Thi*_*sen 7 java spring date formatter
我在使用@DateTimeFormat注释转换实际不存在的日期时遇到问题.
例如,当我设置日期15/10/2017时,我的实体中的注释如下:
@Column(nullable = false)
@NotNull
@DateTimeFormat(pattern = "dd/MM/yyyy")
private Date dataVisita;
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我收到错误:
Failed To Convert Property Value Of Type Java.Lang.String To Required Type Java.Util.Date For Property DataVisita;
Nested Exception Is Org.Springframework.Core.Convert.ConversionFailedException:
Failed To Convert From Type Java.Lang.String To Type @Javax.Persistence.Column @Javax.Validation.Constraints.NotNull @Org.Springframework.Format.Annotation.DateTimeFormat Java.Util.Date For Value 15/10/2017;
Nested Exception Is Java.Lang.IllegalArgumentException:
Cannot Parse "15/10/2017": Illegal Instant Due To Time Zone Offset Transition (America/Sao_Paulo)
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我明白错误告诉我日期15/10/2017 00:00:00实际上并不存在,但我想转换为15/10/2017 01:00:00,忽略这种方式问题并找到相应的日期.
有没有办法让我覆盖@DateTimeFormat注释或将格式化程序指向宽松的方法?
尝试这个:
@Column(可为空 = false)
@NotNull
@DateTimeFormat(模式 = "dd/MM/yyyy HH:mm:ss")
私人日期数据访问;
它似乎做了你想要的(调整 0:00 到 1:00),你需要实现一个验证器,如下所示:
public class PersonValidator implements Validator {
/**
* This Validator validates *just* Person instances
*/
public boolean supports(Class clazz) {
return Date.class.equals(clazz);
}
public void validate(Object obj, Errors e) {
String dateString = (String) obj;
DateTimeFormatter fmt = DateTimeFormat.forPattern("dd/MM/yyyy HH:mm:ss");
DateTime ourDate = null;
try {
fmt.parseDateTime(dateString);
} catch (IllegalArgumentException e) {
// massage your hour here
} finally {
ourDate = fmt.parseDateTime(dateString);
}
}
}
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