## NINTEX Workflow Loop and Loop Exit When Condition Completed SharePoint 2016

I am creating a Nintex workflow for a SharePoint task list that sends a notification between specific date ranges, for example. If the due date of a task is greater than 4 months, a reminder notification is sent monthly. If the due date is less than 4 months but greater than 1 month, another notification is sent every 2 weeks. If the due date is less than one month but more than one week, a notification is sent each week.

After each notification, there is a pause to wait until the job is done.

All this is set up in a series of loops. My problem is finding a way out of the loop.

What I need to understand is how to get the workflow to loop check if the job is done if the job is done. Exit the loop if the job is done, by sending a notification that the job is done.

Inside Nintex, I do not know what action will monitor the element and, once the task is complete, end the loop and send a notification informing it that the task is complete.

Any help is welcome.

Thank you!

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## plot – 3D plot with an inequality condition for parameter values

I'm trying to `ListPlot3D` the next function

$$f = frac {- frac {(t-1) (ds) left (2 cd (d (q-1) + s) -s left (-2 d ^ 2 + d (s + 2) + s ^ 2 right) right)} {2 s ^ 2} + d ^ 2-d sd t + st} {s}$$

against $$c in [0,1]$$ and $$q in [1,2]$$ under the conditions of $$s = 2$$, $$d = 0.8$$, $$t = 0$$, $$0 leq c leq 1$$, and $$1 leq q leq frac {1} {c}$$.

I'm struggling to know how to reflect the last condition, that is to say $$1 leq q leq frac {1} {c}$$ in my Mathematica code. I used `assumption` but did not work.

Here is the code I tried:

``````Block[{S=2d=08T=0}=f(d-d^2-((ds)(2cd(d(1+q)+s)-s(-2d^2+s^2+d(2+s)))(-1+t))/(2s^2)-dt+st)/s;maxn=Flatten[Table[CQF{}{}{C011Q121assumptions}->{0<c<=11[{s=2d=08t=0}f=(d^2-ds-((d-s)(2cd(d(-1+q)+s)-s(-2d^2+s^2+d(2+s)))(-1+t))/(2s^2)-dt+st)/s;maxn=Aplatir[Table[{cqf}{c011}{q121}hypothèses->{0{0 <c <= 11 <= q<= 1/c}], 1];] {ListPlot3D[maxn, AxesLabel -> {"c", "q", "V"}]}
``````

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## Differential equation without solution that satisfies a certain condition

This is the first week that I deal with differential equations and I am stuck at the next question. I do not know how to approach that, and any help would be greatly appreciated.

Let $$L$$ to be a positive number.

Please show that the equation $$( sqrt {x} + sqrt {y}) sqrt {y} dx = xdy$$ There is no solution that satisfies $$lim_ {x to infty} frac {y (x)} {x} = L$$.

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## views – Drupal 8 – Rule: add a condition to the selection by drop-down list

I have a drop-down menu in a node and the id's drop-down menu is "field_b_status".
I want to add a condition to a rule if the value of the drop-down list is equal to one of the "Full" values, then send an e-mail

When I try to add this, I see the error.
Field_b_status.value data selector for the context The data to be compared is not valid. Can not get field_b_status variable, it is not set. Can you help me how to add the dropdown list "field_b_status" in the condition?

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## html – Update the value of the field in MySql using PHP if condition

I have a table in MySql named "Registrationwhere it has the following fields:

• login – Type: primary key
• statusFK – Values ​​1 (On) and 2 (Off) —> If this field is linked to a table called "status", that is to say it is a foreign key in the "cadastre" table

In case I have created an index.php page with an HTML table containing these three fields and that I would like "dataCadastro"is less than the current date, the field statusFK was automatically updated to 2 (disabled).

However, I use the following code but it updates all the values ​​of the "statusFKto 2 (off) which in the case would be only the field with a date lower than today's date:

``````if (date (Y-m-d, strtotime (\$ row['dataCadastro'])) < date('Y-m-d')){

\$connection->query ("UPDATE Register SET statusFK =" 2 ");

}

``````

Below is the connection code to the database where the table Registration is located:

``````

``````

How could I change my code above for such? Thank you

## pde – wave equation with non homogeneous boundary condition

Recently, I encountered the following problem on a domain $$x> 0$$, $$t> 0$$:

$$u_ {tt} = u_ {xx}$$

$$u_x (0, t) = -h (t)$$

$$u (x, 0) = u_t (x, 0) = 0$$

$$h (t)$$ is an arbitrary function. It has been given that the solution has the form $$u (x, t) = F (x-t) + G (x + t)$$. Filling the initial conditions gives you the fact that F and G must be constant for the positive arguments.

But for the negative arguments, it's more difficult – the problem needs to be extended to an infinite domain. I know how to do this for a Dirichlet or Neumann condition, but I can not handle a non-homogeneous boundary condition.

Any help to solve this problem would be highly appreciated!

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## without context – Confusion with CNF's first condition

I had a very short question about CFG (especially the CNF attributes). I've gone through a few examples and I've found a few that baffle me. Here is an example:

S → XA | BB

B → b | SB

X → b

A → a

It is stated that this is well in CNF, but my confusion lies in the fact that, in most of the rules, it is stated that if the starting state S exists in some RHS (in this case B → b | SB), we need to create a rule that: states S →. As this does not exist in this example, why is it considered part of the CNF? I also understand the rules of the CNF, I also see that this example technically satisfies all these rules, so I wonder if that's the reason?

Rules:

A → a

A → BC.

Thanks for the help in advance!

Here is the link to the question for reference (it concerns CFG TO GNF)
https://www.geeksforgeeks.org/converting-context-free-grammar-greibach-normal-form/

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## Mathematical Optimization – Modify / Optimize a Double Sum with a If Condition

I would like to better understand the double summations of which one of the sums depends on the upper limit of the previous sum. This frequently appears in the theory of representation (to the extent of my knowledge) and in some systems of physics. With representations of the form $$[a,b,c]$$ or even of a simpler form, it is tricky – at least for me – as I explain below.

Ideally, I would like to understand how I can effectively implement something like the following example of representations.

Consider the formula:

$$[0,p_1,0] otimes [0,p_2,0] = sum_ {k_1 = 0} ^ {p_1} sum_ {k_2 = 0} ^ {p_1-k_1} [k_1,p_2-p_1+2k_2,k_1]$$
with $$p_1 leq p_2$$.

It is essentially a decomposition of representations; each term $$[a,b,c]$$ is a representation to be more concrete.

Here is a minimal and non-trivial example of what he should give:

$$[0,2,0] otimes [0,2,0] = [0,0,0] oplus [0,2,0] otimes [0,4,0] otimes [2,0,2] otimes [1,0,1] otimes [1,2,1]$$

What I have implemented is the following:

``````break down[p1_, p2_] : =
Module[{x1 = p1, x2 = p2},
If[x1 <= x2,
Sum[Print[k1, x2 - x1 + 2 k2, k1], {k1, 0, x1}, {k2, 0, x1 - k1}

,
Impression["Wrong values for the p1 and the p2"]]]
``````

And to achieve the above example, one must perform the simple

``````break down[2, 2]
``````

If you try to execute this command, the absolute timing is $$0,000328$$- you get the correct decompositions and none are missing. Once the representations are printed, I receive the following message:

``````6 Null
``````

Six is ​​the number of terms; which is great and very useful because I wanted to implement a command m indicating the number of terms that I would get after the decomposition. I understand why I receive the Null. This comes from the way I wrote the sum. However, I have not been able to solve the problem. Namely, if I polish the implementation of the sum, the code does not work.

Punchline: I wish I could run the order and get something like -for the example above-

``````There are 6 channels

000

020

040

101

121

202
``````

We know that on every $$2n$$three-dimensional vector $$V$$, the 2-form $$omega in Lambda ^ {2} (V ^ {*})$$ is symplectic tensor if and only if $$omega ^ {n}$$ to be non-zero.
leave Manifold, is there a condition for $$omega ^ {n}$$ guarantees that $$omega$$ to be a symplectic form? (proximity condition)