I have a requirement where my current table has  id     value 1      newyork 1      boston 1      dallas   I need the following output  id      value 1       newyork, boston, dallas
I'm trying to migrate a MySQL-based app over to Microsoft SQL Server 2005 (not by choice, but that's life).  In the original app, we used almost entirely ANSI-SQL compliant statements, with one significant exception -- we used MySQL's group_concat function fairly frequently.  group_concat, by the way, does this: given a table of, say, employee names and projects...  SELECT empName, projID FROM project_members;   returns:  ANDY   |  A100 ANDY   |  B391 ANDY   |  X010 TOM    |  A100 TOM    |  A510   ... and here's what you get with group_concat:  SELECT      empName, group_concat(projID SEPARATOR ' / ')  FROM      project_members  GROUP BY      empName;   returns:  ANDY   |  A100 / B391 / X010 TOM    |  A100 / A510   So what I'd like to know is: Is it possible to write, say, a user-defined function in SQL Server which emulates the functionality of group_concat?   I have almost no experience using UDFs, stored procedures, or anything like that, just straight-up SQL, so please err on the side of too much explanation :)
Determining the name of the VMware host of a VM guest - from the guest Which is correct: "has died" or "died"? Get the name of an object's type How do I prove that a graph is a topological manifold I'm having problems with that:  Prove that M is a topological manifold.  $f:U \to \mathbb R^k$, U $\subset \mathbb R^{n}$ open, f continuous  M = $\{\left(x,y\right)\in \mathbb R^{n+k} \mid x \in U, y=f(x) \}$  Can anyone help me please? Show that $d_2$ defined by $d_2(x,y)=\frac{|x-y|}{1+|x-y|}$ is a metric Double value returns 0 The internal storage on my phone is almost full. The storage screen tells me    2.77GB / 2.91 GB taken Apps 0.89GB Other 1.89 GB Available 138MB   What is this "other" and how can I clean it up? I already followed the steps of "make more space", it barely makes a difference.   Phone Storage then tells me 43.00MB / 1.09GB is taken.  And on the SD card I have 1.86GB / 14.72GB taken.  I have plenty of storage in various places yet when I try to install apps I'm unable to do so cause of "insufficient storage". How can I fix this? Alignment of axis in tikz I want to show that $$\log N&lt;\sum_{n=1}^{N}\frac{1}{n}&lt;1+\log N.$$ But I don't know how to show this. Assign VNC password using script I am using a script to automatically set up a computer. I need to assign a password to the vnc server for the user, which is normally done using the vncserver command. However, it prompts for the user to enter and re-enter their password, neither of which the script is capable of doing. So, how can I set up the VNC password without an interactive prompt?   Thanks. Sometimes I create Material Groups to be used as appendable reusable assets. Unfortunately sometimes parameters I want to control using a Group Node's slider are not available for routing to a Group Node's input socket. Some examples of these are the Color Ramp and the Mapping Node.  So here's what I want to do:    Can this be done? If so, how?  Thanks! Is the solution to a question regarding four smaller masses on a rotating hollow sphere accurate? I was doing the question :     A uniform 8.40-kg spherical shell, 50.0 cm in diameter has four small   2.00-kg masses attached to its outer surface and equally spaced around it. This combination is spinning about an axis running through the   center of the sphere and two of the small masses (see ). What friction torque is needed to reduce its angular speed   from 75.0 rpm to 50.0 rpm in 30.0 s?   I understand how to get $\alpha$, and how to get $\tau$ from $I$ and $\alpha$. What I do not understand is their computation of the moment of inertia. The question did not provide any information regarding the shapes of the masses, but the solution (which writes $I=\frac{2}{3}MR^2+2mR^2$) seems to assume that the masses are point particles. Thus, the two masses along the rotational axis have a moment of inertia of 0 ($r=0$), and each of the other two masses has a moment of inertia of $I=mr^2 =mR^2$, where $R$ is the radius of the sphere and $r$ is the distance from the axis.  But why are the masses considered to be point particles? I wanted to use the parallel-axis theorem to determine the moment of inertia about the rotational axis; not knowing the shape of the masses made this impossible, since I couldn’t find the moment of inertia about the CM. If I knew $I_{CM}$, then couldn’t I use $I_{axis}=I_{CM}+mR^2$ for the two masses not on the rotational axis and $I_{CM}$ for the two that are? That would make $I_{total}=\frac{2}{3}MR^2+4I_{CM,\:mass}+2mR^2$ Would that not be a better way of solving this problem? Does anyone know the general strategy for summing a series of the form:  $$\sum_{n=1}^{\infty} \frac{1}{n^2 + a^2},$$  where $a$ is a positive integer?  Any hints or ideas would be great! Prove $| \sum_{i \leq n} \frac{\mu(i)}{i} | \leq 1$ So I am trying to prove the above stated formula. I am completely new to number theory so I thank you in advance for baring with me. I start off by using the definition:  $\frac{\phi(n)}{n} = \sum_{d|n} \frac{\mu(d)}{d}$  Where $\phi$ is Eulers function, and $\mu$ is the Mobius mu function. From here, I use the Mobius inversion formula to give:  $\frac{\mu(n)}{n} = \sum_{d|n}d\frac{\phi(\frac{n}{d})}{n}\mu(d)$  Now summing $n$ up to $x$:  $\sum_{n \leq x}\frac{\mu(n)}{n} = \sum_{n \leq x}\sum_{d|n}d\frac{\phi(\frac{n}{d})}{n}\mu(d)$  Now I'm not fully sure on how I can rearrange things inside the two summations on the right hand side, so leaving things as they are; using that $\sum_{d|n} \mu(d) = 1$, if and only if $n =1$. Hence:  $\sum_{n \leq x}\frac{\mu(n)}{n} = \sum_{d|1}d\phi(\frac{1}{d})\mu(d)$  Thus $d=1$. Subbing this in:  $\sum_{n \leq x}\frac{\mu(n)}{n} = \phi(1) = 1$  I am not sure what I have missed (or done wrong) as to have an equality and not a triangle relation. Any suggestions or ideas would be greatly appreciated What's the term in music, where the singer uses words without meaning? I've been searching for the word, which is used for describing parts of lyrics, where the singer sings for example the whole refrain only using words of syllables without any meaning. Maybe you could help? (here's one song with that type of refrain: ) Are there Ubuntu security notices feeds for specific releases I changed permissions of a file (chmod g+w testfile) and running ls -l testfile gives:  -rwxrwxr-x 1 user1 user1 0 2011-01-24 20:36 testfile   I then added a user to that group ("/etc/group" has user1:x:1000:user2 line), but am failing to edit that file as user2. Why is this so? How do you get the current time in milliseconds (long), in VB.Net? Summation of Central Binomial Coefficients divided by even powers of $2$ Whilst working out  the following summation emerged: $$\sum_{m=0}^n\frac 1{2^{2m}}\binom {2m}m$$ The is equivalent to  $$\begin{align} \sum_{m=0}^n \frac {(2m-1)!!}{2m!!}&amp;=\frac 12+\frac {1\cdot3}{2\cdot 4}+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}+\cdots +\frac{1\cdot 3\cdot 5\cdot \cdots \cdot(2n-1)}{2\cdot 4\cdot 6\cdot \cdots \cdot 2n}\\ &amp;=\frac 12\left(1+\frac 34\left(1+\frac 56\left(1+\cdots \left(1+\frac {2n-1}{2n}\right)\right)\right)\right) \end{align}$$  and terms are the same as coefficients in the expansion of $(1-x)^{-1/2}$.  Once the solution  $$ \frac {n+1}{2^{2n+1}}\binom {2n+2}{n+1}$$ is known, the telescoping sum can be easily derived, i.e.    $$\frac 1{2^{2m}}\binom {2m}m=\frac {m+1}{2^{2(m+1)-1}}\binom {2(m+1)}{m+1}-\frac m{2^{2m-1}}\binom {2m}m$$  However, without knowing this a priori, how would we have approached this problem? I received some text that is encoded, but I don't know what charset was used. Is there a way to determine the encoding of a text file using Python?  deals with C#.