使用Spring Tool Suite (STS)创建application.yml文件创建失败


SpringBoot中@Autowired注入bean失败：Error creating bean with name“XXXController”？

# 报错信息： ``` org.springframework.beans.factory.UnsatisfiedDependencyException: Error creating bean with name 'authorizeController': Unsatisfied dependency expressed through field 'userMapper'; nested exception is org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'userMapper' defined in file [E:\STS-WorkSpace\Forum\target\classes\com\Forum\mapper\UserMapper.class]: Invocation of init method failed; nested exception is java.lang.IllegalArgumentException: Property 'sqlSessionFactory' or 'sqlSessionTemplate' are required at org.springframework.beans.factory.annotation.AutowiredAnnotationBeanPostProcessor$AutowiredFieldElement.inject(AutowiredAnnotationBeanPostProcessor.java:643) ~[spring-beans-5.2.2.RELEASE.jar:5.2.2.RELEASE] at org.springframework.beans.factory.annotation.InjectionMetadata.inject(InjectionMetadata.java:116) ~[spring-beans-5.2.2.RELEASE.jar:5.2.2.RELEASE] at org.springframework.beans.factory.annotation.AutowiredAnnotationBeanPostProcessor.postProcessProperties(AutowiredAnnotationBeanPostProcessor.java:399) ~[spring-beans-5.2.2.RELEASE.jar:5.2.2.RELEASE] at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.populateBean(AbstractAutowireCapableBeanFactory.java:1422) ~[spring-beans-5.2.2.RELEASE.jar:5.2.2.RELEASE] at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.doCreateBean(AbstractAutowireCapableBeanFactory.java:594) ~[spring-beans-5.2.2.RELEASE.jar:5.2.2.RELEASE] at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.createBean(AbstractAutowireCapableBeanFactory.java:517) ~[spring-beans-5.2.2.RELEASE.jar:5.2.2.RELEASE] at org.springframework.beans.factory.support.AbstractBeanFactory.lambda$doGetBean$0(AbstractBeanFactory.java:323) ~[spring-beans-5.2.2.RELEASE.jar:5.2.2.RELEASE] at org.springframework.beans.factory.support.AbstractBeanFactory$$Lambda$141/1866850137.getObject(Unknown Source) ~[na:na] at org.springframework.beans.factory.support.DefaultSingletonBeanRegistry.getSingleton(DefaultSingletonBeanRegistry.java:222) ~[spring-beans-5.2.2.RELEASE.jar:5.2.2.RELEASE] at org.springframework.beans.factory.support.AbstractBeanFactory.doGetBean(AbstractBeanFactory.java:321) ~[spring-beans-5.2.2.RELEASE.jar:5.2.2.RELEASE] at org.springframework.beans.factory.support.AbstractBeanFactory.getBean(AbstractBeanFactory.java:202) ~[spring-beans-5.2.2.RELEASE.jar:5.2.2.RELEASE] at org.springframework.beans.factory.support.DefaultListableBeanFactory.preInstantiateSingletons(DefaultListableBeanFactory.java:879) ~[spring-beans-5.2.2.RELEASE.jar:5.2.2.RELEASE] at org.springframework.context.support.AbstractApplicationContext.finishBeanFactoryInitialization(AbstractApplicationContext.java:878) ~[spring-context-5.2.2.RELEASE.jar:5.2.2.RELEASE] at org.springframework.context.support.AbstractApplicationContext.refresh(AbstractApplicationContext.java:550) ~[spring-context-5.2.2.RELEASE.jar:5.2.2.RELEASE] at org.springframework.boot.web.servlet.context.ServletWebServerApplicationContext.refresh(ServletWebServerApplicationContext.java:141) ~[spring-boot-2.2.2.RELEASE.jar:2.2.2.RELEASE] at org.springframework.boot.SpringApplication.refresh(SpringApplication.java:747) [spring-boot-2.2.2.RELEASE.jar:2.2.2.RELEASE] at org.springframework.boot.SpringApplication.refreshContext(SpringApplication.java:397) [spring-boot-2.2.2.RELEASE.jar:2.2.2.RELEASE] at org.springframework.boot.SpringApplication.run(SpringApplication.java:315) [spring-boot-2.2.2.RELEASE.jar:2.2.2.RELEASE] at org.springframework.boot.SpringApplication.run(SpringApplication.java:1226) [spring-boot-2.2.2.RELEASE.jar:2.2.2.RELEASE] at org.springframework.boot.SpringApplication.run(SpringApplication.java:1215) [spring-boot-2.2.2.RELEASE.jar:2.2.2.RELEASE] at com.Forum.ForumApplication.main(ForumApplication.java:19) [classes/:na] Caused by: org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'userMapper' defined in file [E:\STS-WorkSpace\Forum\target\classes\com\Forum\mapper\UserMapper.class]: Invocation of init method failed; nested exception is java.lang.IllegalArgumentException: Property 'sqlSessionFactory' or 'sqlSessionTemplate' are required at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.initializeBean(AbstractAutowireCapableBeanFactory.java:1796) ~[spring-beans-5.2.2.RELEASE.jar:5.2.2.RELEASE] at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.doCreateBean(AbstractAutowireCapableBeanFactory.java:595) ~[spring-beans-5.2.2.RELEASE.jar:5.2.2.RELEASE] at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.createBean(AbstractAutowireCapableBeanFactory.java:517) ~[spring-beans-5.2.2.RELEASE.jar:5.2.2.RELEASE] at org.springframework.beans.factory.support.AbstractBeanFactory.lambda$doGetBean$0(AbstractBeanFactory.java:323) ~[spring-beans-5.2.2.RELEASE.jar:5.2.2.RELEASE] at org.springframework.beans.factory.support.AbstractBeanFactory$$Lambda$141/1866850137.getObject(Unknown Source) ~[na:na] at org.springframework.beans.factory.support.DefaultSingletonBeanRegistry.getSingleton(DefaultSingletonBeanRegistry.java:222) ~[spring-beans-5.2.2.RELEASE.jar:5.2.2.RELEASE] at org.springframework.beans.factory.support.AbstractBeanFactory.doGetBean(AbstractBeanFactory.java:321) ~[spring-beans-5.2.2.RELEASE.jar:5.2.2.RELEASE] at org.springframework.beans.factory.support.AbstractBeanFactory.getBean(AbstractBeanFactory.java:202) ~[spring-beans-5.2.2.RELEASE.jar:5.2.2.RELEASE] at org.springframework.beans.factory.config.DependencyDescriptor.resolveCandidate(DependencyDescriptor.java:276) ~[spring-beans-5.2.2.RELEASE.jar:5.2.2.RELEASE] at org.springframework.beans.factory.support.DefaultListableBeanFactory.doResolveDependency(DefaultListableBeanFactory.java:1287) ~[spring-beans-5.2.2.RELEASE.jar:5.2.2.RELEASE] at org.springframework.beans.factory.support.DefaultListableBeanFactory.resolveDependency(DefaultListableBeanFactory.java:1207) ~[spring-beans-5.2.2.RELEASE.jar:5.2.2.RELEASE] at org.springframework.beans.factory.annotation.AutowiredAnnotationBeanPostProcessor$AutowiredFieldElement.inject(AutowiredAnnotationBeanPostProcessor.java:640) ~[spring-beans-5.2.2.RELEASE.jar:5.2.2.RELEASE] ... 20 common frames omitted Caused by: java.lang.IllegalArgumentException: Property 'sqlSessionFactory' or 'sqlSessionTemplate' are required at org.springframework.util.Assert.notNull(Assert.java:198) ~[spring-core-5.2.2.RELEASE.jar:5.2.2.RELEASE] at org.mybatis.spring.support.SqlSessionDaoSupport.checkDaoConfig(SqlSessionDaoSupport.java:123) ~[mybatis-spring-2.0.1.jar:2.0.1] at org.mybatis.spring.mapper.MapperFactoryBean.checkDaoConfig(MapperFactoryBean.java:73) ~[mybatis-spring-2.0.1.jar:2.0.1] at org.springframework.dao.support.DaoSupport.afterPropertiesSet(DaoSupport.java:44) ~[spring-tx-5.2.2.RELEASE.jar:5.2.2.RELEASE] at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.invokeInitMethods(AbstractAutowireCapableBeanFactory.java:1855) ~[spring-beans-5.2.2.RELEASE.jar:5.2.2.RELEASE] at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.initializeBean(AbstractAutowireCapableBeanFactory.java:1792) ~[spring-beans-5.2.2.RELEASE.jar:5.2.2.RELEASE] ... 31 common frames omitted ``` # Mapper代码： ``` package com.Forum.mapper; import org.apache.ibatis.annotations.Insert; import org.apache.ibatis.annotations.Mapper; import com.Forum.Model.User; @Mapper public interface UserMapper{ @Insert("insert into user (name,account_id,token,gmt_create,gmt_modified) values (#{name},#{accountId}),#{token},#{gmtCreate},#{gmtModified}") public void insert(User user); } ``` # 启动代码： ``` package com.Forum; import org.mybatis.spring.annotation.MapperScan; import org.springframework.boot.SpringApplication; import org.springframework.boot.autoconfigure.SpringBootApplication; import org.springframework.boot.autoconfigure.jdbc.DataSourceAutoConfiguration; import org.springframework.context.annotation.ComponentScan; @SpringBootApplication(exclude = {DataSourceAutoConfiguration.class}) @ComponentScan(basePackages = {"com.Forum.*"}) @MapperScan(basePackages = {"com.Forum.mapper"}) public class ForumApplication { public static void main(String[] args) { SpringApplication.run(ForumApplication.class, args); } } ``` # Controller代码： ```package com.Forum.Controller; import java.util.UUID; import javax.servlet.http.HttpServletRequest; import org.springframework.beans.factory.annotation.Autowired; import org.springframework.beans.factory.annotation.Value; import org.springframework.stereotype.Controller; import org.springframework.web.bind.annotation.GetMapping; import org.springframework.web.bind.annotation.RequestParam; import com.Forum.Dto.AccesstokenDTO; import com.Forum.Dto.GitHubUser; import com.Forum.Model.User; import com.Forum.Provider.GitHubProvider; import com.Forum.mapper.UserMapper; @Controller public class AuthorizeController{ @Autowired private GitHubProvider gitHubProvider; @Value("${github.client.id}") private String clientId; @Value("${github.client.secret}") private String clientSecret; @Value("${github.redirect.uri}") private String redirectUri; @Autowired public UserMapper userMapper; @GetMapping("/callback") public String callback(@RequestParam(name="code") String code, @RequestParam(name="state") String state, HttpServletRequest request ) { AccesstokenDTO accesstokenDTO = new AccesstokenDTO(); accesstokenDTO.setCode(code); accesstokenDTO.setRedirect_uri("redirectUri"); accesstokenDTO.setClient_id(clientId); accesstokenDTO.setClient_secret(clientSecret); accesstokenDTO.setState(state); String accessToken = gitHubProvider.getAccessToken(accesstokenDTO); GitHubUser githubUser = gitHubProvider.getUser(accessToken); if(githubUser != null) { // 登录成功，写cookie和session User user = new User(); user.setToken(UUID.randomUUID().toString()); user.setName(githubUser.getName()); user.setAccountId(String.valueOf(githubUser.getId())); user.setQmtCreate(System.currentTimeMillis()); user.setQmtModified(user.getQmtCreate()); userMapper.insert(user); request.getSession().setAttribute("user", githubUser); return "redirect:/"; // 去掉地址后缀 }else { // 登录失败，重新登录 return "redirect:/"; } } } ```

eclipse spring.xml文件属性值提示失效

eclipse里面，xml中 类路径提示问题 ，我rinzo+sts插件都安装了，还是没用。（标签是可以提示的，就是写class的时候，这个自动提示不了，已经花了5个小时在搞这个问题，还没解决，求大神带走~） 比如： 【温馨提示：JavaSE/EE、SSH/SSM、Hybrid APP、JQ/JS/CSS3/H5等编程爱好者加QQ号：28554482，互相学习，共同进步！】

eclipse 4.5.2安装STS想用spring boot，过程显示成功却找不到spring boot，求解惑

如题如图，我的是e4.5.2  Java版本1.8  一开始是打算在Eclipse Marketplace上下载STS，可是居然找不到  没办法，只能下载资源包，本地安装，看了攻略，我下的是： springsource-tool-suite-3.7.3.RELEASE-e4.5.2-updatesite.zip 然后就Help --> Install New Software,填入自定名字和包位置 --> OK  又点击Available Software Site --> 去掉所有http的打勾项，只留下sts（jar）--> OK  主选项页面只勾选IDE结尾的  而后一路next      发现File --> New --> Other找不到spring和spring boot选项  preference里头也没有spring和spring boot选项  该怎么做，求解

eclipse安装完对应的sts插件后按alt+/却不能提示属性值

求教各位大佬：我eclipse安装了sts插件，配置spring中的bean标签下的property标签时按alt+/能自动提示name，value等，但进一步name=""时按alt+/却不能自动提示name里的东西。eclipse左下角提示：Content Assist not available at the current location

Game Simulator 模拟器的问题

Problem Description “Tractor” is a very popular poker game in China. There are four players in the game. Suppose their names are Alice, Bob, Charles and David, in clockwise order. A judge is needed for this game. The players are divided into two teams, Alice and Charles are in team 1, and the other two are in team 2. The prop they use to play the game are two decks of pokers, including 108 cards in total. A simplified rule of the game is described below. The whole game contains a number of rounds. In each round, one team is called “Declarers” (CT); the other team is called “Defenders” (FT). Each team has a current rank (CR). The goal of the player is to increase his own team's CR as much as possible. A certain round has a Main Suit (Heart - H, Spade - S, Club - C, Diamond - D, or None - no main suit in this round) and its CR. The CR in this round is the CR of the CT, and Main Suit will be given. The Main Suit and CR will be used to determine the order of the cards. Cards ranked 5, 10, King value 5, 10, 10 pts (points) respectively, all other cards value 0 pts. In one round, we only consider the FT's pts. The rules of getting pts for FT will be discussed later. If the FT gets less than 80 pts in one round, they will hold be FT in the next round. This situation is called “make”. Otherwise, they become CT in the next round and the original CT become FT instead. This situation is called “down”. If the FT gets 0 pts, the CR of the current CT will be increased by 3, for example, if the CR of the CT is 9, it will become Queen (12). Otherwise, if the FT gets less than 40 pts, the CR of the CT will be increased by 2. Otherwise, if the FT gets less than 80 pts, the CR of the CT will be increased by 1. Otherwise, if the FT gets not less than 80 + k * 40 pts and less than 120 + k * 40 pts, the CR of the current FT will be increased by k. For example, if the FT gets 255 pts in a round, the CR of the current FT will be increased by 4; and if the FT gets 80 pts, both teams' CR remain unchanged. If a team's CR becomes beyond Ace, this team is considered the WINNER of the whole game. During a round, one of the players in CT is called the dealer. If “make”, the pard (teammate) of the dealer becomes the next round’s dealer. Otherwise (“down”), the player on dealer's right-hand side becomes the dealer of the next round. For example, if the dealer of the current round is Alice and her team (CT) is “down”, the dealer of the next round should be Bob (on Alice's right-hand side). At the start of a round, each of the players except the dealer gets exactly 25 cards; the dealer gets all the remaining 33 cards. After that, the dealer chooses 8 of his cards and gives them to the judge, and these cards are called “hidden cards”. Now each player has exactly 25 cards. A round consists of several tricks. In the first trick, the dealer plays one or more cards (called “lead”), then, in clockwise order, players play the same number of cards as the first player one by one (called “follow”). The winner of the current trick leads cards during the next trick, and so on. If the winner of the current trick is a member of FT, then the FT gets the sum of the cards' pts played in this trick. Now we start to describe how to determine the winner of a trick. After the main suit and the CR of the current round are fixed, we can determine the “trumps” which are cards with main suit or CR (Current Rank), and the Jokers. All other cards are “not-trumps”. We can have an order among all the cards according the following rules: (1) “Trumps” are ordered higher than “not-trumps”. (2) For the trumps, the order are listed below: Red Joker Black Joker card with main suit and CR (if exists) other card with CR other trumps ordered by their ranks(i.e., A, K, Q, J, T, 9, 8, 7, ..., 3, 2) (3) For the “not-trumps”, they are ordered by their ranks. Assume in all the description below, in the current round, the CR of the CT is 7. Suppose the main suit is H, the cards can be arranged in this order (as an example): S2 , C2 , D2 < S3 , C3 , D3 < S4 , C4 , D4 < S5 , C5 , D5 < S6 , C6 , D6 < S8 , C8 , D8 < S9 , C9 , D9 < ST , FT , CT (T - 10) < SJ , CJ , DJ (J - Jack) < SQ , CQ , DQ (Q - Queen) < SK , CK , DK (K - King) < SA , CA , DA (A - Ace) < H2 < H3 < H4 < H5 < H6 < H8 < H9 < HT < HJ < HQ < HK < HA < S7 = C7 = D7 < H7 < BJ (the Black Joker) < RJ (the Red Joker) If “None” during this round, then the pokers can be arranged in this order: H2 , S2 , C2 , D2 < H3 , S3 , C3 , D3 < H4 , S4 , C4 , D4 < H5 , S5 , C5 , D5 < H6 , S6 , C6 , D6 < H8 , S8 , C8 , D8 < H9 , S9 , C9 , D9 < HT , ST , FT , CT < HJ , SJ , CJ , DJ < HQ , SQ , CQ , DQ < HK , SK , CK , DK < HA , SA , CA , DA < H7 = S7 = C7 = D7 < BJ < RJ In these two tables, cards written in italic are “trumps”, and cards written in boldface are “not-trumps”. In each trick, the lead cards (played by the player leading this trick) must be either all “trumps”, or all “not-trumps” with same suit. The possible structures of the cards are listed below (assume the main suit is H and main rank is 7 for the example): Single. A single card, such as D9. Pair. Two same cards, such as D9D9. But D7S7 is not a pair although their orders are the same. Tractor. Two or more consecutive-ordered pairs, satisfying the condition that they are all “trumps”, or all “not-trumps” with same suit, such as SJSJSQSQSKSKSASA, H7H7S7S7HAHA or RJRJBJBJ. But, these are not tractors: S7S7C7C7 (their orders are the same), C7C7C6C6 (they are not consecutive-ordered), DADAD2D2 (Ace is not “one”, so they are not consecutive-ordered), H2H2H4H4, or D2D2D3. Be careful: if “None” in this round, H7H7S7S7HAHA is not a tractor (H7 and S7 are same-ordered because of “None”). Throw. The combination of the structures above, satisfying the condition that they are all “trumps”, or all “not-trumps” with same suit. Each of the Single, Pair or Tractor in a Throw is one of the Throw’s component. In the original tractor game, in some situation, the throw will be rejected. But, to keep the rule simple, we assume in this problem all the throws are accepted. For example, RJRJBJBJH7H7HQHQHJHJH9H9H6H6HAH2 contains six components: two tractors, two pairs and two single cards (RJRJBJBJH7H7-HQHQHJHJ-H9H9-H6H6-HA-H2); CACAC8C8CK contains three components: two pairs and one single card(CACA-C8C8-CK). A throw can be treated as different list of components, for example, H2H2H3H3H4H4H5H5H6H6 can also be treated as H2H2H3H3-H4H4H5H5H6H6, or H2H2-H3H3H4H4H4H5H5-H6H6, and so on. For the lead cards, each time we choose the longest component (choose the one with the highest order to break the tie) to construct a list of components, this list is the structure of the lead cards, also the structure of the trick. So that the structure of the trick is unique. After the first player lead his or her cards, other players follow cards one by one in clockwise order, as mentioned above. An important part of the game is to determine the winner of a trick: If one’s follow cards contain both “trumps” and “not-trumps”, or all “not-trumps” but with different suits, this player can't be winner of this trick. Otherwise, if the lead cards are all “not-trumps” and one’s follow cards contain “not-trumps” with different suit from the lead cards, this follow player can’t be the winner of this trick. Else, if one’s follow cards can't be constructed as the same structure of lead cards, this player can't be the winner of this trick either. Otherwise, if the structure of this trick is not “throw”, the one who played the highest-ordered card wins this trick. If more than one player played the same highest-ordered card, the winner of this trick will be the one who plays the highest-ordered card first. Now let's consider the “throw” situation. We construct the follow cards into the structure of the lead cards, so that the order of the highest-ordered card in all the longest components of the “throw” is as high as possible (this card is called “honor card”). Note that tractor can be treated as several pairs or shorter tractors, and pair can be treated as two single cards. The winner of this trick is the one who plays the highest-ordered “honor card”. Similarly, if more than one player played the same highest-ordered “honor card” the winner of this trick will be the one who plays the highest-ordered “honor card” first. There are many hair-raising rules about lead and follow cards; fortunately, they're not related to this problem, the only thing we care about is: when someone leads a “not-trump” “throw” the only possible way to beat it is to “throw” the same structure of “trumps”. And it’s impossible to beat a leading “trump” “throw”. Special attention on the examples below. In these examples, Alice always leads cards. And assume in all the following examples, the CR is 7, and the main suit is H. There is a special rule about “hidden cards”: if the winner of the last trick of a certain round is a member of FT, then, in addition, the FT gets the sum of the hidden cards' pts, multiplied by 2w (2 to the power of w). When the structure of the final trick is not “throw”, then w is the number of lead cards of the last trick of this round. If the structure is “throw” instead, w is the length of the longest components, in the example RJRJBJBJH7H7HQHQHJHJH6H6HA, the w is 6 because the length of RJRJBJBJH7H7 (the longest components of the “throw”) is 6. To make the problem easier, you are only to write a single-round tractor game simulator. Input Multiple test cases, the number of them T is given in the very first line. For each test case: The first line contains the main suit of this round (H, S, C, D, O; O denotes “None” in this round), the dealer of this round, the CR of team 1, the CR of team 2, separated by single spaces. Each of the rest lines contains 4 strings: the lead cards and the cards played by the second, third and last player. In one string, the cards can be given in any order. Each player will play exactly 25 cards in one round. You may assume the input is always valid. There is a blank line between consecutive test cases, and a blank line also appears between T and the first test case. Output For each test case: The first line contains the case number. The second line contains the pts get by the FT in this round. If a team wins the whole game after this round, output “Winner: Team X”(without quotes, X should be either 1 or 2) in the second line. If no team wins, output the new CR of team 1, the new CR of team 2 after this round, followed by the name of the dealer of the next round, separated by single spaces. See the example for further details. Sample Input 1 O Charles 2 2 S6S6S7S7 SASKSJST STS8S4S4 S3S5SJSQ S9S9 H3D3 S3DT SAD3 DA DQ DK D4 SKS8S5S3 RJC2D2H2 C6C8CJD9 H3CKDTD5 H7H7 H6H4 HJHQ H9H9 DJDJ DKH5 D5D4 D6D6 D8D8 C4C3 HTH5 D9D7 C5C5 C6CT H8HQ C7C4 H8 C7 HA HA H2 RJ BJ CK DA BJ C8 HK S2S2C2 CQCAD2 HTHJHK C9CQCA Sample Output Case #1: 50 3 2 Alice

STS离线环境使用git时无法pull/push,提示如下

STS离线环境使用git时无法pull/push，提示Cannot run program “ssh”:CreateProcess error=2 ,系统找不到指定的文件 ，哪位遇到过？该如何解决？

使用STS-3.7.2怎么声明一个interceptor

使用STS-3.7.2，想在dispatchor的相应文件添加一个interceptor，但是不知道interceptor在这个版本里的格式，很无奈。。。。

sts 隐藏dashboard服务

请问sts 的 dashboard如何隐藏不需要的服务 

为什么eclisep成功安装sts插件，spring的其他选项都有，唯独没有springboot starter poject这个选项


bootstrap fileinput 插件上传文件之后有错误提示.jpg: [object Object]

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