网易云阅读、网易文漫拟以1.5亿元卖身上市公司平治信息



  • 投中教育11月9日讯,平治信息(300571.SZ)11月8日晚间发布公告称,公司拟以1.5亿元收购网易云阅读业务全部核心资产及网易云漫100%股权。

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    根据公告内容显示,网易杭州、妖鹿科技应在不晚于交割日后6个月内将网易云阅读业务全部核心资产均剥离至网易文漫,并将网易云阅读业务对应的供应商和客户资源、关系一并迁移至网易文漫,且网易文漫无需支付对价;从而实现网易文漫在业务、资产、人员、财务、机构等方面实现独立,自主开展经营活动,具备持续盈利能力。已注入网易云阅读业务全部核心资产后的网易文漫100%股权的总对价为人民币1.5亿元。只有在公司经过尽职调查,且双方就收购条款经过充分讨论、协商并达成一致后,各方才能签署正式的、具有法律效力的正式交易文件。

    平治信息成立于2002年,于2016年A股上市,主营业务具体包括移动阅读业务、资讯类业务及其他增值电信业务,其中移动阅读业务为目前着力发展推广的核心业务。

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    文章来源: https://www.toutiao.com/group/6757221441101365774/


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  • I'm working on a deserialization class in .NET, I have to develop a method that provides to me with a variable name that is stored in a string.

    I have a string such as:

    string string_name = "this_is_going_to_be_var_name";

    Now what can I do so that my code dynamically declares a variable named this_is_going_to_be_var_name?

    So to clear things up: There will be a deserialization class that will declare variables of the same names as strings provided as input with their PARENT TYPES as per wish of the Higher Level Programmer/User.

    For Example: In javascript/jQuery, when I fetch JSON by making a request, the interpreter declares variable(s)/array(s) of the same name and assigns values to them. If {"var_name":"var_value"} is a JSON string, the interpreter will create a variable named var_name and will assign "var_value" to it such as json_data_object.var_name.

    No you can't. C# variables are all statically declared.

    The best thing you can do is create a dictionary and use keys instead of variable names.

    // Replace object with your own type Dictionary<string, object> myDictionary = new Dictionary<string, object>(); myDictionary.Add("this_is_going_to_be_var_name", value_of_the_variable); // ... // This is equivalent to foo($this_is_going_to_be_var_name) in PHP foo(myDictionary["this_is_going_to_be_var_name"]);

    This isn't possible, variable names are defined at compile time, not run time. One approach is to create a dictionary or hash table to map string names to objects to sort of achieve what you want.

    Not sure what you meant by

    my code dynamically declares a variable named this_is_going_to_be_var_name?

    but the .Net version of what explode does in PHP is Split:

    string[] zz = "this_is_going_to_be_var_name".Split('_');

    yoel halb

    The only thing that I can think on the moment (I haven't tested it so I have no clue if it is possible), is to have a object of type dynamic and then try to set the fields at runtime using reflection and InvokeMember(), which I can give a chance that it will work since the there is no validation on objects of type dynamic.

    **UPDATE:**I have tested it with ExpendoObject and InvokeMember doesn't appear to work (at least not with the default binder, but I haven't tested it with DynamicObject and allthough I don't give it to much chances to work you might still try it, you can check out http://msdn.microsoft.com/en-us/library/ee461504.aspx on how to use DynamicObject.

    Take a look on Dynamically adding properties to an ExpandoObject which essentially describes a method in which the dynamic object is being cast as a IDictionary and then you can add properties to it by using the standard dictionary access, while they are actually getting properties of the object.

    I tested it in a sample project by having a dynamic object of type ExpendoObject and then I add another variable that referenced it using type IDictionary, and then I tried to set and get properties on both, as in the following example:

    dynamic test = new ExpandoObject(); //reference the object as a dictionary var asDictinary = test as IDictionary<string, Object>; //Test by setting it as property and get as a dictionary test.testObject = 123; Console.Write("Testing it by getting the value as if it was a dictionary"); Console.WriteLine(asDictinary["testObject"]); //Test by setting as dictionary and get as a property //NOTE: the command line input should be "input", or it will fail with an error Console.Write("Enter the varible name, "); Console.Write("note that for the example to work it should the word 'input':"); string variableName = Console.ReadLine(); Console.Write("Enter the varible value, it should be an integer: "); int variableValue = int.Parse(Console.ReadLine()); asDictinary.Add(variableName, variableValue); Console.WriteLine(test.input);//Provided that the command line input was "input"

    (Still however in your case since you will anyway not access the properties directly in code I don't see the need for it and you would probably be able to use a Dictionary directly, and I don't understand why you need them to be properties of the object, which is only needed if you want to reference them at compile time.

    But maybe I am misunderstanding you and you are looking for a dynamic variable and not for a dynamic property [something that is available in PHP using the $$ syntax], if this is the case then note that in c# there are no variables at all as everything is encapsulated in a object).

    You can also take a look in How can I dynamically add a field to a class in C# for more answers.

    来源:https://stackoverflow.com/questions/11888144/name-variable-using-string-net

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  • 殿

    I have a project in which I want to disable the cells of a UITableView at the beginning when the view loads, and after pressing a button, the cells of the UITableView should be enabled and should work properly.

    How can I do this?

    You can use table.allowsSelection = NO; in your viewwillappear method and then make it YES onbutton click event.

    If you want to make entire table disable then u can use:

    table.userInteractionEnabled = FALSE;

    And then on click on the button:

    table.userInteractionEnabled = TRUE; //In UIButton pressed event boolEnabled=YES; //Declare it in header file [yourTableView reloadData]; //reload UITableView - (UITableViewCell *)tableView:(UITableView *)tableView cellForRowAtIndexPath:(NSIndexPath *)indexPath { //After creating a cell if(boolEnabled==YES) //Will enable for any action { cell.userInteractionEnabled=YES; } else { //disable for any action cell.userInteractionEnabled=NO; //Default boolEnabled=NO; } }

    What I would do is have a @property (nonatomic) BOOL tableIsActive

    In my - (UITableViewCell *)tableView:(UITableView *)tableView cellForRowAtIndexPath:(NSIndexPath *)indexPath I would do

    - (UITableViewCell *)tableView:(UITableView *)tableView cellForRowAtIndexPath:(NSIndexPath *)indexPath { if(self.tableIsActive) { //style cells for enabled table } else { //style cells for enabled table } }

    And then in my - (void)tableView:(UITableView *)tableView didSelectRowAtIndexPath:(NSIndexPath *)indexPath I would

    - (void)tableView:(UITableView *)tableView didSelectRowAtIndexPath:(NSIndexPath *)indexPath { if(self.tableIsActive) { //handle selection } else { [tableView deselectRowAtIndexPath:indexPath animated:NO]; } }

    And then I would have a function -(void) enableTable:(id)sender which I bind to my button that does

    -(void) enableTable:(id)sender { if(self.tableIsActive) self.tableIsActive = NO; else self.tableIsActive = YES; }

    And then after enabling call [yourTable reloadData];

    You can do it by using the below code , Just check for the condition in TableView delgate method

    - (UITableViewCell *)tableView:(UITableView *)tableView cellForRowAtIndexPath:(NSIndexPath *)indexPath { static NSString *categoryIdentifier = @"Category"; static NSString *CellIdentifier = @"Cell"; UITableViewCell *cell = [tableView dequeueReusableCellWithIdentifier:CellIdentifier]; if (cell == nil) { cell = [[[UITableViewCell alloc] initWithStyle:UITableViewCellStyleDefault reuseIdentifier:CellIdentifier] autorelease]; } //Way to stop to choose the cell selection if(indexpath.row==0) { cell.selectionStyle = UITableViewCellSelectionStyleNone; Or cell.userIneractionEnabled=False; } //while rest of the cells will remain active, i.e Touchable return cell; }

    来源:https://stackoverflow.com/questions/9974485/how-to-enable-and-disable-cells-in-uitableview

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  • 问题

    This question already has an answer here:

    What exactly is Java EE? 4 answers

    I realize that literally it translates to Java Enterprise Edition. But what I\'m asking is what does this really mean? When a company requires Java EE experience, what are they really asking for? Experience with EJBs? Experience with Java web apps?

    I suspect that this means something different to different people and the definition is subjective.

    回答1:

    136

    Java EE is actually a collection of technologies and APIs for the Java platform designed to support "Enterprise" Applications which can generally be classed as large-scale, distributed, transactional and highly-available applications designed to support mission-critical business requirements.

    In terms of what an employee is looking for in specific techs, it is quite hard to say, because the playing field has kept changing over the last five years. It really is about the class of problems that are being solved more than anything else. Transactions and distribution are key.

    share|improve this answer

    edited Aug 2 '17 at 14:23

    Gab - reinstate Monica 6,37422 gold badges2828 silver badges5858 bronze badges

    answered Sep 20 '08 at 2:43

    Toby HedeToby Hede 34.2k2626 gold badges122122 silver badges159159 bronze badges

    What exactly goes into this collection and in what version of Java EE is nicely presented on this Java EE Version History page. – Johnny Baloney Apr 19 '16 at 3:44

    1

    Jakarta EE is the new name for Java EE. – Viraj Mar 7 '18 at 17:00

    1

    The term "mission-critical" reeks of defense contractor – Kolob Canyon Apr 23 '18 at 15:01

    well java EE is just like ruby on rails EE or javascript EE chose any language today and just call it EE that it sounds niceer....no other language needs some Marketing buzzwords like enterprise edition just to say hey i can do web and db. "I know java" "wait, you know Java or Java EE" .. WTF its like braging i can do http request in javascript or acces db in rails.. wow magic. but business ppl love the words ENTERPRISE EDITION .... drop the EE Java! – H.R. May 14 at 15:07

    add a comment |

    回答2:

    224

    Java EE is a collection of specifications for developing and deploying enterprise applications.

    In general, enterprise applications refer to software hosted on servers that provide the applications that support the enterprise.

    The specifications (defined by Sun) describe services, application programming interfaces (APIs), and protocols.

    The 13 core technologies that make up Java EE are:

    JDBC JNDI EJBs RMI JSP Java servlets XML JMS Java IDL JTS JTA JavaMail JAF

    The Java EE product provider is typically an application-server, web-server, or database-system vendor who provides classes that implement the interfaces defined in the specifications. These vendors compete on implementations of the Java EE specifications.

    When a company requires Java EE experience what are they really asking for is experience using the technologies that make up Java EE. Frequently, a company will only be using a subset of the Java EE technologies.

    share|improve this answer

    edited May 15 '16 at 14:06

    BalusC 898k314314 gold badges32803280 silver badges33053305 bronze badges

    answered Sep 20 '08 at 19:04

    user19685user19685 2,32911 gold badge1313 silver badges77 bronze badges

    34

    +1 Nice list. BTW, I believe JPA (Java Persistence API) should also be in the list. – sleske May 17 '11 at 19:48

    3

    The last point added much strength to you r answer thanks +1 – Faisal Naseer Jun 23 '15 at 8:50

    1

    According to Oracle's website, JNDI is included in the Java SE Platform. (docs.oracle.com/javase/tutorial/jndi/software/index.html#JDK). – ROMANIA_engineer Nov 19 '15 at 11:38

    1

    JDBC is part of Java SE actually. – Tiny Dec 27 '15 at 15:05

    1

    According to this picture ( docs.oracle.com/javase/8/docs/index.html 😞 JDBC, JNDI and RMI are in Java SE. – ROMANIA_engineer Jan 6 '16 at 19:42

    | show 4 more comments

    回答3:

    35

    J(2)EE, strictly speaking, is a set of APIs (as the current top answer has it) which enable a programmer to build distributed, transactional systems. The idea was to abstract away the complicated distributed, transactional bits (which would be implemented by a Container such as WebSphere or Weblogic), leaving the programmer to develop business logic free from worries about storage mechanisms and synchronization.

    In reality, it was a cobbled-together, design-by-committee mish-mash, which was pushed pretty much for the benefit of vendors like IBM, Oracle and BEA so they could sell ridicously over-complicated, over-engineered, over-useless products. Which didn't have the most basic features (such as scheduling)!

    J2EE was a marketing construct.

    share|improve this answer

    answered Oct 23 '08 at 22:43

    oxbow_lakesoxbow_lakes 124k5050 gold badges297297 silver badges436436 bronze badges

    I don't know any development tool / software / platform / framework more over-complicated than that. If you need something simple but very easy then use PHP. If you need something powefull and complex but still easy then use .Net. – Eduardo Sep 5 '14 at 2:42

    This is a good post. No doubt IBM WebSphere Commerce / Oracle ATG – TheBlackBenzKid Dec 15 '14 at 15:44

    1

    I know this is an old answer, but as people are still reading it: it's a looooong time ago that Java EE's main selling point was distributed transactional systems. Even in '08, but surely today it's about REST APIs (JAX-RS), Validation (Bean Validation) easy persistence (JPA) MVC web frameworks (JSF, MVC) and a coherent extendible bean model (CDI). – dexter meyers Jan 11 '16 at 13:44

    add a comment |

    回答4:

    22

    There are 2 version of the Java Environments, J2EE and Se. SE is the standard edition, which includes all the basic classes that you would need to write single user applications. While the Enterprise Edition is set up for multi-tiered enterprise applications, or possible distributed applications. If you'd be using app servers, like tomcat or websphere, you'd want to use the J2EE, with the extra classes for n-tier support.

    share|improve this answer

    answered Sep 20 '08 at 2:37

    stephenbayerstephenbayer 10.9k1313 gold badges5555 silver badges9898 bronze badges

    9

    Don't forget J2ME! – Dave L. Sep 20 '08 at 3:50

    6

    I guess Tomcat is still a Servlet Container and not an App server like Jboss/ Weblogic. – Soumya Feb 5 '13 at 10:58

    add a comment |

    回答5:

    12

    It's meaning changes all the time. It used to mean Servlets and JSP and EJBs. Now-a-days it probably means Spring and Hibernate etc.

    Really what they are looking for is experience and understanding of the Java ecosystem, Servlet containers, JMS, JMX, Hibernate etc. and how they all fit together.

    Testing and source control would be an important skills too.

    share|improve this answer

    answered Sep 20 '08 at 2:43

    macarthymacarthy 3,04211 gold badge1919 silver badges2424 bronze badges

    add a comment |

    回答6:

    7

    Yes, experience with EJB, Web Apps ( servlest and JSP ), transactions, webservices, management, and application servers.

    It also means, experience with "enteprise" level application, as opposed to desktop applications.

    In many situations the enterprise applications needs to connect to with a number of legacy systems, they are not only "web pages", and with the features availalble on the "edition" of java that kind of connectivity can be solved.

    share|improve this answer

    answered Sep 20 '08 at 2:38

    OscarRyzOscarRyz 172k100100 gold badges351351 silver badges528528 bronze badges

    add a comment |

    回答7:

    3

    J2EE traditionally referred to products and standards released by Sun. For example if you were developing a standard J2EE web application, you would be using EJBs, Java Server Faces, and running in an application server that supports the J2EE standard. However since there is such a huge open source plethora of libraries and products that do the same jobs as well as (and many will argue better) then these Sun offerings, the day to day meaning of J2EE has migrated into referring to these as well (For instance a Spring/Tomcat/Hibernate solution) in many minds.

    This is a great book in my opinion that discusses the 'open source' approach to J2EE http://www.theserverside.com/tt/articles/article.tss?l=J2EEWithoutEJB_BookReview

    share|improve this answer

    answered Sep 20 '08 at 3:27

    PeterPeter 27.1k1515 gold badges7777 silver badges114114 bronze badges

    add a comment |

    回答8:

    3

    I would say that J2EE experience = in-depth experience with a few J2EE technologies, general knowledge about most J2EE technologies, and general experience with enterprise software in general.

    share|improve this answer

    answered Sep 20 '08 at 18:25

    James A. N. StaufferJames A. N. Stauffer 2,54333 gold badges2020 silver badges3131 bronze badges

    add a comment |

    回答9:

    0

    Seems like Oracle is now trying to do away with JSPs (replace with Faces) and emulate Spring's REST (JAX-RS) and DI.

    ref: https://docs.oracle.com/javaee/7/firstcup/java-ee001.htm

    Table 2-1 Web-Tier Java EE Technologies

    JavaServer Faces technology

    A user-interface component framework for web applications that allows you to include UI components (such as fields and buttons) on a XHTML page, called a Facelets page; convert and validate UI component data; save UI component data to server-side data stores; and maintain component state

    Expression Language

    A set of standard tags used in Facelets pages to refer to Java EE components

    Servlets

    Java programming language classes that dynamically process requests and construct responses, usually for HTML pages

    Contexts and Dependency Injection for Java EE

    A set of contextual services that make it easy for developers to use enterprise beans along with JavaServer Faces technology in web applications

    share|improve this answer

    edited Mar 17 '16 at 20:59

    user393219

    answered Mar 25 '15 at 18:37

    killjoykilljoy 59466 silver badges1515 bronze badges

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