我有一個具有即時類型的 JSON 對象，其值2019-10-17T15:10:33.639Z但是當我解析它時，我遇到了這個錯誤：org.springframework.http.converter.HttpMessageNotReadableException: JSON parse error: Unexpected token (START_OBJECT), expected one of [VALUE_STRING, VALUE_NUMBER_INT, VALUE_NUMBER_FLOAT] for java.time.Instant value; nested exception is com.fasterxml.jackson.databind.exc.MismatchedInputException: Unexpected token (START_OBJECT), expected one of [VALUE_STRING, VALUE_NUMBER_INT, VALUE_NUMBER_FLOAT] for java.time.Instant value
at [Source: (PushbackInputStream); line: 1, column: 28] (through reference chain: com.bendiciones.buenas.noches.Bendiciones["estimatedResolutionDate"])
1 回答
慕蓋茨4494581
TA貢獻1850條經(jīng)驗 獲得超11個贊
在您的 Instant 屬性上添加@JsonFormat，執(zhí)行如下操作
?@JsonFormat(shape?=?JsonFormat.Shape.STRING,?pattern?=?"yyyy-MM-dd'T'HH:mm:ss.SSSZ") ?private?Instant?currentdDate;
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