java – Continue the flow after throwing the exception

In my use case, I'm going through a map and checking if a particular key is present in a list. If it is present, I have to discard it and, if not, continue with the execution.

Map myMap = new HashMap();
// code to populate values ​​in myMap
List myList = new ArrayList();
// code to fill values ​​in myList
for (Map.Entry eachElementInMap: myMap.entrySet ()) {
if (myList.contains (eachElementInMap: myMap.getKey ())) {
// Release exception
launch a new MyCustomizedException ("someString");

// the code continues

In the previous example, if there are 3 elements in the map (myMap) in which 1 key is present in the list (myList), I want to throw the exception for one and should continue executing other lines of code for the other two. . Am I using the wrong design to achieve this? Any help or suggestion is appreciated! Thank you

flow – workflow reducing balance

I have a workflow that follows a vacation application form, once approved it reduces the Days used for 1 because I can do that.

However, I want to reduce the Days used For the number of days requested in the application.

Do I need to set a variable or use a function?

This is my first publication, so yell if I need to provide more information.

My workflow is shown below with all the details I have:

Workflow diagram

Design of a data flow for an advanced search of products.

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What software is used for the flow of users?

Does anyone know what software is used to make this user flow?

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I have the Google photo.

mongodb – Handling the renewable message flow pattern (queue)

My question is actually quite generic, although I will use a particular technology Mongodb Change Streams.

Exchange flows in Mongo allow you to obtain a flow of messages of the changes requested in a collection of db, each message contains resume token this allows, in case it is necessary, to interrupt the processes (failure / restart, etc.), restart the process of handling messages and receive all the message that came after this token of resumption.

Just to handle all messages sequentially, the process will probably like that:

  • in a new flow message start to handle it (asynchronously)
  • if a new message arrives while the previous one is being handled, it is placed in an "internal waiting queue"
  • after the message handled, resume token The message handled is stored somewhere (from where we will load it in case of restarting the process)
  • then message taken from the "internal waiting queue" or waiting for a message from the transmission

More complex case if I want to handle some messages not sequentially (for example, sequentially for each aggregate ID) but in parallel for different instances added. As I understand it, a more complex model will be needed to store the managed message resumption tokens (in case of reboot) to resume the handling correctly and not to omit messages and not to handle the same messages several times.

So, the question: is the way I described it is a common pattern for handling resumable message flows? And where is it possible to read about such patterns and cases to clarify things?

Vortex motion equation for flow flow instability

I am solving the equation of motion of an isolated vortex oscillation in a superconductor. I am assuming that the driving force is generated by an RF current that oscillates at 1.3 GHz.
The main point of this calculation is to see what the effect of the instability of the flow of flow in the RF unit is.
The code that I wrote is reported below.

f = 1.3 10 ^ 9; (* Hz *)
[Omega] = 2. Pi f;
Trf = 1./f;

Tc = 9.25; (* K *)
T = 1.5; (* K *)

[Mu]0 = 4. Pi 10 ^ -7; (* H / m *)
[Phi]0 = 2.07 10 ^ -15; (* Wb *)
n = 5.56 10 ^ 28; (* m ^ -3 *)
e = 1.6 10 ^ -19; (*DO*)
m = 9.1 10 ^ -31; (* kg *)

[Tau] = 0.5 10 ^ -9; (* s *)
vf = 1.37 10 ^ 6; (*Mrs*)
Bc2 = 410. 10 ^ -3; (* T *)
l = 100 10 ^ -9; (*subway*)
[Lambda] = 49.5 10 ^ -9; (*subway*)
[Xi] = 28 10 ^ -9; (*subway*)
[Kappa] = [Lambda]/ [Xi];

[Gamma] = ([Phi]0) / ([Mu]0 [Lambda] );

v0 = Sqrt[(l vf Sqrt[14 Zeta[3] (1 - T / Tc)]) / (3 Pi [Tau])];

[Rho] = (m vf) / (n e ^ 2 l);
[Eta] = ([Phi]0 Bc2) / [Rho];

g = Log[[[[[Kappa]]+ 0.5 + Exp.[-04-08Log[[-04-08Log[[-04-08Log[[-04-08Log[[Kappa]]-0.1 (Log[[[[[Kappa]]) ^ 2];
[Epsilon] = ([Phi]0 ^ 2 g) / (4. Pi [Mu]0 [Lambda]^ 2);

Tmax = 3. Trf;
Zmax = 1. 10 ^ -6; (*subway*)

B = 100. 10 ^ -3; (* T *)

sol = NDSolve[{
(v0 ^ 2 [Eta]) / (v0 ^ 2 + D[x[t, z], t]^ 2) D[x[t, z], t]==[Epsilon] re[x[t, z], z, z]+ [Gamma] B Cos[[[[[Omega] t]Exp[-(z/[-(z/[-(z/[-(z/[Lambda])],
X[0., z] == 0.,
X[t, Zmax] == 0.,
(RE[x[t, z], z]/. z -> 0) == 0.
}, x, {t, 0., Tmax}, {z, 0., Zmax},
AccuracyGoal -> 13,
PrecisionGoal -> 2,
MaxSteps -> Infinity];
you[t_, z_] = Evaluate[x[t, z] /. Sun][[1]];
du[t_, z_] = D[u[t, z], t];

phaseSpace =
Parallel table[{Re[u[t, 0.]]10 ^ 6, Re[du[t, 0.]]10 ^ -3}, {t, Trf,
Tmax, Tmax / 500.}];
ListPlot[phaseSpace, Joined -> True, PlotRange -> All]

I can get an almost correct result from this calculation only if I fix the domain of the solution to 1 micron (Zmax = 1. 10 ^ -6). However, this is not correct because the vortex should be able to oscillate more than that within the volume, therefore, literally, I am cutting my solution using too small a domain.

If I fix Zmax to a larger number (for example, 5 um), I need to correct MaxStepSize to a small number to avoid errors and the calculation would take more than 2 days on my PC (actually I stopped it after 2 days because the PC froze !). Does anyone know how to solve this problem?

Also, I was wondering if there is any way to solve the equation when the velocity of the vortex is greater than v0 (the start of the velocity for the instability of the flow of flow). If I fix B to say 200 mT, then the velocity of the vortex would be greater than v0 and the calculation will not yield a correct solution, just a series of peaks. Is there a method I can use to solve this problem?

Please let me know!
Thanks in advance.


Entropy information of zero flow even with given range?

Suppose we have a task Y = X and that the rank of X is [0,255].

I want to know how many bits of information happened in the assignment given that Y> 80 using entropy

If I declare that Y = 80 I get that all 8 bits of X flowed towards Y. On the other hand, if I declare that Y> 80 I still get 8 bits of flow, since it does not matter what the following equation is: H (X | Y = j) = 0.

When I calculate entropy I get a sum of all H (X | Y = j) which is still 0

How is it possible if we now do not know the exact value of Y that the number of bits that flow is the same? I understand that the bits have passed to Y, but how do I take into account the fact that Y is not a specific value in the case of Y> 80?

micropaymentchannels: Does Lightning Network have a concept of continuous value flow?


  • Car insurance: $ 1000 per year.
  • Legal consultation: $ 500 per hour.
  • Visiting a website: $ 100 per hour.

Assuming 1 BTC = $ 8000

1 hour = $ 100 = 0.0125 BTC = 1,250,000 sat.

1 second = 1,250,000 sat / 3600 = 3472.22 sat

I could generate a new invoice and pay it every second.

Continuous flow of value?

I was thinking about something else:

Until further notice, please charge me 3472.22 sat per second

To keep me from disconnecting, there could be 10 seconds of heartbeat.

Is it too complicated?

Due to the speed of light limit and the propagation of the network, 1s The resolution is good enough, without the need for a continuous flow of value.

Issuing a new invoice every second is a simple solution and it is not necessary to complicate excessively.

Or maybe it exists natively?

sharepoint online – Flow – The activator & # 39; For a selected file & # 39; does not fill the dynamic content

Previously, I created several streams that are activated from SharePoint libraries, all starting in the same way. The trigger is "For a selected file" immediately followed by a "Get file properties" action. In the past, the dynamic content of the trigger was available in the action, so I could use the ID of the selected file to get the properties of the file.

Now, when I try to create a new stream, dynamic content is missing in the Get file properties action. If I access one of the Flows that I had previously created, the Dynamic Content is also "missing" there. (Fortunately, the field is still there, and the flows are still working).

Also, if I choose a different action (for example, send me an email notification), the dynamic content of the trigger is filled in perfectly.

So, how can I get the properties of a selected file? Do I have to use a different combination of trigger / action? Can I use an expression of some kind as an alternative solution? Any suggestion is appreciated!


Other action (same flow):

Previously created flow (DC used to be there, but not anymore):
Old flow

dg.differential geometry – A vector field not disappeared in $ S ^ 3 $ whose flow does not retain any transverse foliation

Is there a vector field that has not disappeared? $ X $ in $ S ^ 3 $ Which does not support a transversal. $ 2 $ dimensional foliation? If the answer is negative, is there a vector field that has not disappeared? $ X $ in $ S ^ 3 $ which does not admit a transversal foliation whose leaves are invariant under the flow of $ X $? If the answer is positive, is there an example of this situation with the additional assumption that $ X $ is invariant under the obvious action of $ S ^ 1 $ in $ S ^ 3 $?

The motivation is described in the following publication and the linked document in that publication. (The property "invariable flow leakage" is frequently used in the document and the closed orbits of the flow are related to a trace formula consisting of the cohomology of Leaf De Rham)

Irrational closed orbits of vector fields in $ S ^ 2 $ (limit cycles and trace formula)