无法从 VALUE_STRING 令牌中反序列化 java.util.ArrayList 的实例
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原文地址: http://stackoverflow.com/questions/40147304/
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Can not deserialize instance of java.util.ArrayList out of VALUE_STRING token
提问by Harleen
I have looked for other post in stack overflow but none of them work for me. Here is the piece of code:
我已经在堆栈溢出中寻找其他帖子，但没有一个对我有用。这是一段代码：
public class Forward implements Serializable {
private List<String> freq;
public List<String> getFreq() {
System.out.println("Print Freq::: --> " + freq);
return freq;
}
public void setFreq(List<String> freq) {
this.freq = freq;
}
}
The JSON string is:
JSON 字符串是：
{"forward":[{"freq":"78000000"}]}
My mapper is:
我的映射器是：
ObjectMapper mapper = new ObjectMapper();
mapper.configure(DeserializationFeature.ACCEPT_SINGLE_VALUE_AS_ARRAY, true);
String jsonString = mapper.writeValueAsString(result);
If I remove List freq and change to String freq it works but my JSON can contain one or more freq so I need to create a List.I get exception as:
如果我删除 List freq 并更改为 String freq 它可以工作，但我的 JSON 可以包含一个或多个 freq，所以我需要创建一个 List。我得到异常：
Can not deserialize instance of java.util.ArrayList out of VALUE_STRING token at [Source: org.glassfish.jersey.message.internal.ReaderInterceptorExecutor$UnCloseableInputStream
采纳答案by Andremoniy
DeserializationFeature.ACCEPT_SINGLE_VALUE_AS_ARRAYworks fine in order to deserialize "freq":"78000000"fragment as value of List<String> freqlist.
DeserializationFeature.ACCEPT_SINGLE_VALUE_AS_ARRAY工作正常，以便将"freq":"78000000"片段反序列化为List<String> freq列表的值。
But you have another problem: your jsoncontains explicit array of forward. In order to deserialize this entire jsonyou need to have some kind of wrapper class, say:
但是您还有另一个问题：您json包含显式的forward. 为了反序列化这整个json你需要有某种包装类，说：
public class ForwardWrapper {
private List<Forward> forward;
public List<Forward> getForward() {
return forward;
}
public void setForward(List<Forward> forward) {
this.forward = forward;
}
}
In this case
在这种情况下
ForwardWrapper fw = mapper.readValue("{\"forward\":[{\"freq\":\"78000000\"}]}", ForwardWrapper.class);
will deserialize it perfectly.
将完美地反序列化它。
