使用 Spring RestTemplate 解析 Json 文件答案

【问题标题】：Parsing Json file using Spring RestTemplate使用 Spring RestTemplate 解析 Json 文件
【发布时间】：2017-05-27 18:01:15
【问题描述】：

我有一个基本的 SpringBoot 应用程序。使用 Spring Initializer、嵌入式 Tomcat、Thymeleaf 模板引擎，并打包为可执行 JAR。

我连接到 URL 并打印响应：

BufferedReader br = new BufferedReader(new InputStreamReader(
                (conn.getInputStream())));

        StringBuilder response = new StringBuilder("");
        String output;
        while ((output = br.readLine()) != null) {
            response.append(output);
        }

结果如下：

{"data":[{"device":"18AE63","time":1494516023,"data":"3235","snr":"36.72","linkQuality":"GOOD"},{"device":"18AE63","time":1494515750,"data":"484f4c41","snr":"35.69","linkQuality":"GOOD"}],"paging":{}}

但是当我使用 RestTemplate 时

RestTemplate restTemplate = new RestTemplate();

         List<HttpMessageConverter<?>> messageConverters = new ArrayList<HttpMessageConverter<?>>();        
         MappingJackson2HttpMessageConverter converter = new MappingJackson2HttpMessageConverter();

         converter.setSupportedMediaTypes(Arrays.asList(MediaType.ALL));         
         messageConverters.add(converter);  
         restTemplate.setMessageConverters(messageConverters);  

         MessageList messageList = 
                    restTemplate.getForObject(url, MessageList.class);

我收到了这个错误：

Exception in thread "main" org.springframework.http.converter.HttpMessageNotReadableException: Could not read JSON document: Unexpected character ('<' (code 60)): expected a valid value (number, String, array, object, 'true', 'false' or 'null')
 at [Source: java.io.PushbackInputStream@1d119efb; line: 1, column: 2]; nested exception is com.fasterxml.jackson.core.JsonParseException: Unexpected character ('<' (code 60)): expected a valid value (number, String, array, object, 'true', 'false' or 'null')
 at [Source: java.io.PushbackInputStream@1d119efb; line: 1, column: 2]
    at org.springframework.http.converter.json.AbstractJackson2HttpMessageConverter.readJavaType(AbstractJackson2HttpMessageConverter.java:234)
    at org.springframework.http.converter.json.AbstractJackson2HttpMessageConverter.read(AbstractJackson2HttpMessageConverter.java:219)
    at org.springframework.web.client.HttpMessageConverterExtractor.extractData(HttpMessageConverterExtractor.java:95)
    at org.springframework.web.client.RestTemplate.doExecute(RestTemplate.java:655)
    at org.springframework.web.client.RestTemplate.execute(RestTemplate.java:613)
    at org.springframework.web.client.RestTemplate.getForObject(RestTemplate.java:287)
    at com.tdk.web.controllers.restful.client.RestTemplateGETExample.main(RestTemplateGETExample.java:49)
Caused by: com.fasterxml.jackson.core.JsonParseException: Unexpected character ('<' (code 60)): expected a valid value (number, String, array, object, 'true', 'false' or 'null')
 at [Source: java.io.PushbackInputStream@1d119efb; line: 1, column: 2]
    at com.fasterxml.jackson.core.JsonParser._constructError(JsonParser.java:1702)
    at com.fasterxml.jackson.core.base.ParserMinimalBase._reportError(ParserMinimalBase.java:558)
    at com.fasterxml.jackson.core.base.ParserMinimalBase._reportUnexpectedChar(ParserMinimalBase.java:456)
    at com.fasterxml.jackson.core.json.UTF8StreamJsonParser._handleUnexpectedValue(UTF8StreamJsonParser.java:2689)
    at com.fasterxml.jackson.core.json.UTF8StreamJsonParser._nextTokenNotInObject(UTF8StreamJsonParser.java:878)
    at com.fasterxml.jackson.core.json.UTF8StreamJsonParser.nextToken(UTF8StreamJsonParser.java:772)
    at com.fasterxml.jackson.databind.ObjectMapper._initForReading(ObjectMapper.java:3834)
    at com.fasterxml.jackson.databind.ObjectMapper._readMapAndClose(ObjectMapper.java:3783)
    at com.fasterxml.jackson.databind.ObjectMapper.readValue(ObjectMapper.java:2922)
    at org.springframework.http.converter.json.AbstractJackson2HttpMessageConverter.readJavaType(AbstractJackson2HttpMessageConverter.java:231)
    ... 6 more

POJO：

@JsonInclude(JsonInclude.Include.NON_NULL)
@JsonPropertyOrder({
"data",
"paging"
})
public class MessageList {

    @JsonProperty("data")
    private List<Message> data = null;
    @JsonProperty("paging")
    private Paging paging;
    @JsonIgnore
    private Map<String, Object> additionalProperties = new HashMap<String, Object>();

    @JsonProperty("data")
    public List<Message> getData() {
    return data;
    }

    @JsonProperty("data")
    public void setData(List<Message> data) {
    this.data = data;
    }

    @JsonProperty("paging")
    public Paging getPaging() {
    return paging;
    }

    @JsonProperty("paging")
    public void setPaging(Paging paging) {
    this.paging = paging;
    }

    @JsonAnyGetter
    public Map<String, Object> getAdditionalProperties() {
    return this.additionalProperties;
    }

    @JsonAnySetter
    public void setAdditionalProperty(String name, Object value) {
    this.additionalProperties.put(name, value);
    }

    @Override
    public String toString() {
        return "MessageList [data=" + data + ", paging=" + paging + ", additionalProperties=" + additionalProperties
                + "]";
    }
}

【问题讨论】：

Spring boot 将设置所有标准消息转换器，如果您只需要 json 反序列化，您只需要一个与 json 响应匹配的正确构造的 POJO。你可以发布你的pojo吗？
消息说，您的响应具有意外字符 'restTemplate 试图将响应解析为 JSON。但是你提供了一个有效的 json。我想你错过了什么......

标签： json spring rest jackson restful-url

【解决方案1】：

您得到的响应是 HTML，而不是 JSON。

JsonParseException: Unexpected character ('<' (code 60))

这通常意味着您的 HTTP 调用收到了 HTML 错误响应（想想<html><h1>404 Not Found</h1>...</html>），因此 JSON 解析器抱怨响应不是有效的 JSON。

【讨论】：