class

Description: Cannot determine embedded database driver class for database type NONE 解决

由 北城以北 提交于 2020-01-08 23:51:03
环境eclipse 4.7 jdk1.8 搭建springcloud子父项目报错 出现了这两种报错 解决 加上下面两个注解 @ComponentScan(basePackages = {"com.wf.zhang.springcloud.dao"}) @EnableAutoConfiguration(exclude={DataSourceAutoConfiguration.class}) @ComponentScan(basePackages = {"com.wf.zhang.springcloud.dao"}) @EnableAutoConfiguration(exclude={DataSourceAutoConfiguration.class}) @SpringBootApplication public class DeptProvider8001_App { public static void main(String[] args) { SpringApplication.run(DeptProvider8001_App.class, args); } } 来源： https://www.cnblogs.com/wf-zhang/p/12169140.html

栅格

由 橙三吉。 提交于 2020-01-08 23:19:54
<!DOCTYPE html> <html> <meta charset="UTF-8"> <meta name="viewport" content="width=device-width,initial-scale=1"> <title></title> <link rel="stylesheet" href="bootstrap.min.css"> <script src="jquery.min.js"></script> <style type="text/css"> .row{ margin-bottom: 20px; } [class*="col-"]{ padding-top: 15px; padding-bottom: 15px; background-color: #66afe9; border: 1px silver solid; } </style> <body> <div class="container"> <div class="row"> <!--最多一行12个 内容高度会根据内容自适应--> <div class="col-md-1">col-md-1</div> <div class="col-md-1">col-md-1</div> <div class="col-md-1">col-md-1</div> <div class="col-md-1

Java: How can I access a class's field by a name stored in a variable?

由 |▌冷眼眸甩不掉的悲伤 提交于 2020-01-08 14:32:29
问题 How can I set or get a field in a class whose name is dynamic and stored in a string variable? public class Test { public String a1; public String a2; public Test(String key) { this.key = 'found'; <--- error } } 回答1: You have to use reflection: Use Class.getField() to get a Field reference. If it's not public you'll need to call Class.getDeclaredField() instead Use AccessibleObject.setAccessible to gain access to the field if it's not public Use Field.set() to set the value, or one of the

Java: How can I access a class's field by a name stored in a variable?

由 帅比萌擦擦* 提交于 2020-01-08 14:32:04
问题 How can I set or get a field in a class whose name is dynamic and stored in a string variable? public class Test { public String a1; public String a2; public Test(String key) { this.key = 'found'; <--- error } } 回答1: You have to use reflection: Use Class.getField() to get a Field reference. If it's not public you'll need to call Class.getDeclaredField() instead Use AccessibleObject.setAccessible to gain access to the field if it's not public Use Field.set() to set the value, or one of the

Why do attribute references act like this with Python inheritance? [duplicate]

由 感情迁移 提交于 2020-01-08 12:28:12
问题 This question already has answers here : How to avoid having class data shared among instances? (7 answers) Closed last year . The following seems strange.. Basically, the somedata attribute seems shared between all the classes that inherited from the_base_class . class the_base_class: somedata = {} somedata['was_false_in_base'] = False class subclassthing(the_base_class): def __init__(self): print self.somedata first = subclassthing() {'was_false_in_base': False} first.somedata['was_false_in

Type1Font 是什么？

由 倾然丶 夕夏残阳落幕 提交于 2020-01-08 10:07:05
LoadSimpleType1Font https://blog.csdn.net/zhouschina/article/details/12162641?utm_source=blogxgwz6 class CPDF_TrueTypeFont; class CPDF_Type1Font; class CPDF_Type3Font; class CFX_SubstFont; class CPDF_CIDFont; 来源： https://www.cnblogs.com/hshy/p/12164939.html

获取java路径的几种方法

由 我们两清 提交于 2020-01-08 06:32:32
很多情况下，我们需要在 JavaBean 、 Servlet 中获得当前的目录路径，比如载入配置文件，上传文件到服务器等，这是各种获得路径的方法的总结，备忘。 1. 载入 jdbc.properties 1 ） ClassLoader 的 getResourceAsStream("XXX") InputStream in=XXX. class .getClassLoader().getResourceAsStream("jdbc.properties"); getResourceAsStream() 会到 classes 目录下找 jdbc.properties 文件，所以在工程项目中，我们一般把它放在 src 目录下。 2 ） Class.getResourceAsStream("/XXX") XXX. class .getResourceAsStream("/jdbc.properties"); 两种方式的共同点就是： jdbc.properties 都应该在 classes 下，也就是项目的 src 目录 不同点就是： ClassLoader.getResourceAsStream(“XXX”) 中参数不以 "/" 开头，而 Class.getResourceAsStream("/XXX") 终参数以 "/" 开头 2. 在 Javabean 中获得路径 　　1) 获得当前类

Python class attribute inconsistency

由 巧了我就是萌 提交于 2020-01-08 04:52:46
问题 I am trying to understand how class attributes work in Python. I have confusion based on following example. #!/usr/bin/python3 class Bag: val = 100 items = [] def add(self, x): self.items.append(x) print(self.items) b1 = Bag() b1.add('book') b1.val += 1 print(b1.val) b1.add('pen') b1.val += 1 print(b1.val) b2 = Bag() b2.add('sketches') b2.val += 1 print(b2.val) b2.add('text') b2.val += 1 print(b2.val) b2.add('canvas') b2.val += 1 print(b2.val) Output expected: ['book'] 101 ['book', 'pen'] 102

C++ classes getting value using pointers and strcpy [closed]

由 一曲冷凌霜 提交于 2020-01-08 01:25:34
问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 5 years ago . I am trying to understand C++. Can anybody explain what this code does exactly, I understood that it is some type of setter and getter in Java but I am not sure. Comm::Comm(const char* id) { strcpy(this->id, id); } char* Comm::getId() { return id; } 回答1: What does this code do? It burns the eyes of children. The

ES6 class（基本语法+方法）

由 只愿长相守 提交于 2020-01-07 23:42:19
静态属性与静态方法 1. 不会被类实例所拥有的属性与方法 只是类自身拥有 2. 只能通过类调用 静态方法与普通方法重名，不会冲突 static 关键字（静态方法） 静态属性 类名.属性名 = 属性值; 静态属性应用举例： //职业类 class Profession{ } class Character { constructor(pfs) { this.pfs = pfs; } } // 静态属性做配置 Character.config = { profession: { '咒术师': 1, '弓箭手': 2 } } // 创建类的实例 new Character(Character.config.profession['咒术师']); 静态方法应用举例 class Person { // 静态方法 static format(programmer) { programmer.haveGirlFriend = true; programmer.hair = true; } } // 程序员类 class Programmer { constructor() { this.haveGirlFriend = false; this.hair = false; } } // 将程序员类的实例转为了普通类 const programmer = new Programmer();
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