我正在链接多个MapReduce作业,并希望传递/存储一些元信息(例如配置或原始输入的名称)和结果.至少文件"_SUCCESS"以及目录"_logs"中的任何内容都会被忽略.
是否有任何文件名模式默认忽略InputReader?或者这只是一个固定的有限列表?
我的 Mapper 有问题。我正在使用 mapstruct-processor 来构建 Maven 项目。我一直收到警告:警告:(15, 16)java:未映射的目标属性:“从,到”。警告:(13, 13) java:未映射的目标属性:“clientFrom,clientTo”。我想用它做什么?类 Client 运行良好并创建了一个客户端。
@Entity
public class Message {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private Long id;
private String message;
@ManyToOne(cascade = {CascadeType.ALL})
@JoinColumn(name = "client_from")
private Client clientFrom;
@ManyToOne
@JoinColumn(name = "client_to")
private Client clientTo;
public Message(){}
public Long getId() {
return id;
}
public void setId(Long id) {
this.id = id;
}
public String getMessage() {
return message;
}
public void setMessage(String message) {
this.message = message;
}
public Client getClientFrom() …Run Code Online (Sandbox Code Playgroud) 我从这个来源下载了应用程序https://github.com/springframeworkguru/spring5-mvc-rest/tree/vendor-api 而且我的MapStruct有问题。
@Mapper
public interface CategoryMapper {
CategoryMapper INSTANCE = Mappers.getMapper(CategoryMapper.class);
CategoryDTO categoryToCategoryDTO(Category category);
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}
@Data
public class CategoryDTO {
private Long id;
private String name;
}
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域类:
@Data
@Entity
public class Category {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
private String name;
}
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服务等级:
@Service
public class CategoryServiceImpl implements CategoryService {
private final CategoryMapper categoryMapper;
private final CategoryRepository categoryRepository;
public CategoryServiceImpl(CategoryMapper categoryMapper, CategoryRepository categoryRepository) {
this.categoryMapper = categoryMapper;
this.categoryRepository = categoryRepository;
}
}
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在pom.xml依赖项中,我仅粘贴了两个:
<properties>
<project.build.sourceEncoding>UTF-8</project.build.sourceEncoding>
<project.reporting.outputEncoding>UTF-8</project.reporting.outputEncoding> …Run Code Online (Sandbox Code Playgroud) 我有一个班(A并且B是我的对象)
public A {
private List<B> b;
private int c;
}
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我有一个String变量temp.所以我会在这个变量中保存我的对象的JSON字符串.怎么样?
我试过这个2方法,但没有结果.
public void save() {
ObjectMapper mapper = new ObjectMapper();
String temp = mapper.writeValueAsString(a);
}
public void read(String temp) {
ObjectMapper mapper = new ObjectMapper();
A a = mapper.readValue(temp, A.class);
}
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我该如何解决?
在解决以下代码时,我得到了下面给出的例外情况.
List<Map<String, Object>> obj = mapper.readValue(result.getBody(), new TypeReference<Map<String,Object>>(){});
for (Map<String, Object> map : obj) {
for(Map.Entry<String, Object> entry : map.entrySet()){
System.out.println("Key : "+entry.getKey()+" Value is: "+entry.getValue());
}
}
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Stacktrace:java.lang.ClassCastException: java.util.LinkedHashMap cannot be cast to java.util.List
我试图从API中获取一些Json并将它们解析为一些POJO以与它们一起工作但我有这种情况我可以获得一个简单的String或一个字符串的数组.
Json看起来像这样:
{
"offerDisplayCategoryMapping": [
{
"offerKey": "EUICC_BASE_ACTIVATION_V01",
"categoriesKeys": {
"categoryKey": "Included"
}
},
{
"offerKey": "EUICC_BASE_ACTIVATION_V02",
"categoriesKeys": {
"categoryKey": "Included"
}
},
{
"offerKey": "EUICC_BASE_ACTIVATION_V03",
"categoriesKeys": {
"categoryKey": [
"Option",
"Included"
]
}
}]
}
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我正在使用Spring Rest从API中获取结果.我创建了一个POJO代表categoriesKeys一个List<String>定义categoryKey,在我RestTemplate定义的ObjectMapper地方我启用DeserializationFeature.ACCEPT_SINGLE_VALUE_AS_ARRAY了简单字符串的情况,但这不起作用!!
有什么建议?
我的 flutter 应用程序目前通过 USB 在我的手机上运行..但是当我运行我的应用程序时,它显示W/Gralloc3(12347): mapper 3.x is not supported我该如何解决这个问题?
我有以下课程:
public class A{
List<AA> aaList;
public A(List<AA> aaList){
this.aaList = aaList;
}
//getters and setters + default constructor
public class AA {
String aaString;
public AA(String aaString){
this.aaString = aaString;
}
//getters and setters + default constructor
}
}
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我希望有两个同类的对象,让我们说:
A a = new A(Arrays.asList(new A.AA(null)));
A a2 = new A(Arrays.asList(new A.AA("test")));
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当我映射a到时a2,a2应该保留,test因为a有一个null.
我该如何配置Orika?
我尝试过类似的东西:
mapperFactory.classMap(A.AA.class, A.AA.class)
.mapNulls(false)
.byDefault()
.register();
mapperFactory.classMap(A.class, A.class)
.mapNulls(false)
.customize(new CustomMapper<A, …Run Code Online (Sandbox Code Playgroud) 我的代码中有这个:
ObjectMapper oMapper = new ObjectMapper();
Map<String, Object> objectMap = oMapper.convertValue(test, Map.class);
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我收到以下警告:
Type safety: The expression of type Map needs unchecked conversion to conform to Map<String,Object>
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我该如何解决这个问题?
解决方案:
import com.fasterxml.jackson.core.type.TypeReference;
ObjectMapper oMapper = new ObjectMapper();
Map<String, Object> objectMap = oMapper.convertValue(prueba, new TypeReference<Map<String, Object>>(){});
Run Code Online (Sandbox Code Playgroud) 是否可以从application.properties文件中的字段设置值?
我正在寻找类似的东西
@Mapping(target="version", expression="${application.version}")
StateDto stateToStateDto(State state);
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其中application.version=v1来自application.properties文件。