instance

ambiguous class with namespace names in 2 dlls

由 流过昼夜 提交于 2019-12-03 15:18:18
问题 I've imported 2 dlls to my application (third party) Now both of them have a namespace with same name. For example A.B and in both of them there is a class again with a same name. Now I want to create an instance of one of them, but because the namespace and class names are same, the compiler goes ambiguous. How can I specify witch dll used in the place? 回答1: Let's suppose that you have 2 assemblies ( ClassLibrary1.dll and ClassLibrary2.dll ) that both define the same class in the same

codeigniter instance of model class

由 给你一囗甜甜゛ 提交于 2019-12-03 12:37:02
问题 I'm developing a site with codeigniter. Now, normally when you use a class in codeigniter, you basically use it as if it were a static class. For example, if I head a model called 'user', I would first load it using $this->load->model('user'); and than, I could invoke methods on that user class like $this->user->make_sandwitch('cheese'); in the application that I'm building, I would like to have one UserManagement class, which uses a class called 'user'. so that, for example I could $this-

Does it make sense to have a non static method which does not use an instance variable?

由 霸气de小男生 提交于 2019-12-03 11:35:34
问题 The compiler does not let a static method call a non static method. I understand it does this because a not static method usually ends up using an instance variable. But does it make sense to have a non static method which does not use an instance variable. If we have a behavior which does not affect or isn't affected by the instance state , shouldn't such a method be marked as static. 回答1: Often times, no. If the method doesn't touch any instance state, there's no reason to tie it to an

Why does my AWS EC2 Instance terminates when stopped?

由 喜你入骨 提交于 2019-12-03 09:44:56
问题 Everytime I stop my AWS EC2 instance, it terminates automatically with a few seconds. Additionally a new instance is created. Can anyone suggest why this would be happening? 回答1: Because it is part of an Auto Scaling group. According to AWS Support, it is normal behavior to Terminate an instance when you tell it to Stop if it is part of an Auto Scaling group. My Termination Protection was set to Stop but apparently it overrides this setting. In order to Stop an instance that's part of a Auto

How to test same object instance in Javascript?

由 ﹥＞﹥吖頭↗ 提交于 2019-12-03 09:43:19
Say I have the following objects in Javascript: var a = { xxx: 33 }; var b = { xxx: 33 }; var c; c = a; What is the Javascript test that will tell me whether I am dealing with the same object instance? In other words, it should return false for a and b, b and c, but true for a and c. You just need this if(c == a) { // same instance } a == b and b == c will return false Just a standard equality test: ( a == c ) // true ( a == b ) // false 来源： https://stackoverflow.com/questions/26214966/how-to-test-same-object-instance-in-javascript

Multiple docker containers for mysql or one instance with multiple databases

由 …衆ロ難τιáo~ 提交于 2019-12-03 08:21:52
问题 I have a question regarding best practices using docker containers. I need one database for each application I develop. Now my question is whether I should use one mysql docker instance with multiple databases inside or should I create one instance for each database. The disadvantage I see with creating one instance per database is that I can't have a user which has access to all database. I know this is a pro for security reasons but when I try to backup from a client than I need to go in

__init__ and arguments in Python

由 折月煮酒 提交于 2019-12-03 06:32:29
问题 I want to understand arguments of the constructor __init__ in Python. class Num: def __init__(self,num): self.n = num def getn(self): return self.n def getone(): return 1 myObj = Num(3) print myObj.getn() RESULT: 3 I call the getone() method: print myObj.getone() RESULT: Error 'getone()' takes no arguments (1given). So I replace: def getone(): return 1 with def getone(self): return 1 RESULT:1 This is OK. But getone() method needs no arguments. Do I have to use meaningless argument? 回答1: In

Example of an instance method? (Java)

由 最后都变了- 提交于 2019-12-03 06:27:31
I'm still learning about methods in Java and was wondering how exactly you might use an instance method. I was thinking about something like this: public void example(String random) { } However, I'm not sure if this is actually an instance method or some other type of method. Could someone help me out? If it's not a static method then it's an instance method. It's either one or the other. So yes, your method, public void example(String random) { // this doesn't appear to do anything } is an example of an instance method. Regarding and was wondering how exactly you might use an instance method

How to access a static method from a instance method in mongoose?

由 半城伤御伤魂 提交于 2019-12-03 05:40:55
How to access a static method from a instance method in mongoose? I have a job model named Job. From a instance method job.start I want to call the static method Job.someStatic(). How do I get the reference to the Job, from the "this" in the instance method? thanks The only way I've found to do that generically (without just calling Job.someStatic() ) is: this.model(this.constructor.modelName).someStatic(); Update thanks to @numbers1311407: I don't know if it's always been the case, but as of at least Mongoose 3.6.11, you can shorten this to: this.constructor.someStatic(); Mongoose 4.x Update

Local Declaration “x” hides instance variable xcode warning

由 感情迁移 提交于 2019-12-03 04:29:07
I've been have trouble understand this problem. If I change the variable name fifthViewController the error goes away but the view controller doesn't load. Lost. Once again it's probably something simple. Thanks in advance. Here is the code: { FifthViewController *fifthViewController = [[FifthViewController alloc] initWithNibName:@"FifthView" bundle:nil]; fifthViewController.transactionID = transactionID; [self.navigationController pushViewController:fifthViewController animated:NO]; [fifthViewController release]; } This means the variable fifthViewController is named the same as something
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