fasterxml

Json (fasterxml) stackoverflow exception

由 北城以北 提交于 2020-01-01 11:26:12
问题 When trying to serialize a Category I get a stackoverflow. Exception Warning: StandardWrapperValve[dispatcher]: Servlet.service() for servlet dispatcher threw exception java.lang.StackOverflowError at java.lang.ClassLoader.defineClass1(Native Method) at java.lang.ClassLoader.defineClass(ClassLoader.java:760) at org.apache.felix.framework.BundleWiringImpl$BundleClassLoader.findClass(BundleWiringImpl.java:2279) at org.apache.felix.framework.BundleWiringImpl.findClassOrResourceByDelegation

com.fasterxml.jackson.databind.JsonMappingException: Jackson version is too old 2.5.3

由 六月ゝ 毕业季﹏ 提交于 2020-01-01 09:22:38
问题 My OS is OS X 10.11.6. I'm running Spark 2.0, Zeppelin 0.6, Scala 2.11 When I run this code in Zeppelin I get an exception from Jackson. When I run this code in spark-shell - no exception. val filestream = ssc.textFileStream("/Users/davidlaxer/first-edition/ch06") com.fasterxml.jackson.databind.JsonMappingException: Jackson version is too old 2.5.3 at com.fasterxml.jackson.module.scala.JacksonModule$class.setupModule(JacksonModule.scala:56) at com.fasterxml.jackson.module.scala

@XMLElementWrapper issue with com.fasterxml.jackson JacksonJaxbJsonProvider

由 非 Y 不嫁゛ 提交于 2019-12-31 03:18:08
问题 I am trying to get the correct JSON for public class MyTestResponse { @XmlElementWrapper(name = "data") @XmlElement(name = "values") public List<String> test = Arrays.asList("Sidney"); } I now get "values": [ "Sidney" ], instead of "data":{ "values": [ "Sidney" ] }, So the wrapper element "data" is not present. I am using com.fasterxml.jackson stack (2.8.6) inside ServiceMix 7 M3. My JSON provider extends com.fasterxml.jackson.jaxrs.json.JacksonJaxbJsonProvider: import com.fasterxml.jackson

Spring Boot + JSON 处理

由 大城市里の小女人 提交于 2019-12-30 00:57:40
【推荐】2019 Java 开发者跳槽指南.pdf(吐血整理) >>> Spring Boot 之使用 Json 详解 1. 前台不显示属性的空值：NULL 类上增加注解：@JsonSerialize(include=JsonSerialize.Inclusion.NON_NULL) 2. 前后台显示属性名称不一致 属性增加注解： @JsonProperty("myName") 3. 注解：不显示属性 com.fasterxml.jackson.annotation.JsonIgnore OR @JSONField(serialize = false)>来自com.alibaba.fastjson.annotation 4. pom.xml增加 <dependency> <groupId>com.fasterxml.jackson.datatype</groupId> <artifactId>jackson-datatype-jsr310</artifactId> </dependency> 来源： oschina 链接： https://my.oschina.net/u/248712/blog/3055599

How to enable 'ALLOW_NUMERIC_LEADING_ZEROS' feature to allow leading zeroes in JSON Request Body?

由 空扰寡人 提交于 2019-12-24 13:33:09
问题 As per JSON specification, I am aware that leading zeroes are not allowed in integers in JSON. But as per Jackson documentation, there is a property in Jackson library i.e. ALLOW_NUMERIC_LEADING_ZEROS which when enabled, does not throw exceptions when leading zeroes are found. I enabled the property ALLOW_NUMERIC_LEADING_ZEROS by setting following property and still I am getting error: Leading zeroes not allowed . spring.jackson.parser.ALLOW_NUMERIC_LEADING_ZEROS=true Relevant Logs: Caused by

XMLWriter extends attribute name with zdef?

由 你说的曾经没有我的故事 提交于 2019-12-24 00:14:26
问题 I try to serialize some config classes to xml config files using the XMLMapper. but I have some trouble with the attribute generation. Actually the generated XML is perfect, but XMLMapper adds sometimes a prefix to my attribute names. e.g. <Config zdef-2031720317:value="0"> instead of <Config value="0"> This is really bad because i can't process the xml Structure afterwards with XOM anymore :( Where is this effect coming from? I found already the fact that the xml generator seems to auto fix

Gradle import com.fasterxml.jackson.datatype cannot be resolved

由 筅森魡賤 提交于 2019-12-22 17:46:53
问题 import com.fasterxml.jackson.datatype cannot be resolved As the screen shot shows, import com.fasterxml.jackson.datatype cannot be resolved. I have imported jackson-datatype-jsr310', version: '2.9.3 in the build.gradle like this, buildscript { ext { springBootVersion = '1.5.9.RELEASE' } repositories { mavenCentral() } dependencies { classpath("org.springframework.boot:spring-boot-gradle-plugin:${springBootVersion}") } } apply plugin: 'java' apply plugin: 'eclipse' apply plugin: 'org

fasterxml serialize using toString and deserialize using String constructor

由 那年仲夏 提交于 2019-12-20 20:40:46
问题 I have a POJO which looks something like this: public class Thing { private final int x; private final int y; private final int z; public Thing(String strThing) { // parse strThing which is in some arbitrary format to set x, y and z } @Override public String toString() { // return a string representation of thing // (same format as that parsed by the constructor) } @Override public boolean equals(Object obj) ... @Override public int hashCode() ... } and I want to use it as a key for a map (e

jaxrs could not find my custom (de)serializers for joda.money type

由 放肆的年华 提交于 2019-12-20 04:51:33
问题 I have written my custom (de)serializer for joda.money.Money type. I register them with Object Mapper. But when I deploy my war file, it says could not find serializers for joda.money.Money type. import org.joda.money.Money; import com.fasterxml.jackson.core.JsonParser; import com.fasterxml.jackson.core.JsonProcessingException; import com.fasterxml.jackson.databind.DeserializationContext; import com.fasterxml.jackson.databind.deser.std.StdDeserializer; public class MoneyDeserializer extends

Dynamic addition of fasterxml Annotation?

由 不想你离开。 提交于 2019-12-20 04:27:18
问题 Is there a way to set @JsonProperty annotation dynamically like: class A { @JsonProperty("newB") //adding this dynamically private String b; } or can I simply rename field of an instance? If so, suggest me an idea. Also, in what way an ObjectMapper can be used with serialization? 回答1: Assume that your POJO class looks like this: class PojoA { private String b; // getters, setters } Now, you have to create MixIn interface: interface PojoAMixIn { @JsonProperty("newB") String getB(); } Simple
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