我有一个非常简单的Sping Boot 应用程序,我有application.properties,现在我想转到application.yml
下面是应用程序.yml文件:
spring:
datasource:
url: jdbc:mysql://localhost:3306/employee_directory
username: student
password: Password123
字符串
因此,我删除了application.properties文件,并编写了应用程序.yml文件。如果我运行应用程序,我会得到此异常:
20:31:22.446 [restartedMain] ERROR org.springframework.boot.SpringApplication - Application run failed
java.lang.IllegalStateException: Failed to load property source from location 'classpath:/application.yml'
at org.springframework.boot.context.config.ConfigFileApplicationListener$Loader.load(ConfigFileApplicationListener.java:524)
at org.springframework.boot.context.config.ConfigFileApplicationListener$Loader.loadForFileExtension(ConfigFileApplicationListener.java:473)
at org.springframework.boot.context.config.ConfigFileApplicationListener$Loader.load(ConfigFileApplicationListener.java:443)
at org.springframework.boot.context.config.ConfigFileApplicationListener$Loader.lambda$null$6(ConfigFileApplicationListener.java:425)
at java.lang.Iterable.forEach(Iterable.java:75)
at org.springframework.boot.context.config.ConfigFileApplicationListener$Loader.lambda$load$7(ConfigFileApplicationListener.java:425)
at java.lang.Iterable.forEach(Iterable.java:75)
at org.springframework.boot.context.config.ConfigFileApplicationListener$Loader.load(ConfigFileApplicationListener.java:422)
at org.springframework.boot.context.config.ConfigFileApplicationListener$Loader.load(ConfigFileApplicationListener.java:321)
at org.springframework.boot.context.config.ConfigFileApplicationListener.addPropertySources(ConfigFileApplicationListener.java:202)
at org.springframework.boot.context.config.ConfigFileApplicationListener.postProcessEnvironment(ConfigFileApplicationListener.java:186)
at org.springframework.boot.context.config.ConfigFileApplicationListener.onApplicationEnvironmentPreparedEvent(ConfigFileApplicationListener.java:176)
at org.springframework.boot.context.config.ConfigFileApplicationListener.onApplicationEvent(ConfigFileApplicationListener.java:164)
at org.springframework.context.event.SimpleApplicationEventMulticaster.doInvokeListener(SimpleApplicationEventMulticaster.java:172)
at org.springframework.context.event.SimpleApplicationEventMulticaster.invokeListener(SimpleApplicationEventMulticaster.java:165)
at org.springframework.context.event.SimpleApplicationEventMulticaster.multicastEvent(SimpleApplicationEventMulticaster.java:139)
at org.springframework.context.event.SimpleApplicationEventMulticaster.multicastEvent(SimpleApplicationEventMulticaster.java:127)
at org.springframework.boot.context.event.EventPublishingRunListener.environmentPrepared(EventPublishingRunListener.java:75)
at org.springframework.boot.SpringApplicationRunListeners.environmentPrepared(SpringApplicationRunListeners.java:53)
at org.springframework.boot.SpringApplication.prepareEnvironment(SpringApplication.java:340)
at org.springframework.boot.SpringApplication.run(SpringApplication.java:304)
at org.springframework.boot.SpringApplication.run(SpringApplication.java:1213)
at org.springframework.boot.SpringApplication.run(SpringApplication.java:1202)
at com.dgs.springboot.SpringBootRESTJPA.SpringBootRestJPAApplication.main(SpringBootRestJPAApplication.java:10)
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62)
at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
at java.lang.reflect.Method.invoke(Method.java:498)
at org.springframework.boot.devtools.restart.RestartLauncher.run(RestartLauncher.java:49)
Caused by: org.yaml.snakeyaml.scanner.ScannerException: while scanning for the next token
found character '\t(TAB)' that cannot start any token. (Do not use \t(TAB) for indentation)
in 'reader', line 2, column 1:
datasource:
^
at org.yaml.snakeyaml.scanner.ScannerImpl.fetchMoreTokens(ScannerImpl.java:419)
at org.yaml.snakeyaml.scanner.ScannerImpl.checkToken(ScannerImpl.java:227)
at org.yaml.snakeyaml.parser.ParserImpl$ParseBlockMappingValue.produce(ParserImpl.java:586)
at org.yaml.snakeyaml.parser.ParserImpl.peekEvent(ParserImpl.java:158)
at org.yaml.snakeyaml.parser.ParserImpl.checkEvent(ParserImpl.java:148)
at org.yaml.snakeyaml.composer.Composer.composeNode(Composer.java:124)
at org.yaml.snakeyaml.composer.Composer.composeValueNode(Composer.java:236)
at org.yaml.snakeyaml.composer.Composer.composeMappingChildren(Composer.java:227)
at org.yaml.snakeyaml.composer.Composer.composeMappingNode(Composer.java:215)
at org.yaml.snakeyaml.composer.Composer.composeNode(Composer.java:144)
at org.yaml.snakeyaml.composer.Composer.getNode(Composer.java:85)
at org.yaml.snakeyaml.constructor.BaseConstructor.getData(BaseConstructor.java:123)
at org.yaml.snakeyaml.Yaml$1.next(Yaml.java:547)
at org.springframework.beans.factory.config.YamlProcessor.process(YamlProcessor.java:160)
at org.springframework.beans.factory.config.YamlProcessor.process(YamlProcessor.java:134)
at org.springframework.boot.env.OriginTrackedYamlLoader.load(OriginTrackedYamlLoader.java:75)
at org.springframework.boot.env.YamlPropertySourceLoader.load(YamlPropertySourceLoader.java:49)
at org.springframework.boot.context.config.ConfigFileApplicationListener$Loader.loadDocuments(ConfigFileApplicationListener.java:542)
at org.springframework.boot.context.config.ConfigFileApplicationListener$Loader.load(ConfigFileApplicationListener.java:497)
... 28 common frames omitted
型
我想我需要对我的代码进行其他修改以使用此应用程序.yml文件,但我不知道该怎么办?
这是pom.xml:
<?xml version="1.0" encoding="UTF-8"?>
<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
<modelVersion>4.0.0</modelVersion>
<parent>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-parent</artifactId>
<version>2.1.6.RELEASE</version>
<relativePath/> <!-- lookup parent from repository -->
</parent>
<groupId>com.dgs.springboot</groupId>
<artifactId>SpringBootRESTHibernate</artifactId>
<version>0.0.1-SNAPSHOT</version>
<name>SpringBootRESTHibernate</name>
<description>Demo project for Spring Boot</description>
<properties>
<java.version>1.8</java.version>
<!-- Add work around for Eclipse bug -->
<maven-jar-plugin.version>3.1.1</maven-jar-plugin.version>
</properties>
<dependencies>
<dependency>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-data-jpa</artifactId>
</dependency>
<dependency>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-web</artifactId>
</dependency>
<dependency>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-devtools</artifactId>
<scope>runtime</scope>
<optional>true</optional>
</dependency>
<dependency>
<groupId>mysql</groupId>
<artifactId>mysql-connector-java</artifactId>
<scope>runtime</scope>
</dependency>
<dependency>
<groupId>org.projectlombok</groupId>
<artifactId>lombok</artifactId>
</dependency>
<dependency>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-test</artifactId>
<scope>test</scope>
</dependency>
</dependencies>
<build>
<plugins>
<!-- This is used for packaging and running our app -->
<plugin>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-maven-plugin</artifactId>
</plugin>
</plugins>
</build>
</project>
型
任何反馈都将不胜感激。谢谢!
9条答案
按热度按时间yhqotfr81#
我构建了一个在线转换工具(https://env.simplestep.ca/),它可以在spring Boot yaml、属性和环境变量之间进行转换--使用spring Boot 宽松的绑定规则
x1c 0d1x的数据
6ioyuze22#
您可以使用命令行工具props2yaml进行从属性到yaml的通用自动(正确)转换。
t0ybt7op3#
对于
intellij-IDEA
:去文件>设置>插件然后搜索一个插件称为:Properties to YAML Converter
.安装它,然后重新启动您的IDE,然后右键单击application.properties文件,你会得到一个选项称为“转换为yaml”ijxebb2r4#
检查是否使用制表符进行缩进。这是YAML规范不允许的:
为了保持可移植性,不能在缩进中使用制表符,因为不同的系统对制表符的处理是不同的。请注意,大多数现代编辑器都可以配置为按下制表符键会插入适当数量的空格。
sg2wtvxw5#
如果您使用Eclipse或STS,则非常简单。
只需打开
.properties
文件的上下文菜单,然后选择“转换为Yaml文件”。该文件将自动转换。如果您的上下文菜单上没有此功能,您需要从安装Yaml插件
字符串
bvjveswy6#
我认为你的
datasource.url
值需要在引号中,因为':'。试试这个:字符串
此外,正如上面所指出的,您应该使用空格进行缩进;通常,每级两个空格。
祝你好运!
2w3rbyxf7#
这只是一个额外的答案,可能会帮助其他开发人员。
我发现了一个更好,更容易的工具来转换属性文件到YML文件。convert from properties to yml
您必须上传.properties文件,然后选择ToYML文件选项,它将生成一个非常好的yml文件。该工具支持许多其他格式,如Yml到XML,Yml到Properties,CSV,JSON等。
hrysbysz8#
文件
application.yml
必须位于根文件夹resources
中。62lalag49#
对于其他寻找类似答案的人来说,在IntelliJ中运行的Spring项目中,它只需要将application.properties的扩展名更改为application.yml。在我的情况下,文件是空的,所以我不必更改任何内容,我可以正确地使用application.yml。