest path which will pass through the water. To prevent that you need to remove the impossible connections. Your graph should look something like this:
To achieve that you cam simply remove the unwanted connections from your graph using GKGraphNode.removeConnections(to:bidirectional:)
This could be achieved by a function like this one:
QUESTION
I have a map that has some overlapping points. I'm using a quadtree and turf.js to determine, when I click on a point, how many other points are within a 30 mile radius.
What I'd like to do (if there's more than one point in that radius) is use d3.forceSimulation to equally spread out the overlapping points.
This is a pretty close example of what I want to do, but uses d3v3 and google maps: http://bl.ocks.org/cdmahoney/raw/9876525/?raw=true
I have included d3.forceSimulation and when I click on a place with more than one point within the 30 mile radius, the points do take on the force-- but they move up to the left hand corner of the page.
How can I get the points to push out in even distances from where I click on the map, like this:
Help very much appreciated!!
...ANSWER
Answered 2021-Dec-05 at 06:03While I'd be tempted to not use a force layout for this, I'll work with the code you have here (though the question of the lines connecting the circles to their original location is not addressed here) and quickly address why the circles do not behave as you expect.
A force layout will create the appropriate properties on a node if they don't exist. For position of a node, these properties are d.x and d.y. Your data does not have x or y properties, so when you create the force, the nodes are initialized with values around the origin, [0,0], which is why they migrate to the top left corner. This problem can be solved by creating x and y properties:
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