相关疑难解决方法(0)

我将我的项目从 spring-boot 2.1.9 移到了 2.2.0。

在启动项目时，我面临以下error消息。

Caused by: org.springframework.beans.factory.NoUniqueBeanDefinitionException: No qualifying bean of type 'org.springframework.plugin.core.PluginRegistry<org.springframework.hateoas.client.LinkDiscoverer, org.springframework.http.MediaType>' available: expected single matching bean but found 17: modelBuilderPluginRegistry,modelPropertyBuilderPluginRegistry,typeNameProviderPluginRegistry,syntheticModelProviderPluginRegistry,documentationPluginRegistry,apiListingBuilderPluginRegistry,operationBuilderPluginRegistry,parameterBuilderPluginRegistry,expandedParameterBuilderPluginRegistry,resourceGroupingStrategyRegistry,operationModelsProviderPluginRegistry,defaultsProviderPluginRegistry,pathDecoratorRegistry,apiListingScannerPluginRegistry,relProviderPluginRegistry,linkDiscovererRegistry,entityLinksPluginRegistry


Caused by: org.springframework.beans.factory.UnsatisfiedDependencyException: Error creating bean with name 'linkDiscoverers' defined in class path resource [org/springframework/hateoas/config/HateoasConfiguration.class]: Unsatisfied dependency expressed through method 'linkDiscoverers' parameter 0; nested exception is org.springframework.beans.factory.NoUniqueBeanDefinitionException: No qualifying bean of type 'org.springframework.plugin.core.PluginRegistry<org.springframework.hateoas.client.LinkDiscoverer, org.springframework.http.MediaType>' available: expected single matching bean but found 17: modelBuilderPluginRegistry,modelPropertyBuilderPluginRegistry,typeNameProviderPluginRegistry,syntheticModelProviderPluginRegistry,documentationPluginRegistry,apiListingBuilderPluginRegistry,operationBuilderPluginRegistry,parameterBuilderPluginRegistry,expandedParameterBuilderPluginRegistry,resourceGroupingStrategyRegistry,operationModelsProviderPluginRegistry,defaultsProviderPluginRegistry,pathDecoratorRegistry,apiListingScannerPluginRegistry,relProviderPluginRegistry,linkDiscovererRegistry,entityLinksPluginRegistry


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APPLICATION FAILED TO START
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Description:


Parameter 0 of …

由于与我正在处理的项目相关的原因,我希望将JSON文件的整个查询保存为字符串,例如,$.store.book[*].title(而不是必须将文档的每个级别临时存储为单独的对象) .

我目前正在使用JsonPath(版本0.8.0,这是我能找到的最新版本),这基本上正是我正在寻找的,但我得到的例外情况如下所示.我只是使用JsonPath谷歌代码页上给出的示例JSON,使用他们的一个示例查询.

我在这做错了什么？或者,如果没有解决方案,Java中是否有JsonPath的替代方案？我希望能够将整个查询作为字符串传递,并且它必须是Java.

功能:

public void testJsonPath() throws Exception
{
    String query = "$.store.book[*].title";
    List toRet = (List) JsonPath.read(practiceJson, query, new Filter[0]);
    System.out.println(toRet.toString());
}

例外:

java.lang.NoClassDefFoundError: net/minidev/json/parser/ParseException
at com.jayway.jsonpath.spi.JsonProviderFactory$1.create(JsonProviderFactory.java:27)
at com.jayway.jsonpath.spi.JsonProviderFactory.createProvider(JsonProviderFactory.java:32)
at com.jayway.jsonpath.JsonPath.read(JsonPath.java:202)
at com.jayway.jsonpath.JsonPath.read(JsonPath.java:307)
at net.windward.datasource.test.TestJsonDataSource.testJsonPath(TestJsonDataSource.java:119)

练习JSON:

private String practiceJson = "{\n" +
        "    \"store\": {\n" +
        "        \"book\": [ {\n" +
        "            \"category\": \"reference\",\n" +
        "            \"author\": \"Nigel Rees\",\n" +
        "            \"title\": \"Sayings of the Century\",\n" +
        "            \"price\": 8.95\n" +
        "        }, {\n" + …