action

How to pass parameters to action in Flutter

由 久未见 提交于 2021-01-28 09:50:39
问题 I have this action Future<void> signUpAction(Store<AppState> store) async { try { // ... } catch (e) { // .. } } And I dispatch it like this store.dispatch(signUpAction); Now, if I want to pass two paramters, how would I do that? Since there is already one parameter there. I tried this Future<void> signUpAction(Store<AppState> store, email, password) async { try { // ... } catch (e) { // .. } } but then on dispatching, if I do store.dispatch(signUpAction("some@email.com", "somEPa55word!"));

woocommerce_new_order action not returning all order details

由 扶醉桌前 提交于 2021-01-27 18:47:06
问题 I'm using the woocommerce_new_order action to send order details to a Mailchimp list. // Add user to mailchimp list function add_user_to_mailchimp($order_id) { $order = get_post_meta($order_id); write_log($order); } add_action( 'woocommerce_new_order', 'add_user_to_mailchimp', 1, 1 ); An important part to keep in mind is that the user may not necessarily register an account. The most important part is retrieving the the customers email and sending that through to Mailchimp. I want to achieve

Dispatch action on the callback of socket.on()

由 故事扮演 提交于 2021-01-27 12:46:33
问题 So basically I got this socket, that is working correctly sending me 'new order' message. I'm using redux, and i want to dispatch an action, than a reducer would get it and my store would be updated. but this code doesn't do anything! socket.on('new order', (order) => { return (dispatch) => { dispatch(socketNewOrder(order)); } }); And here is my action, which is located at the same file: export const socketNewOrder = (order) => { return { type: 'SOCKET_NEW_ORDER', payload: order } } I also

public Event in abstract class

由 陌路散爱 提交于 2021-01-27 03:53:35
问题 I have event declared in abstract class: public abstract class AbstractClass { public event Action ActionEvent; } public class MyClass : AbstractClass { private void SomeMethod() { //Want to access ActionEvent-- Not able to do so if (ActionEvent != null) { } } } I wanted to access this base class event in derived. Further I wanted to access this event in some other derived class of MyClass: MyClass.ActionEvent += DerivedMethod() Please help me understand how to work with event defined in

public Event in abstract class

由 本小妞迷上赌 提交于 2021-01-27 03:53:25
问题 I have event declared in abstract class: public abstract class AbstractClass { public event Action ActionEvent; } public class MyClass : AbstractClass { private void SomeMethod() { //Want to access ActionEvent-- Not able to do so if (ActionEvent != null) { } } } I wanted to access this base class event in derived. Further I wanted to access this event in some other derived class of MyClass: MyClass.ActionEvent += DerivedMethod() Please help me understand how to work with event defined in

public Event in abstract class

由 余生颓废 提交于 2021-01-27 03:53:25
问题 I have event declared in abstract class: public abstract class AbstractClass { public event Action ActionEvent; } public class MyClass : AbstractClass { private void SomeMethod() { //Want to access ActionEvent-- Not able to do so if (ActionEvent != null) { } } } I wanted to access this base class event in derived. Further I wanted to access this event in some other derived class of MyClass: MyClass.ActionEvent += DerivedMethod() Please help me understand how to work with event defined in

How can I force a throw to be a statement and not an expression (in a lambda expression)?

由 ≡放荡痞女 提交于 2021-01-20 04:20:19
问题 Starting from C# 7.0 the throw keyword can be used both as an expression and as a statement, which is nice. Though, consider these overloads public static void M(Action doIt) { /*use doIt*/ } public static void M(Func<int> doIt) { /*use doIt*/ } When invoking like this M(() => throw new Exception()); or even like this (with a statement lambda) M(() => { throw new Exception(); }); the M(Func<>) overload is selected by the compiler indicating that the throw is here considered as an expression.

How can I force a throw to be a statement and not an expression (in a lambda expression)?

由 泪湿孤枕 提交于 2021-01-20 04:19:17
问题 Starting from C# 7.0 the throw keyword can be used both as an expression and as a statement, which is nice. Though, consider these overloads public static void M(Action doIt) { /*use doIt*/ } public static void M(Func<int> doIt) { /*use doIt*/ } When invoking like this M(() => throw new Exception()); or even like this (with a statement lambda) M(() => { throw new Exception(); }); the M(Func<>) overload is selected by the compiler indicating that the throw is here considered as an expression.

How can I force a throw to be a statement and not an expression (in a lambda expression)?

由 拈花ヽ惹草 提交于 2021-01-20 04:15:28
问题 Starting from C# 7.0 the throw keyword can be used both as an expression and as a statement, which is nice. Though, consider these overloads public static void M(Action doIt) { /*use doIt*/ } public static void M(Func<int> doIt) { /*use doIt*/ } When invoking like this M(() => throw new Exception()); or even like this (with a statement lambda) M(() => { throw new Exception(); }); the M(Func<>) overload is selected by the compiler indicating that the throw is here considered as an expression.

《Spring In Action》第三版中文版 Chapter 1 Piece 1

由 半腔热情 提交于 2020-11-19 04:49:34
版权与免责声明：本博客文章翻译自《Spring in action》第三版，原书版权归原作者及其出版单位所有，本人仅出于个人兴趣，翻译其中内容。本人不以盈利为目的，放置在博客中仅为了方便网友阅读原作遇到困难时参考。本人严正声明，本人并不拥有翻译该作品之正当权力，因此严禁他人转载，亦不保证作品之正确性；如侵犯您的权益，望告知，即行关闭。 翻译声明：本人目前计划每天翻译两页，且不做任何工期承诺，之前未在网上发现类似内容，如已有相似内容，望告知，也省得我翻译了。 Chapter 1 Piece 1 本章涵盖： 浏览Spring的核心模块 对应用对象进行解耦 利用AOP管理横向关注点 Spring的bean容器 事情还得从一颗豆子 (bean)说起 。 1996 年，当时的 Java 还只是一个新兴的、激动人心的、即将流行的平台。大量的开发者便涌向这门语言，因为他们见识到了如何用 applets 创建丰富和动态的 web 应用。这些开发者很快发现，除了能让杂技动画人物活动起来之外，这门陌生的新语言还可以做很多事。与之前任何语言都不同的是， Java 让编写复杂的、由松散部件组成的应用成为可能。他们因为 applets 而接触 Java ，但是让他们最终决定留下来的是 Java 的组件特性。 在这一年的 12 月， Sun 公司发布了 JavaBeans 1.00-A 规范。
订阅 action