bane | better AppArmor profile generator for Docker containers | Continuous Deployment library
kandi X-RAY | bane Summary
kandi X-RAY | bane Summary
AppArmor profile generator for docker containers. Basically a better AppArmor profile, than creating one by hand, because who would ever do that. "Reviewing AppArmor profile pull requests is the bane of my existence" Jess Frazelle.
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QUESTION
I need to create a nested array using the response received from different API. I have this flat array, with all the data. What would be the best approach to create a nested array where the children are nested to its parent based on its result status.
...Input JSON flat array as follows:
ANSWER
Answered 2021-Jun-11 at 12:36A quick and easy way could be as follows:
Script
QUESTION
I have a JBOSS server (7.0) running an application that uses ServiceWorkers, which requires an HTTPS connection. I was able to update the standalone.xml and Eclipse launch configuration to bind my JBOSS server to my local IP (I'll worry about port forwarding later). Connecting to http://192.168.0.197:8080/[application] works just fine, except that ServiceWorkers won't start because it isn't an HTTPS connection. If I try https://192.168.0.197:8080/[application], the connection fails with the browser reporting "unable to connect".
I've researched several documentation sources and can't figure out what needs to be updated. Please forgive any terminology errors - my background is with application programming and networking tends to be the bane of my existence.
This is the pertinent standalone.xml configuration:
...ANSWER
Answered 2021-Jun-10 at 15:15It's there in your configuration:
QUESTION
I have trouble returning results from Entry in tkinter. I do not use it immediately after creating tne Entry:
...ANSWER
Answered 2021-May-22 at 15:02The problem lies in the fact that you don't remember the Entry widgets. You just store them to a local variable in function enchwind()
and don't store them anywhere. They are created but you lose all reference.
You could use a list as an attribute of window
to store them:
In the definition of your window, add the line:
QUESTION
Below is the simple function I am working on.
When I execute the 3 tests called out as examples in the specification, I get an error on the 3rd test (crane to bane).
I know that the error is specifically in the after
variable as I would expect that to just be ane
and instead, I am seeing rane
printed when I run the test.
I have tried the following and each time, these changes have caused unintentional errors in the first two tests called out in the specification.
...ANSWER
Answered 2021-Mar-07 at 21:46The reason why your code is not working for crane
to bane
is because you are not taking into account the length of the input of a
(in this case cr
, therefore length of 2) here:
QUESTION
Have set up fail2ban service on CentOS 8 by this tutorial: https://www.cyberciti.biz/faq/how-to-protect-ssh-with-fail2ban-on-centos-8/.
I have set up settings similiarly according to tutorial above like this:
...ANSWER
Answered 2021-Mar-03 at 15:22A persistent banning is not advisable - it simply unnecessarily overloads your net-filter subsystem (as well as fail2ban)... It is enough to have a long ban.
If you use v.0.11, you can use bantime increment feature, your config may looks like in this answer - https://github.com/fail2ban/fail2ban/discussions/2952#discussioncomment-414693
QUESTION
(I'm sorry if the question involves too much Mathematics, that's the bane of Mathematical programmation !)
My goal is to visualize finite radii balls in the Diestel-Leader graph DL(p,q). Here is an article by W. Woess which describes with pictures the construction (starting from page 3). I will however give it here with less mathematics to capture the idea.
It is constructed by taking two infinite trees Tp and Tq, each of respective valence p+1 and q+1. We consider height functions h:Tp→Z and h':Tq→Z, meaning that each vertex x with height k has exactly one parent vertex with height k-1 and exactly p (resp. q) children with height k+1.
Now, the vertex set of DL(p,q) is defined to be {(x,y)∈Tp×Tq : h(x)+h'(y)=0}, and two vertices (x,y) and (x',y') are adjacent in DL(p,q) if and only if x and x' are adjacent in Tp and y and y' are adjacent in Tq.
It turns out DL(p,q) is not a tree ! Balls whose radii are greater than 4 always have non-contractible loops.
I am fixing arbitrary elements o∈Tp and o'∈Tq whose heights are zero, and I am willing to plot the sub-graph of DL(p,q) consisting of elements whose distance to (o,o') is at most a certain radius R (that is the ball of radius R). This means that I must find all elements of distance at most R to o in Tp, and of distance at most R to o' in Tq, and then somehow get the adjacencies between couples of such vertices.
Also, I am mainly concerned about the case p=2 and q=3, that is DL(2,3). However, an algorithm in the general case would be more than welcome !
So far, I first tried finding a representation for elements in Tp. I tried implementing the representation given in the paragraph (2.1) Tree and sequences. in the article I linked, but even that seemed doomed. Then, I thought about taking a vertex at height -R, and represent all its children up to the generation R (that is, I would have 2R+1 generations of vertices). To represent a child of a vertex, this corresponds to labelling all the downwards edges with an integer from 0 to p-1. I first used a dec2base
conversion algorithm :
ANSWER
Answered 2020-Dec-05 at 18:51It turns out I was very close to finishing ! It boiled down to writing the definition of the edge set for the Diestel-Leader graph :
E[DL(p,q)]={(e1,e2)∈E[Tp]×E[Tq] : h(i(e1))+h'(i(e2))=h(t(e1))+h'(t(e2))=0}
with i(e) denoting the initial vertex of the edge e and t(e) the terminal edge. Therefore, I simply looped through all edges of both trees and checked for that condition :
QUESTION
I am working on a random name feature for naming items, Now i have everything working, Except one piece in the code, Where it displays a const as NaN, Here is my code;
...ANSWER
Answered 2020-Nov-10 at 14:20There are many things wrong with your code
- You declare
nameOneCompleted
inside yourif
so it's undefined elsewhere. Same fornameTwoCompleted
. You should define them in the upper scope - Your if condition is always truthy thus useless
- Use const and let instead of the deprecated var
- You don't need to do 10 ifs, just define an array and pick the n-th element
- Use camel case when naming js variables: its
nameDecode2
, notNameDecode2
nameDecode2
was a string of several characters, you probably meant to only take the first one, you don't need to slicenameDecode1
for that btw, just use string index. Also convert it to a number before indexing the string.- Don't slice when building substrings, just use substring, it's much easier
- I don't know what you tried to do with the ellipsis but it was wrong, also be aware that the CSS
text-overflow: ellipsis
might come more handy than trimming the string on the JS side
That should be good:
QUESTION
Hello I was wondering how I can add in my bots kick command and the rest, the server name or (Guild) they were kicked/baned/warned. From or in...
This is in DM's and the channel the command was used...
My Code:
...ANSWER
Answered 2020-Nov-02 at 06:12Should work
QUESTION
Let's say I have 3 arrays of objects:
...ANSWER
Answered 2020-Oct-25 at 10:20You would need to get two random numbers - the first to decide which group to pick and the second to pick that an item from that. To give one group a higher chance of being picked, the first random number needs to be from a range 1-100 (to allow for percentages) and the group is then determined by a range within that.
For example, to give the fruits group an 80% chance of being picked:
QUESTION
The error code it shows is:
...ANSWER
Answered 2020-Oct-25 at 06:47If spellname
is not in spell.spelldictionary
, i
becomes spell.spelldictionary.length
after the for loop, and the execution of msg.channel.send(spell.spelldictionary[i][1])
causes the error.
You can avoid it by moving msg.channel.send
before break
in the for loop, so that no message will be sent in this case. Also, it is better to explicitly declare i
before using it, or it may cause some unexpected bugs after your code becomes more complicate.
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