I have a production database running on MSSQL 2019 on a hosted platform. I am wanting to create a sandbox copy of the db with all of the data in a local MSSQL server running 2014.  Everything I've tried  (backup / restore , generating scripts, export data) has error-ed out.  Does anyone have an easy button I am overlooking?
What are my options if I need to move a database from SQL Server 2012(32bit) to SQL Server 2005(64bit)?  I know I cannot:   restore a backup of the database on SQL Server 2005 detach &amp; attach   I know I can:   use import data wizard, and I tried it on one database, but it only moves data, and even that was troublesome as I needed to do a lot of work creating temp tables to maintain identity columns, recreating all FKs, indexes etc.    Is there an easier option?
Catching poisoned arrows When catching an arrow tipped with poison (like the ones used by yuan-ti) with an ability like the one the monk has or the one given by arrow catching gloves: does the ability reduce the poison damage? If so, is the total damage or the piercing one caused by the arrow only the damage that must be downed to 0 to catch the missile? How to display a block with menu child items programmatically? I keep reading in economics journals about a particular result used in random utility models. One version of the result is: if $\epsilon_i \sim_{iid}, $ Gumbel($\mu, 1), \forall i$, then:   $$E[\max_i(\delta_i + \epsilon_i)] = \mu + \gamma + \ln\left(\sum_i \exp\left\{\delta_i \right\} \right),  $$  where $\gamma \approx 0.52277$ is the Euler-Mascheroni constant.  I've checked that this makes sense using R, and it does.  The CDF for the Gumbel$(\mu, 1)$ distribution is:  $$G(\epsilon_i) = \exp(-\exp(-(\epsilon_i - \mu)))$$  I'm trying to find a proof of this and I've had no success.  I've tried to prove it myself but I can't get past a particular step.   Can anyone point me to a proof of this?  If not, maybe I can post my attempted proof up to where I get stuck. Joined pipelines Considering a routine such as this one:  alpha() { echo a b c |tr ' ' '\n'; }   which outputs a stream, I would like to take the output stream, transform it, and paste it with the original output stream.  If I take use upcasing as a sample transformation, I can achieve what I want with:  $ mkfifo p1 p2 $ alpha | tee p1 &gt;( tr a-z A-Z &gt; p2) &gt;/dev/null &amp; $ paste p1 p2 a       A b       B c       C   My question is, is there a better way to do this, preferably one not involving named pipes? Picturing the how the covariant derivative acts on vector fields Where did they go / do they went? Where did they go?   Where do they went?   Can you please advise which of above forms is correct?  I'm pretty sure that first one is fine but wondering about second one.   Can I use a second form as well or it's incorrect? How to prove $|a+b|^k \leq 2^{k-1} (|a|^k+|b|^k)$? Assume that you have any two real numbers $a$ and $b$, and $1\leq k &lt;\infty$, $k \in \mathbb{R}$.  How would you prove the inequality $|a+b|^k \leq 2^{k-1} (|a|^k+|b|^k)$? What identifies an action-reaction pair of forces? This is a question on a worksheet that I'm giving to a student:  A man of weight $W$ steps into the lift. The lift then moves downwards with a constant speed. If $R$ represents the force acting on the man by the lift floor, which of the following statements is/are correct? (3) $R$ and $W$ are an action-reaction pair according to Newton's third law  Why isn't this statement (3) correct? After the downfall of Voldemort, what happened to the Death Eaters?  They committed many crimes, including mass murder and torture. Nothing in the books made mention of their punishments. What happened? Suppose $\sqrt6 = \frac pq$ where $p$ and $q$ have no common factors.  $$6 = \frac {p^2}{q^2}$$  $$6p^2 = q^2$$  So $q^2$ and therefore $q$ is divisible by $6$.   $$p^2 = \frac {q^2}{6}$$  So $p^2$ and therefore $p$ is divisible by $6$.   S0, $p$ and $q$ have a common factor $6$. Contradiction.  Therefore $\sqrt6$ is irrational.  Does the proof work? The two sentences here both indicate that, at some point in the past, I performed some work:   I did work I worked   What is the difference between these two sentences? Does constructing one with did and the other with a simple past change the meaning? Draw a vector with mathematica How to reinstall yum? Why is feed-forward mechanism used in hash functions? The compression function of SHA-1 when used in Davies-Meyer mode adds its input to the chaining values at the final step. For the first message block, the IV is used as the input and in the next step, the previous block's output (as specified by Merkle-Damgård) is used. These values are fed to the last round of the compression function.  Why would removing the feed-forward part (i.e.,the final step where the input value is added to the result to get the final output) make the hash function insecure? Apart from the fact that the compression function would not satisfy desired randomness properties (being a permutation for each fixed message). Let $G$ be a simply connected algebraic group defined over $C$.   Note : I see the definition of simply connected as "Every isogeny to $G$ is an isomorphism" as given in Hochschild's  "Basic Theory of Algebraic Groups and Lie algebras".   $C$ is an algebraically closed field.   Please help me realise the follwoing things provided they are true:   $G_C$ (the $C$-points of $G$) is simply connected as well. $Lie(G) \cong Lie(G_C)$    Thanks ! is_active_sidebar() Always Returns False I am calculating the class group of $\mathbb Q(\sqrt 6)$. My working is as follows:  The Minkowski bound is $\lambda(6)=\sqrt 6&lt;3$ so we only need to look at prime ideals of norm $2$. $2$ divides the discriminant, $24$, of $\mathbb Q(\sqrt 6)$, so $2$ ramifies, and so for any prime ideal $\mathfrak p$ of norm $2$, $\mathfrak p^2$ is principal. But $(2, \sqrt 6)^2=(2)$, so $\mathfrak p=(2, \sqrt 6)$ has norm $2$. If I can show $\mathfrak p$ is not principal, then I can conclude that $Cl(\mathbb Q(\sqrt 6))\cong C_2.$ But if there was an element of $\mathcal O_{\mathbb Q(\sqrt 6)}$ of norm 2, this would correspond to a solution of the diophantine equation $x^2-6y^2=2$. Clearly $x$ is even, so writing $x=2n$ we get $2n^2-3y^2=1$. Now reducing modulo $3$, we get $2n^2\equiv 1 \pmod 3$ which has no soultions. Hence $\mathfrak p$ is not principal and $Cl (\mathbb Q(\sqrt 6)) \cong C_2.$  However,  says that the class group is trivial. Where have I gone wrong? What am I misunderstanding? Is $(2, \sqrt 6)$ principal? What make is my BMX frame? I have a BMX forsale, but I would like to know what make the frame is so I can determine how much it is worth. Does anyone recognise this symbol on the frame?  Thanks for your help Jon An alternative software to Live Writer? Boolean expression (grammar) parser in c++