views – Debugging Pager when it isn’t showing

I have been looking for tips to debug the Pager when it doesn’t show and I have not really found anything as of yet. We have only 1 view that is not displaying the pager in our view no matter what we do. When we test for $pager it is not present at all in the view.

We have the pager turned on on the view but it does not show. Our query returns the results we expect and if we change the url to have page=1, page=2 the results change, however like I have said the pager does not show.

So if anyone has any advice on how to do debug the pager it would be great as at this point I am totally lost.

Ubuntu 18.04 Desktop on VirtualBox – not showing desktop or anything just black screen after odoo installation

I have installed Ubuntu 18.04 Desktop edition on VirtualBox, working fine after guest edition installations from virtualbox too. I installed Odoo 14 using a guide, after successul it showing screen in browser to create first database. at that time I shutdown machine and then start it again but after booting (showing Ubuntu and some dots blow) it is not showing anything on screen, leave it for 10 minutes then used mouse to click, pressing enter key or any other key nothing happened. I have checked on 3 laptop machines and the same behaviour on all ( 1 is Lenovo and other 2 are different versions of HP ). Please help, no screenshot no log files etc. to submit here.
regards

dynamic php menu with hidden sub-categories only showing when the category name is clicked

This is for a wordpress store and I want to show the categories names automatically, but with the sub-categories hidden, the code below works perfectly with the hover effect, now I wish to have the click effect instead. The $cats variable is to use the get_terms() function, but for here to work everywhere I had included a few categories, the categories only go so far as the “third generation”.

<?php $cats = array(array('first', 'sub-first', 'sub-sub-first'),
                    array('second', 'sub-second'),
                    array('third', 'sub-third', 'sub-sub-third')); ?>
<h3>Categories</h1>
<ul class="ul1"> <?php
  foreach($cats as $cat):
    if (isset($cat(0))): ?>
      <li class="li1" onclick="test()"><a href="#"><?php echo $cat(0) ?></a> <?php
        if (isset($cat(1))):?>
          <ul class="ul2">
            <li class="li2"><a href="#"><?php echo $cat(1) ?></a> <?php
              if (isset($cat(2))):?>
                <ul class="ul3">
                  <li class="li3"><a href="#"><?php echo $cat(2) ?></a> </li>
                </ul> <?php
              endif; ?>
            </li>
          </ul> <?php
        endif; ?>
      </li> <?php
    endif;
  endforeach; ?>
</ul>

<style media="screen">
  ul{
    list-style: none;
  }
  a{
    text-decoration: none;
    color:#fff;
  }
  a:hover{
    color:#EF7522;
  }
  .li1{
    background:#1B2332;
    width:200px;
    border:1px solid lightgreen;
    padding:10px;
  }
  .ul2, .ul3{
    display:none;
  }
  .li1:hover .ul2{
    display:block;
  }
  .ul2:hover .ul3{
    display:block;
  }
</style>

for the click effect I had remove from the style the last two hover effects and add the javascript function test();

<script type="text/javascript">
    function test(){
        $(".ul2").css({
            "display":"block"
        });
    } 
</script>

What is happening is that I click a category and all sub-categories are showing, What I wanted is the same as the hover effect, that if I click the first category, I have only the sub-categories of that category and so on.

If someone could help me, I have been twisting my mind over this for a few days already!

partial order – Showing that $F$ is a monotone function

I am currently studying the textbook Principles of Program Analysis by Flemming Nielson, Hanne R. Nielson, and Chris Hankin. Chapter 1.3 Data Flow Analysis says the following:

The least solution. The above system of equations defines the twelve sets
$$text{RD}_text{entry}(1), dots, text{RD}_{text{exit}}(6)$$
in terms of each other. Writing $overrightarrow{RD}$ for this twelve-tuple of sets we can regard the equation system as defining a function $F$ and demanding that:
$$overrightarrow{RD} = F(overrightarrow{RD})$$
To be more specific we can write
$$F(overrightarrow{RD}) (F_text{entry}(1)(overrightarrow{RD}), F_text{exit}(1)(overrightarrow{RD}), dots, F_text{entry}(6)(overrightarrow{RD}), F_text{exit}(6)(overrightarrow{RD}))$$
where e.g.:
$$F_text{entry}(3)(dots, overrightarrow{RD}_text{exit}(2), dots, overrightarrow{RD}_text{exit}(5), dots) = overrightarrow{RD}_text{exit}(2) cup overrightarrow{RD}_text{exit}(5)$$
It should be clear that $F$ operates over twelve-tuples of sets of pairs of variables and labels; this can be written as
$F : (mathcal{P}(mathbf{text{Var}_star times mathbf{text{Lab}_star}))}^{12} to (mathcal{P}(mathbf{text{Var}_star times mathbf{text{Lab}_star}))}^{12}$
where it might be natural to take $mathbf{text{Var}_star} = mathbf{text{Var}}$ and $mathbf{text{Lab}_star} = mathbf{text{Lab}}$. However, it will simplify the presentation in this chapter to let $mathbf{text{Var}_star}$ be a finite subset of $mathbf{text{Var}}$ that contains the variables occurring in the program $mathbf{S_star}$ of interest and similarly for $mathbf{text{Lab}_star}$. So for the example program we might have $mathbf{text{Var}_star} = { x, y, z }$ and $mathbf{text{Lab}_star} = { 1, dots, 6, ? }$.

It is immediate that $(mathcal{P}(mathbf{text{Var}_star times mathbf{text{Lab}_star}))}^{12}$ can be partially ordered by setting
$$overrightarrow{text{RD}} sqsubseteq overrightarrow{text{RD}}^prime text{iff} forall i : text{RD}_i subseteq text{RD}_i^prime$$
where $overrightarrow{text{RD}} = (text{RD}_1, dots, text{RD}_{12})$ and similarly $overrightarrow{text{RD}}^prime = (text{RD}_1^prime, dots, text{RD}_{12}^prime)$. This turns $(mathcal{P}(mathbf{text{Var}_star times mathbf{text{Lab}_star}))}^{12}$ into a complete lattice (see Appendix A) with least element
$$overrightarrow{emptyset} = (emptyset, dots, emptyset)$$
and binary least upper bounds given by:
$$overrightarrow{text{RD}} sqcup overrightarrow{text{RD}}^prime = (text{RD}_1 cup text{RD}_1^prime, dots, text{RD}_{12} cup text{RD}_{12}^prime)$$

It is easy to show that $F$ is in fact a monotone function (see Appendix A) meaning that:
$$overrightarrow{text{RD}} sqsubseteq overrightarrow{text{RD}}^prime text{implies} F(overrightarrow{text{RD}}) sqsubseteq F(overrightarrow{text{RD}})^prime$$
This involves calculations like
$$text{RD}_text{exit}(2) subseteq text{RD}_text{exit}^prime(2) text{and} text{RD}_text{exit}(5) subseteq text{RD}_text{exit}^prime(5)$$
imply
$$text{RD}_text{exit}(2) cup text{RD}_text{exit}(5) subseteq text{RD}^prime_text{exit}(2) cup text{RD}_text{exit}^prime(5)$$
and the details are left to the reader.

Appendix A gives the following definition for monotone function:

The function $f$ is monotone (or isotone or order-preserving) if
$$forall l, l^prime in L_1 : l sqsubseteq_1 l^prime Rightarrow f(l) sqsubseteq_2 f(l^prime)$$

I am trying to do as the author said, and show that $F$ is a monotone function. However, I have so far been unable to make progress. It seems to me that such a proof should proceed by showing that, for some arbitrary element of the set of elements $F(overrightarrow{text{RD}})$, if we use the fact that $overrightarrow{text{RD}} sqsubseteq overrightarrow{text{RD}}^prime$, then we can deduce that said arbitrary element is also an element of the set $F(overrightarrow{text{RD}})^prime$, and so $F(overrightarrow{text{RD}}) sqsubseteq F(overrightarrow{text{RD}})^prime$. However, it seems to me that the textbook is very poorly written, and so it is difficult for me to even understand what said arbitrary elements of the set $F(overrightarrow{text{RD}})^prime$ even are (they seem to be some kind of cartesian product, but I get very confused when trying to figure out precisely what they are). So how is it shown that $F$ is a monotone function?

Google Authenticator on two devices showing different codes after scanning the same QR code

When I scan my QR code to enable my 2FA security on some of my trading accounts, I do it on 2 devices (my old phone, and new phone) for safety in case I lose my new phone. My issue is that the two devices then show two different codes and only one of them is correct/works.

I have updated all the apps on both so that should not be the issue.

Has anyone else encountered this problem or got any advice on it?

music.app – Music 1.0.5.10 on MacOS Catalina 10.15.7 not showing ‘Up Next’ list in Mini Player

I click the ‘up next’ button, I get history. I don’t see ‘up next’. I see nothing in view, preferences or window menus. Available docs refer to this as the ‘up next’ button and do not indicate how to switch from/to history. Is there a mystery meat UI control I am missing?

In IOS I note that the UI provides a way to switch from history and up next: after scrolling to the bottom of the history list, the ‘up next’ list appears beneath it.

plotting – VectorPlot and VectorPlot3D showing blank output

I am trying to plot the following simple vector field
$$sqrt{4-x^2-y^2-z^2}*(y,-x,0)$$
but when I plug this field into the functions VectorPlot and VectorPlot3D, the output graph is completely blank. The MWE im using is

ClearAll(expr)
expr(x_, y_, z_) := 
 N@Piecewise({{Sqrt(4 - (x^2 + y^2 + z^2)) {y, -x, 0}, 
     4 > (x^2 + y^2 + z^2)}}, {0, 0, 0})
VectorPlot(expr(x, y, 0), {x, -4, 4}, {y, -4, 4})
VectorPlot3D(
 Evaluate@expr(x, y, z), {x, -5, 5}, {y, -5, 5}, {z, -2, 2}, 
 VectorPoints -> Fine)

with output:

enter image description here enter image description here

which is no good :(.

I tried defining my vector field using the ConditionalExpression function:

ClearAll(expr)
field = Sqrt(4 - (x^2 + y^2 + z^2)) {y, -x, 0}
expr(a_, b_, c_) := 
 ConditionalExpression(field, Im(field) == {0, 0, 0}) /. {x -> a, 
   y -> b, z -> c}
expr(4, 4, 4)
VectorPlot3D(expr(x, y, z), {x, -5, 5}, {y, -5, 5}, {z, -2, 2}, 
 VectorPoints -> Fine)

Now, this actually produces something, but its largely nonsense (even though it contains what Im looking for):

Undefined

enter image description here

I entered expr(4,4,4) to verify it returns undefined, as it should. However, VectorPlot3D is still plotting something. I don’t even know what it could be plotting in the area where it should be empty.
Perhaps Mathematica is doing something strange and somehow evaluating the imaginary part and then plotting it? I really don’t know.

I appreciate any help!

2019 – SharePoint people picker columns not showing as hyperlinks in modern view

In SharePoint modern experience, you get hover experience for person or group field.

When you hover over a column value, it opens a pop-up showing group or user details like below:

enter image description here

Workaround:

You can use column formatting (JSON formatting) to format the list view and generate a link which will redirect to group settings. you can refer below sample JSON to get started with:

{
  "$schema": "https://developer.microsoft.com/json-schemas/sp/column-formatting.schema.json",
  "elmType": "div",
  "children": (
    {
      "elmType": "a",
      "txtContent": "@currentField.title",
      "attributes": {
        "href": "=@currentWeb + '/_layouts/15/people.aspx?MembershipGroupId=' + @currentField.id",
        "target": "_blank",
        "title": "@currentField.title"
      }
    }
  )
}

Note:

  1. This might not work when you add office 365 group or person to your field. I have only tested this with SharePoint groups.
  2. If you have allowed multiple values for your column then you need to enhance this code accordingly using iterators (see below reference no.1)

References:

  1. Multi-Person Facepile
  2. Use column formatting to customize SharePoint

showing $mu(limsup_{ntoinfty} E_n) = 0.$

Let $(X,mathcal{M},mu)$ be a measure space. Suppose $E_nin mathcal{M}$ such that $$sum_{n=1}^infty mu(E_n) < infty$$ show $mu(limsup_{ntoinfty} E_n) = 0.$

Also, how can I prove or give a counterexample of the conclusion if the hypothesis is replaced with $$sum_{n=1}^infty mu(E_n)^2 < infty$$

Can I prove this by Fatou’s Lemma? Thank you