So I had to switch SharePoint from one server to another.  Our network guy installed SharePoint onto a new server and I backed up and uploaded my stuff into the new server (with the same name as the original one).  Everything seems to be working alright except for the Business Data Connectivity Service.  I keep getting this error when I click on it:  The load balancer is not provisioned. Verify the load balancer was provisioned successfully before retrieving endpoint addresses.    I have tried stopping and starting the service, deleting and re-adding the service, changing the app pools and users it uses, restarting IIS, rebooting the entire server...  I'm at a loss right now and I'm not sure what to do.  Are there any other options or ideas out there that I can try?  EDIT  When I try installing the hotfix, it says that it is not applicable
When I click on 'Business Data Connectivity Services' link on 'Manage Service Application' page of Central Administration, I get the following error : "The load balancer is not provisioned. Verify the load balancer was provisioned successfully before retrieving endpoint addresses. " I got a solution from  link, but as per instruction I am unable to install the hotfix for SharePoint 2013. Is there any solution available for this problem? Thanks in advance
How can you tell when a sunset is going to be spectacular? I have a spot that I like to take sunset pictures. I find that the most spectacular moments last for only maybe 5-10 minutes. What I would like to do is know about a half hour before that a sunset is going to be spectacular, so I can grab my camera, tripod, walk 5-10 minutes to my sunset spot, set up, and have a few minutes to spare in case something happens. However, I've often missed really cool sunsets, because I just don't know when to photograph them. How can you tell beforehand when a sunset is going to be spectacular? Thanks! Companion affinity change notification Is it possible to match encrypted documents using user-defined search terms? Suppose I am storing a number of encrypted documents in a database. I would like to make it possible to identify the subset of documents whose contents match user-specified search terms without a) maintaining a plaintext index or b) decrypting documents on the fly. 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So in the the table of content there will be a line saying List of figures ... 19 for example? How to control your iPad with your Mac remotely? I would like to know, is there a way to control your iPad with your Mac remotely (be able to see and manipulate iPad's desktop on your Mac)? I know that with the TeamViewer or other programs you can control your Mac with your iPad but not vice versa. Can a coin with an unknown bias be treated as fair? Finding the missing digits of $23!$ It is given $23!=2585201xy38884976640000$. Now it is required to find the value of $x$ and $y$. I know I could find it by using divisibility rules and solving simultaneous equations. Is there any other way to solve it (without computing it by a calculator)? This question is just out of curiosity. 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Login with Facebook using Stack Exchange Android App I've got two tables:  TableA ------ ID, Name  TableB ------ ID, SomeColumn, TableA_ID (FK for TableA)   The relationship is one row of TableA - many of TableB.  Now, I want to see a result like this:  ID     Name      SomeColumn  1.     ABC       X, Y, Z (these are three different rows) 2.     MNO       R, S   This won't work (multiple results in a subquery):  SELECT ID,        Name,         (SELECT SomeColumn FROM TableB WHERE F_ID=TableA.ID) FROM TableA   This is a trivial problem if I do the processing on the client side. But this will mean I will have to run X queries on every page, where X is the number of results of TableA.   Note that I can't simply do a GROUP BY or something similar, as it will return multiple results for rows of TableA.   I'm not sure if a UDF, utilizing COALESCE or something similar might work? Proving that $m+n\sqrt{2}$ is dense in $\mathbb R$ The divisibility rule for $3$ is well-known: if you add up the digits of $n$ and the sum is divisible by $3$, then $n$ is divisible by three.  This is quite helpful for determining if really large numbers are multiples of three, because we can recursively apply this rule:  $$1212582439 \rightarrow 37 \rightarrow 10\rightarrow 1 \implies 3\not\mid 1212582439$$ $$124524 \rightarrow 18 \rightarrow 9 \implies 3\mid 124524$$  This works for as many numbers as I've tried.  However, I'm not sure how this may be proven.  Thus, my question is:     Given a positive integer $n$ and that $3\mid\text{(the sum of the digits of $n$)}$, how may we prove that $3\mid n$? Error Code: 1215. Cannot add foreign key constraint (foreign keys)