follow | Very stable , very reliable , NodeJS CouchDB
kandi X-RAY | follow Summary
kandi X-RAY | follow Summary
Follow (upper-case F) comes from an internal Iris Couch project used in production for over a year. It works in the browser (beta) and is available as an NPM module.
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Top functions reviewed by kandi - BETA
- Make a request
- Main entry point for changes .
- Search listeners for listeners .
- response callback function
- grow the listener tree
- handle the feed
- Exports a file to the target module
- Base64 encode string
- Handle a change event .
- Handle pending changes
follow Key Features
follow Examples and Code Snippets
def follow(username):
form = EmptyForm()
if form.validate_on_submit():
user = User.query.filter_by(username=username).first()
if user is None:
flash(_('User %(username)s not found.', username=username))
def click_to_follow(browser):
# my_follow_btn_xpath = "//a[contains(text(), 'Follow')][not(contains(text(), 'Following'))]"
# my_follow_btn_xpath = "//*[contains(text(), 'Follow')][not(contains(text(), 'Following'))]"
my_follow_btn_xpath
Community Discussions
Trending Discussions on follow
QUESTION
I have been trying to learn about functional programming, but I still struggle with thinking like a functional programmer. One such hangup is how one would implement index-heavy operations which rely strongly on loops/order-of-execution.
For example, consider the following Java code:
...ANSWER
Answered 2022-Mar-07 at 21:17This is not an index-heavy operation, in fact you can do this with a one-liner with scanl1 :: (a -> a -> a) -> [a] -> [a]
:
QUESTION
In my flutter project, I have made some updates of plugins and then used flutter upgrade. After that, whenever I am running my flutter project it is showing following error-
...ANSWER
Answered 2021-Dec-16 at 11:49For me, cleaning and getting the packages didn't work. This error started after I upgraded flutter. I was on the master channel, a quick fix for me was to switch to stable.
QUESTION
I am getting this create React app error again and again even after doing the uninstall part.
npm uninstall -g create-react-app
up to date, audited 1 package in 570ms
found 0 vulnerabilities
npx create-react-app test-app
...Need to install the following packages: create-react-app Ok to proceed? (y) y
You are running
create-react-app
4.0.3, which is behind the latest release (5.0.0).We no longer support global installation of Create React App.
Please remove any global installs with one of the following commands:
- npm uninstall -g create-react-app
- yarn global remove create-react-app
The latest instructions for creating a new app can be found here: https://create-react-app.dev/docs/getting-started/
ANSWER
Answered 2022-Jan-01 at 22:34You will have to clear the npx cache to make it work.
You can locate the location of the folder where create-react-app is installed using npm ls -g create-react-app
.
Also, to clear the cache, refer to this answer in How can I clear the central cache for `npx`?
QUESTION
When I publish my ABP project I get the following error:
...ANSWER
Answered 2022-Jan-13 at 21:59Issue:
The issue raises after .NET 6 migration. There's a new feature that blocks multiple files from being copied to the same target directory with the same file name. See https://docs.microsoft.com/en-us/dotnet/core/compatibility/sdk/6.0/duplicate-files-in-output
Solution #1 (workaround):
You can add the following build property to all your publishable (*.Web) projects' *.csproj files. This property will bypass this check and works as previously, in .NET5.
QUESTION
In earlier versions, we had Startup.cs class and we get configuration object as follows in the Startup file.
...ANSWER
Answered 2021-Oct-26 at 12:26WebApplicationBuilder
returned by WebApplication.CreateBuilder(args)
exposes Configuration
and Environment
properties:
QUESTION
When trying to build my project I am getting the following error:
Could not GET
'https://jcenter.bintray.com/androidx/lifecycle/lifecycle-common/maven-metadata.xml'.
Received status code 502 from server: Bad Gateway
- In my build.gradle repositories I don't have JCenter, so this error I'm getting is from dependencies that are still pointing to JCenter.
- Gradle offline mode is not the solution I'm expecting.
- I know that JCenter is down and that we should all move to Maven Central (I already did)
Is there a workaround?
...ANSWER
Answered 2022-Jan-30 at 23:31It's a global outage in JCenter. You can monitor status at https://status.gradle.com. It replaces the bintray status page which seems is now fully sunset and returns a 502 error.
UPDATE Jan 13, 06:35 UTC
JCenter is now back online, and systems are fully operational.
UPDATE Jan 20
Gradle Plugin resolution outage postmortem
https://blog.gradle.org/plugins-jcenter
Following this incident, the Gradle Plugin Portal now uses a JCenter mirror hosted by Gradle instead of JCenter directly. This should shield users from short JCenter outages for libraries that have been cached by the mirror. We saw another short outage of JCenter over the weekend and this did not appear to impact Gradle Plugin Portal users.
QUESTION
I'm studying for the final exam for my introduction to C++ class. Our professor gave us this problem for practice:
...Explain why the code produces the following output:
120 200 16 0
ANSWER
Answered 2021-Dec-13 at 20:55It does not default to zero. The sample answer is wrong. Undefined behaviour is undefined; the value may be 0, it may be 100. Accessing it may cause a seg fault, or cause your computer to be formatted.
As to why it's not an error, it's because C++ is not required to do bounds checking on arrays. You could use a vector and use the at
function, which throws exceptions if you go outside the bounds, but arrays do not.
QUESTION
Following a previous question of mine, most comments say "just don't, you are in a limbo state, you have to kill everything and start over". There is also a "safeish" workaround.
What I fail to understand is why a segmentation fault is inherently nonrecoverable.
The moment in which writing to protected memory is caught - otherwise, the SIGSEGV
would not be sent.
If the moment of writing to protected memory can be caught, I don't see why - in theory - it can't be reverted, at some low level, and have the SIGSEGV converted to a standard software exception.
Please explain why after a segmentation fault the program is in an undetermined state, as very obviously, the fault is thrown before memory was actually changed (I am probably wrong and don't see why). Had it been thrown after, one could create a program that changes protected memory, one byte at a time, getting segmentation faults, and eventually reprogramming the kernel - a security risk that is not present, as we can see the world still stands.
- When exactly does a segmentation fault happen (= when is
SIGSEGV
sent)? - Why is the process in an undefined behavior state after that point?
- Why is it not recoverable?
- Why does this solution avoid that unrecoverable state? Does it even?
ANSWER
Answered 2021-Dec-10 at 15:05When exactly does segmentation fault happen (=when is SIGSEGV sent)?
When you attempt to access memory you don’t have access to, such as accessing an array out of bounds or dereferencing an invalid pointer. The signal SIGSEGV
is standardized but different OS might implement it differently. "Segmentation fault" is mainly a term used in *nix systems, Windows calls it "access violation".
Why is the process in undefined behavior state after that point?
Because one or several of the variables in the program didn’t behave as expected. Let’s say you have some array that is supposed to store a number of values, but you didn’t allocate enough room for all them. So only those you allocated room for get written correctly, and the rest written out of bounds of the array can hold any values. How exactly is the OS to know how critical those out of bounds values are for your application to function? It knows nothing of their purpose.
Furthermore, writing outside allowed memory can often corrupt other unrelated variables, which is obviously dangerous and can cause any random behavior. Such bugs are often hard to track down. Stack overflows for example are such segmentation faults prone to overwrite adjacent variables, unless the error was caught by protection mechanisms.
If we look at the behavior of "bare metal" microcontroller systems without any OS and no virtual memory features, just raw physical memory - they will just silently do exactly as told - for example, overwriting unrelated variables and keep on going. Which in turn could cause disastrous behavior in case the application is mission-critical.
Why is it not recoverable?
Because the OS doesn’t know what your program is supposed to be doing.
Though in the "bare metal" scenario above, the system might be smart enough to place itself in a safe mode and keep going. Critical applications such as automotive and med-tech aren’t allowed to just stop or reset, as that in itself might be dangerous. They will rather try to "limp home" with limited functionality.
Why does this solution avoid that unrecoverable state? Does it even?
That solution is just ignoring the error and keeps on going. It doesn’t fix the problem that caused it. It’s a very dirty patch and setjmp/longjmp in general are very dangerous functions that should be avoided for any purpose.
We have to realize that a segmentation fault is a symptom of a bug, not the cause.
QUESTION
In the following code, I create two lists with the same values: one list unsorted (s_not), the other sorted (s_yes). The values are created by randint(). I run some loop for each list and time it.
...ANSWER
Answered 2021-Nov-15 at 21:05Cache misses. When N
int objects are allocated back-to-back, the memory reserved to hold them tends to be in a contiguous chunk. So crawling over the list in allocation order tends to access the memory holding the ints' values in sequential, contiguous, increasing order too.
Shuffle it, and the access pattern when crawling over the list is randomized too. Cache misses abound, provided there are enough different int objects that they don't all fit in cache.
At r==1
, and r==2
, CPython happens to treat such small ints as singletons, so, e.g., despite that you have 10 million elements in the list, at r==2
it contains only (at most) 100 distinct int objects. All the data for those fit in cache simultaneously.
Beyond that, though, you're likely to get more, and more, and more distinct int objects. Hardware caches become increasingly useless then when the access pattern is random.
Illustrating:
QUESTION
In C++20, we are now able to constrain the auto
keyword to only be of a specific type. So if I had some code that looked like the following without any constraints:
ANSWER
Answered 2021-Nov-01 at 18:38A constraint on the deduced auto
type doesn't mean it needs to be a specific type, it means it needs to be one of a set of types that satisfy the constraint. Note that a constraint and a type are not the same thing, and they're not interchangeable.
e.g. a concept like std::integral constrains the deduced type to be an integral type, such as int
or long
, but not float
, or std::string
.
If I really need a
std::integral
datatype, couldn't I just omit theauto
completely?
In principle, I suppose you could, but this would at the minimum lead to parsing difficulties. e.g. in a declaration like
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