javascript - Typescript: Create interface based on a class -

-

i got class called myclass. has bunch of properties. want create interface contains options of class. many options have same name , typing myclass, not all.

edit: options properties optional.

the goal duplicate least possible.

right now, i'm using dummy object avoid code duplication. there cleaner solution this?

class myclass {     callback:(p0:number,p1:string,p2:string[]) => number[];     myattr:number;     composed:string;      constructor(options:myclassoptions){         if(options.callback !== undefined)             this.callback = options.callback;         if(options.myattr !== undefined)             this.myattr = options.myattr;         if(options.composedp1 !== undefined && options.composedp2 !== undefined)             this.composed = options.composedp1 + options.composedp2;     } }  var dummy = <myclass>null;  interface myclassoptions {     callback?:typeof dummy.callback;     myattr?:typeof dummy.myattr;     composedp1:string;     composedp2:string; }

many options have same name , typing myclass, not all.

if case, there's not option.

if want have exactly same members, , class doesn't have private or protected properties/methods, can write

interface myclassoptions extends myclass {   extraprop: string; }

one thing make base class options want share, make interface there "real" class use being derivative of base class.


Comments

Post a Comment

Popular posts from this blog

mysql - Dreamhost PyCharm Django Python 3 Launching a Site -

-
i'm new python/django , created simple django website using pycharm. site works fine running on computer, i'm lost when comes taking site onto host (dreamhost). i'm able upload files correct directories fine, , believe made correct changes database on mysql. i'm not sure go there though. right now, page shows files rather site. direction appreciated. much! you should not run django application using python manage.py runserver on production. how run application on django depends on server (apache, nginx) want use for apache - https://docs.djangoproject.com/en/1.10/howto/deployment/wsgi/modwsgi/ for nginx - https://www.digitalocean.com/community/tutorials/how-to-serve-django-applications-with-uwsgi-and-nginx-on-ubuntu-14-04 the mysql database should, of course, exist , have correct credentials. or, if choose use sqlite, file should accessible, readable , writable www-user (may different depending on linux distributive)
Read more

java - Sending SMS with SMSLib and Web Services -

-
i have made rest web service send smss using hsdpa usb modem. using smslib in java send smss. every time web service invoked, create gateway start service, send message, stop service , remove gateway. takes 20 seconds each message. found starting service takes lot of time. part of code use send smss service.getinstance().addgateway(gateway); service.getinstance().startservice(); outboundmessage msg = new outboundmessage(phonenumber, message); if (service.getinstance().sendmessage(msg)) { result = "message sent successfully!!"; } else { result = "could not send message."; } service.getinstance().stopservice(); service.getinstance().removegateway(gateway);//remove gateway is there way can start service once if not started , use send message when ever web service invoked? why dont group messages , send in 1 go? service.getinstance().sendmessages(messa...
Read more

java - How to resolve The method toString() in the type Object is not applicable for the arguments (InputStream) -

-
i'm trying read json file local machine. @path("/") public class jsonparsing { file f = new file("file.json"); if (f.exists()){ inputstream = new fileinputstream("file.json"); string jsontxt = ioutils.tostring(is); system.out.println(jsontxt); jsonobject json = new jsonobject(jsontxt); string = json.getstring("1000"); system.out.println(a); } } but i'm getting error in tostring() method. possible read .txt file containing json object local machine? if possible, how done? firstly: whenever ask question- add imports well. secondly: yes possible read .txt file containing json object. steps: 1. add maven dependency in pom.xml - <dependency> <groupid>com.googlecode.json-simple</groupid> <artifactid>json-simple</artifactid> <version>1.1.1</version> </dependency> or add jar of dependency. 2.import...
Read more