We have an AOAG named with "space" and I need to rename it to not have "spaces". It was created before my time. The 3rd party vendor tool does not work with AOAG that has spaces. There are about 40 databases in the AOAG (SharePoint).  The property is greyed out in SSMS. Is there another option? As this is production, I am looking for a solution with minimal downtime.  Thanks much, Debbie
I have two questions for you guys. I really appreciate your help.  1.- Is there a way to rename an Availability Group?  2.- Is there a way to get the T-SQL statements in order to re create a Availability Group in case that I need to drop it and create it again?  If there's information that you want me to share, please let me know and I will provide it.
How to install latest HPLIP on my Ubuntu to support my HP printer and/or scanner? Let's assume that one has some modern HP printer and/or scanner which is not supported by HPLIP package from official repository. Currently Ubuntu versions have :  trusty (14.04LTS): 3.14.3-0ubuntu3.4 xenial (16.04LTS): 3.16.3+repack0-1 bionic (18.04LTS): 3.17.10+repack0-5 focal (20.04LTS): 3.20.3+dfsg0-2 groovy (20.10): 3.20.5+dfsg0-3build1 hirsute (21.04): 3.21.2+dfsg1-2   What can one do if printer is supported in newer version of HPLIP (checked this in )? Transpose/Unzip Function (inverse of zip)? After using gdalwarp to project and align-to-grid (via -tap) a number of rasters I noticed that the output rasters were significantly larger than the original rasters. A fairly thorough web search turned up :  Frank Warmerdam explained the reason: &quot;On careful review, the difference in the file in question is because gdal_translate uses the TIFFWriteScanline() interface to write the output file from within GTiffDataset::CreateCopy?(), and this only writes as much of the final 'strip' of the file as is required to complete image area. But gdalwarp goes through the blockio interface which writes the complete final strip, even the portion that falls off the end of the file.&quot;  This Trac issue is ~7 years old, however, and I know some changes to the GDAL utilities, including gdalwarp have been made since. I'd like to know if the above reasoning still holds and if the file size inflation I'm seeing is &quot;normal.&quot; The word &quot;normal&quot; here might be taken to mean unsurprising or expected but, more importantly: is there anything that can be done to mitigate the effects i.e. reduce the output raster file size? Below is a table of the file size inflation I'm experiencing. Input File Size (bytes)     Output File Size (bytes)    Inflation 1437380431                  1698334217                   18% 1428001178                  1698334433                   19%   41683165                   137036637                  228%  The input TIFF files were created in ArcGIS and thus have external Worldfiles, XML and DBF files but these do not make up the difference in file size. Here is a sample gdalwarp call as I've used it in all of these cases; the actual execution was handled by a Python subprocess (subprocess.Popen): $ gdalwarp -tap -tr 30 30 -t_srs &quot;+proj=aea +lat_1=20 +lat_2=60 +lat_0=40 +lon_0=-96 +x_0=0 +y_0=0 +ellps=GRS80 +datum=NAD83 +units=m +no_defs&quot; -co &quot;COMPRESS=LZW&quot; input_file.tif output_file.tif  I understand that in rare cases compression makes a larger file, but the effect is the same without the LZW compression. The ratios in the table are with LZW compression. In  it states that      [xkcd] means calling the Ackermann function with Graham's number as the arguments just to horrify mathematicians.   $A(g_{64},g_{64})$  In  it states that      Even simply making $g_{65}$ dwarfs the number $A( A(g_{64}, g_{64}), A(g_{64}, g_{64}))$ into insignificance.   So:  Is $g_{65}$ greater than $A(g_{64}, g_{64})$?  And is it greater than $A( A(g_{64}, g_{64}), A(g_{64}, g_{64}))$? What is the exact difference between a 'terminal', a 'shell', a 'tty' and a 'console'? I think these terms almost refer to the same thing, when used loosely:   terminal shell tty console   What exactly does each of these terms refer to? The standard files/tools that report memory seem to have different formats on different Linux distributions. For example, on Arch and Ubuntu.   Arch  $ free               total        used        free      shared  buff/cache   available Mem:        8169312     3870392     2648348       97884     1650572     4110336 Swap:      16777212      389588    16387624   $ head /proc/meminfo  MemTotal:        8169312 kB MemFree:         2625668 kB MemAvailable:    4088520 kB Buffers:          239688 kB Cached:          1224520 kB SwapCached:        17452 kB Active:          4074548 kB Inactive:        1035716 kB Active(anon):    3247948 kB Inactive(anon):   497684 kB  Ubuntu  $ free              total       used       free     shared    buffers     cached Mem:      80642828   69076080   11566748    3063796     150688   58358264 -/+ buffers/cache:   10567128   70075700 Swap:     20971516    5828472   15143044   $ head /proc/meminfo  MemTotal:       80642828 kB MemFree:        11565936 kB Buffers:          150688 kB Cached:         58358264 kB SwapCached:      2173912 kB Active:         27305364 kB Inactive:       40004480 kB Active(anon):    7584320 kB Inactive(anon):  4280400 kB Active(file):   19721044 kB    So, how can I portably (across Linux distros only) and reliably get the amount of memory—excluding swap—that is available for my software to use at a particular time? Presumably that's what's shown as "available" and "MemAvailable" in the output of free and cat /proc/meminfo in Arch but how do I get the same in Ubuntu or another distribution? How do I update the GUI from another thread? Which is the simplest way to update a Label from another Thread?   I have a Form running on thread1, and from that I'm starting another thread (thread2).  While thread2 is processing some files I would like to update a Label on the Form with the current status of thread2's work.   How could I do that? PHP & MySQL: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given Find all continuous functions $f$ satisfying   $$\int_0^xf=(f(x))^2+C$$ for some constant $C \neq 0$, assuming that $f$ has at most one $0$.  I have a question about the solution to this problem. It says that   Clearly $f^2$ is differentiable everywhere, its derivative at $x$ is $f(x)$ (I'm assuming that this is because the left hand side is differentiable, the right hand side must be differentiable as well). So $f$ is differentiable at $x$ whenever $f(x) \neq 0$, and  $$f(x)=2f(x)f'(x),$$ so $f'(x) = \frac{1}{2}$ at such points. Thus, $f= \frac{1}{2}x+b$ for some $b$ on any interval where it is non-zero; if $f$ has a zero, this gives two possible solutions, with two possible values of $b$, but since $f$ is assumed continuous, they must be the same. So we need $$\int_0^x(\frac{1}{2}t+b)dt=(\frac{1}{2}x+b)^2+C$$  so we must have $b=\sqrt{-C}$ or $-\sqrt{-C}$, leading to two solutions for $C \lt 0$.  I don't understand the bolded statements. First, how does the differentiability of $f^2$ guarantee that $f$ is differentiable, and why only at $x$ where $f(x) \neq 0$?  Next, why are there two possible solutions if $f$ has a zero, but since $f$ is continuous they must be the same?  Finally, does the last statement mean that if $C \gt 0$, then there is no solution, since we're restricted to real numbers here?  I would appreciate it if anyone answers my questions. Setting knots in natural cubic splines in R I have data with many correlated features, and I want to start by reducing the features with a smooth basis function, before running an LDA. I'm trying to use natural cubic splines in the splines package with the ns function.  How do I go about assigning the knots?  Here's the basic R code:  library(splines) lda.pred &lt;- lda(y ~ ns(x, knots=5))   But I have no idea about how to chose the knots in ns. Other ways to say "I have a bad hunch" I'm looking for ways to say "I'm having a bad hunch", or more like a bad feeling about something upcoming. The gut-wrenching feeling that something bad will happen. chpasswd apply changes in the chroot directory Add a dependency in Maven Software needed to scrape data from graph Anybody have any experience with software (preferably free, preferably open source) that will take an image of data plotted on cartesian coordinates (a standard, everyday plot) and extract the coordinates of the points plotted on the graph?  Essentially, this is a data-mining problem and a reverse data-visualization problem. What is the norm of this bounded linear functional? Let $a$, $b$ be two arbitrary but fixed real numbers such that $a &lt; b$, let $C[a,b]$ denote the normed space of all continuous real (or complex) valued functions defined on $[a,b]$ with the maximum norm, and let $f$ be the functional defined on $C[a,b]$ as $$f(x):= \int_{a}^{b} x(t) x_0(t) dt $$ for all $x$ in $C[a,b]$, where $x_0$ is a fixed element of $C[a,b]$.   Now $f$ is evidently linear and bounded. How to compute the norm of $f$? What is this norm?   And what is the norm of the bounded linear functional $g$ defined as follows: $$g(x):= \int_{a}^{(a+b)/2} x(t) \cdot dt - \int_{(a+b)/2}^{b} x(t) \cdot dt$$ for all $x$ in $C[a,b]$? How to set metadata for music files? In windows music files have metadata like artist name, album name associated with them. How is this metadata stored. Is it stored in file system like other file attributes or is it part of MP3 data that Windows Explorer extracts using a MP3 reader plugin?  Is it possible to read/change this information in Linux? Is it possible with commandline?  (I know these attributes can be set using Rhythymbox or some music player but I'm curious if these could also be set directly. Like in Windows the windows itself shows you artist information if you right click the file and open properties dialog) Proof of -(-v)=v in a vector space The exercise is: Prove that $-(-v)=v$ for every $v \in V$  Proof Suppose  $v \in V$ and $V$ is a vector space. Then $-(-v) \in V$ as result of the scalar multiplication property and $-(-v)=-(-1\cdot v)=-1 \cdot(-1\cdot v) =(-1 \cdot-1)\cdot v = 1 \cdot v = v $ The desired result $-(-v)=v$ holds.  The solution manual gives the proof:  Proof   I just wanted to make sure that the way I did the proof isn't missing anything?Thanks When I slam my hand onto my keyboard, only a maximum of 6 keys will be registered, I've tested this for the past few minutes and was unable to get more than 6. Is this the same on all keyboards? If not, what would be the distinguishing factor, If yes: why 6 specifically?  For clarification: I do not have a usecase for this, I am simply curious Difference between "while" and "whilst" Possible Duplicate:          In terms of construction of sentences, can the two words while and whilst be used interchangeably? Is my model any good, based on the diagnostic metric ($R^2$/ AUC/ accuracy/ RMSE etc.) value? I've fitted my model and am trying to understand whether it's any good. I've calculated the recommended metrics to assess it ($R^2$/ AUC / accuracy / prediction error / etc) but do not know how to interpret them. In short, how do I tell if my model is any good based on the metric? Is an $R^2$ of 0.6 (for example) sufficient to let me proceed to draw inferences or base scientific/business decisions?     This question is intentionally broad, to cover a wide variety of situations that members frequently encounter; such questions could be closed as duplicates of this one. Edits to broaden the scope beyond the metrics mentioned here are welcome, as are additional answers - particularly those that offer insight about other classes of metrics.