java.sql.Time : 正确插入当前时间

Question

我正在尝试将当前时间插入到 java.sql.Time 字段中，但出现此异常：

default message [Failed to convert property value of type 'java.lang.String' to required type 'java.sql.Time' for property 'entryDate'; nested exception is org.springframework.core.convert.ConversionFailedException: Failed to convert from type [java.lang.String] to type [java.sql.Time] for value ''; nested exception is java.lang.IllegalArgumentException]

这是代码：

firm.setEntryDate(Time.valueOf(java.time.LocalDate.now().toString()));

任何人都可以告诉我做错了什么吗？谢谢！

更新：Firm.class

@JsonProperty("webshow")
private String webShow;
private Time entryDate;

public String getWebShow () {
    return webShow;
}

public void setWebShow (String webShow) {
    this.webShow = webShow;
}

public Time getEntryDate () {
    return entryDate;
}

public void setEntryDate (Time entryDate) {
    this.entryDate = entryDate;
}

完全异常：

 Resolved exception caused by Handler execution: org.springframework.validation.BindException: org.springframework.validation.BeanPropertyBindingResult: 1 errors
Field error in object 'firm' on field 'entryDate': rejected value []; codes [typeMismatch.firm.entryDate,typeMismatch.entryDate,typeMismatch.java.sql.Time,typeMismatch]; arguments [org.springframework.context.support.DefaultMessageSourceResolvable: codes [firm.entryDate,entryDate]; arguments []; default message [entryDate]]; default message [Failed to convert property value of type 'java.lang.String' to required type 'java.sql.Time' for property 'entryDate'; nested exception is org.springframework.core.convert.ConversionFailedException: Failed to convert from type [java.lang.String] to type [java.sql.Time] for value ''; nested exception is java.lang.IllegalArgumentException]

Answer 1

试试这个：

firm.setEntryDate(java.sql.Time.valueOf(LocalTime.now()));

Answer 2

使用时间戳

私有时间戳入口日期；

java.sql.Time : 正确插入当前时间

java.sql.Time : insert current time correctly

java

spring

java-time