## How to use group rights access

I don’t understand why I can’t write into my file. It has write access for a group I’m member of

``````\$ ll my_file
-r--rw-r-- 1 myusername editorial 0 Oct 23 14:31 my_file
\$echo toto >> my_file
zsh: permission denied: my_file
\$ groups
``````

Any ideas ?

## Rotate group of duplicate bones in edit mode

This is the first time rigging anything in Blender. I’m attempting to rig a spider. I have created a set of 6 bones for one leg. Now I simply want to duplicate those bones and place and rotate them about 20 degrees over the next leg. I can duplicate the bones and translate them, but I can’t figure out how to rotate them as a unit to align them with the next leg without destroying their relative alignment and positions to one another. I’m looking for a shortcut to rigging each leg. Is this possible?

## Pull back group cohomology onto handle decomposition

A construction encountered in the Dijkgraaf-Witten invariant uses the following ingredients:

• An oriented, (assumed here to be smooth) manifold $$M^n$$
• A finite group $$G$$ (and a field, chosen to be $$mathbb{C}$$ here)
• An $$n$$-th cohomology class $$[omega] in H^n(G, U(1))$$ (where $$U(1)$$ is considered as a discrete group here)
• A homomorphism $$phicolon pi_1(M) to G$$

There is a canonical (up to homotopy) map $$ccolon M to Bpi_1(M)$$.
We can then abstractly construct the cohomology class $$c^* phi^* ([omega]) in H^n(M, U(1))$$.

I’m interested describing this cohomology class very concretely. Assume I have given:

• $$M$$ as a handle decomposition (and consequently we consider Morse cohomology, where the $$k$$-th grade of the complex is generated by the $$k$$-handles)
• $$phi$$ as an assignment of group elements to each 1-handle of $$M$$ (satisfying the relations for 2-handles)
• $$[omega]$$ represented through a concrete cocycle $$omegacolon G^n to U(1)$$, i.e. a cocycle in the simplicial cohomology of $$BG$$

How can I explicitly pull back $$[omega]$$ along $$phi circ c$$?

Some remarks:

• A handle decomposition of $$M$$ also gives a CW-complex, I think. So $$ccolon M to Bpi_1(M)$$ can be chosen canonically as a cellular map via the Postnikov construction, and it’s the pullback on cohomology can be derived from the pullback on the level of complexes.
• The trouble seems to be that $$phi$$ isn’t obviously cellular on higher cells if we describe $$BG$$ as a simplicial complex (which we might have to at some point because $$[omega]$$ is given that way).

## Group Policy Confusion (Printers configured under users?)

I’m befuddled. Active directory usually makes a bit more sense than this.

The Problem: Referencing the images below (with private info removed), it appears that Printer Connections are being defined beneath User ConfigurationPoliciesWindows Settings, but “Printer Connections” doesn’t exist in that location in the editor view of the policy. I’ve been asked to remove the printer definitions from the User Configuration section of the policy, and I’m stuck.

Additional Info: The main reason for removing the printer definitions from User configuration is to reduce boot time & log in time for VPN users. The policy being altered is the Default Domain Policy as is visible in the images. We have other Policies in other OUs (everything else has been blanked for security purposes), but this one is the relevant policy. The functional level of the domain is Windows Server 2008. The domain controllers are both Server 2008 Enterprise without Hyper-V. Note: Server 2008 R2 is not implemented. If I left out something pertinent, leave a comment & I’ll comment/respond or edit accordingly.

Question: Where is the correct location to configure user-based printer configuration and/or printer definition deployment?

Bonus points: Why is it displaying in this manner? This just doesn’t make sense to me.

## MySQL, would like to get both a field and a count inside GROUP_CONCAT

I can get a list of statuses that a tracking ticket has had, as well as the id of those statuses, as follows:

``````SELECT
t.id, t.name, GROUP_CONCAT( CONCAT(s.id, ':', s.name) ORDER BY s.id) AS
all_statuses_history
FROM tracking t LEFT JOIN changelog c ON t.ID = c.tracking_id
LEFT JOIN statuses s ON c.changed_to = s.id
``````

That will have duplicates (status can bounce around in the real world), so I change the `GROUP_CONCAT` with `DISTINCT` keyword:

``````GROUP_CONCAT(DISTINCT CONCAT(s.id, ':', s.name) ORDER BY s.id) AS .. (etc)
``````

That’s better, but what I’d REALLY like is the id, status name, and count of instances, looking something like this:

``````1:Requested(1), 2:Pending(3), 3:On Hold(2), 4:Completed(1)
``````

You’d read that as “the ticket bounced around between pending and on hold a bit and was finally completed”.

I can’t put `COUNT(*)` or apparently `GROUP BY` inside of a GROUP_CONCAT function. Is it possible to get all of this information inside of a GROUP_CONCAT?

## Group data for subtotals with (technically) inconsistent categories

I found a tutorial online (link > german-tutorial) which explains how to group data in a table and most importantly how to make subtotals.

This is all fine when the data I have is equal to another, for example:

Clearly this can be grouped because there’s equal value’s that excel can detect.

As a human I can still group this to the categories Data A, B and C, but how do I tell excel that these cells correspond to a group, which I need to make subtotals of?

While writing this I had the idea that I might need to “trim away” some of the text and put the variants in the next cell of the row, which then again would make subtotals easier again because the first column contains the similar data again. I don’t know how to do this dynamically though. Here’s a handwritten example anyhow.

Trimming the variants into another column like that makes subtotals great again, but in that case I don’t know the formula to put in that cuts away the last 2 characters in column A and putting them in column B. I could only get the following result by manually deleting the -1 and -2 and putting it in column B.

I would like to know how I can achieve the above either by not having to alter the variants (“trimming”) or dynamically trimming away the variants and putting them in another column, since I don’t really know a formula that does this.

## How to assign customer group automatically based on address information

How to assign customer group automatically following by address information

When customer register on our website, customer write the country where they live.
If customer live in United State, we would set the standard group(just default group).
If not like live in Jamaica, Spain, we’d like to assign to international group automatically.

So, it would show like this,

It seems like if-else statement on any languages but i didn’t touch the code because i fix it myself one by one.

How can i do it?
app/code/core/Mage/Customer/Model
have the information about that i guess.

## What is the automorphism group of the additive group of the p-adic integers?

Sorry if this is an easy one, I’m a little rusty on my group theory. My first guess was that it’s simply the inverse limit of the Aut(\$mathbb{Z}/p^imathbb{Z})\$, with the map when \$ileq j\$ given by taking \$sigmain\$ Aut\$(mathbb{Z}/p^jmathbb{Z})\$ to the map \$tilde{sigma}:mathbb{Z}/p^imathbb{Z}rightarrowmathbb{Z}/p^imathbb{Z}\$ defined by solving \$phicircsigma=tilde{sigma}circphi\$ where \$phi:mathbb{Z}/p^jmathbb{Z}rightarrowmathbb{Z}/p^imathbb{Z}\$ is the reduction map, but that seems too optimistic – I couldn’t think of any reason \$tilde{sigma}\$ would be well-defined, much less be an automorphism of \$mathbb{Z}/p^imathbb{Z}\$.

Also, barring a full answer to my question, I would be interested in whether Aut\$(mathbb{Z}_p)\$ is a \$p\$-group. If not, what can we say about the elements \$sigmain\$ Aut\$(mathbb{Z}_p)\$ with order a power of \$p\$?

## ReadOnly Access for members of Active Directory Group

I have a Linux host that runs on CentOS 7, joined to domain through `realm`.

My goal is to provide ReadOnly access to one of the AD groups say “LinuxRO-USR”, the users of this group should only be able to sign on using their AD credentials and access their home directory along with other non-sudo commands.

I know adding the following to the sudoers file will also give Sudo access to the users of that group.
`%LinuxRO-USR@example.com ALL=(ALL) ALL`.

Whereas, I want to restricate the access of LinuxRO-USR group such that if one of it’s users tries to run any sudo command like `sudo fdisk -l`, he should not be provided with the output of it. How can I achieve this?