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首页 > 解决方案 > 如何使用错误响应休息模板处理 200 状态码？

java - 如何使用错误响应休息模板处理 200 状态码？

问题描述

我想用错误响应处理 200

final ResponseEntity<ResponseType> responseEntity = restTemplate.postForEntity(url, requestEntity, ResponseType.class);

这里我直接使用了rest模板中的响应模型对象，下面是我的catch块

catch (final RestClientException | HttpMessageConversionException ex) {
                throw new CustomException(message, ex);
            }

在这里我添加了HttpMessageConversionException因为我得到了 200 错误响应，并且在将其转换为我在 rest 模板中的成功响应类型类时它会抛出，因为在我的成功响应类构造函数中具有字段验证。

下面是我在休息模板中使用的响应类

@JsonDeserialize(builder = ResponseType.Builder.class)
@JacksonXmlRootElement(localName = "Test")
public class ResponseType {

    private final String details;

    private ResponseType(final Builder builder) {
        this.details = OptionalCheck.checkPresent(builder.details, "details");
    }

    public static Builder builder() {
        return new Builder();
    }

    //getter

    public static final class Builder {
        private Optional<String> details = Optional.empty();

        @JacksonXmlProperty(localName = "details")
        public Builder withDetails(final String theDetails) {
            this.details = Optional.ofNullable(theDetails);
            return this;
        }


        public ResponseType build() {
            return new ResponseType(this);
        }
    }
}

下面是堆栈跟踪

org.springframework.http.converter.HttpMessageConversionException: Type definition error: [simple type, class com.test.ResponseType$Builder]; nested exception is com.fasterxml.jackson.databind.exc.InvalidDefinitionException: Cannot construct instance of `com.test.ResponseType$Builder`, problem: details must be present.
 at [Source: (ByteArrayInputStream); line: 5, column: 1]
    at org.springframework.http.converter.json.AbstractJackson2HttpMessageConverter.readJavaType(AbstractJackson2HttpMessageConverter.java:242)
    at org.springframework.http.converter.json.AbstractJackson2HttpMessageConverter.read(AbstractJackson2HttpMessageConverter.java:227)
    at org.springframework.web.client.HttpMessageConverterExtractor.extractData(HttpMessageConverterExtractor.java:102)
    at org.springframework.web.client.RestTemplate$ResponseEntityResponseExtractor.extractData(RestTemplate.java:994)
    at org.springframework.web.client.RestTemplate$ResponseEntityResponseExtractor.extractData(RestTemplate.java:977)
    at org.springframework.web.client.RestTemplate.doExecute(RestTemplate.java:737)
    at org.springframework.web.client.RestTemplate.execute(RestTemplate.java:670)
    at org.springframework.web.client.RestTemplate.postForEntity(RestTemplate.java:445)


    Caused by: java.lang.IllegalArgumentException: details must be present.
    at com.test.OptionalCheck.lambda$checkPresent$0(OptionalCheck.java:46)
    at java.base/java.util.Optional.orElseThrow(Optional.java:408)

有没有其他方法来处理这个？

标签： javaspringspring-bootresttemplatespring-rest

解决方案

为了获得更好的帮助，您应该向我们展示一个良好响应和错误响应的示例。

但我会举一个例子来说明你应该如何做。

假设一个好的响应如下所示：

{
    "details": {
        ...
    }
}

错误响应如下所示：

{
    "error": {
        ...
    }
}

然后您必须将响应类型定义为具有 2 个字段的类：details和error，然后检查哪个存在并根据需要做出反应，例如

DetailsOrErrorResponse response = restTemplate.getForObject(url, DetailsOrErrorResponse.class);
if (response.getError() != null) {
    // handle error here
} else if (response.getDetails() == null) {
    throw new IllegalArgumentException("Invalid response: Missing `details` or `error`");
} else {
    // handle good response here
}

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