如何在Spring中使用LocalDateTime RequestParam?我收到“无法将String转换为LocalDateTime的信息"； [英] How to use LocalDateTime RequestParam in Spring? I get "Failed to convert String to LocalDateTime"

问题描述

我使用Spring Boot并将jackson-datatype-jsr310包含在Maven中:

I use Spring Boot and included jackson-datatype-jsr310 with Maven:

<dependency>
    <groupId>com.fasterxml.jackson.datatype</groupId>
    <artifactId>jackson-datatype-jsr310</artifactId>
    <version>2.7.3</version>
</dependency>

当我尝试使用Java 8日期/时间类型的RequestParam时，

When I try to use a RequestParam with a Java 8 Date/Time type,

@GetMapping("/test")
public Page<User> get(
    @RequestParam(value = "start", required = false)
    @DateTimeFormat(iso = DateTimeFormat.ISO.DATE_TIME) LocalDateTime start) {
//...
}

并使用以下URL进行测试:

and test it with this URL:

/test?start=2016-10-8T00:00

我收到以下错误:

{
  "timestamp": 1477528408379,
  "status": 400,
  "error": "Bad Request",
  "exception": "org.springframework.web.method.annotation.MethodArgumentTypeMismatchException",
  "message": "Failed to convert value of type [java.lang.String] to required type [java.time.LocalDateTime]; nested exception is org.springframework.core.convert.ConversionFailedException: Failed to convert from type [java.lang.String] to type [@org.springframework.web.bind.annotation.RequestParam @org.springframework.format.annotation.DateTimeFormat java.time.LocalDateTime] for value '2016-10-8T00:00'; nested exception is java.lang.IllegalArgumentException: Parse attempt failed for value [2016-10-8T00:00]",
  "path": "/test"
}

推荐答案

TL; DR -您可以仅使用@RequestParam将其捕获为字符串，也可以让Spring额外解析该字符串也可以通过参数上的@DateTimeFormat转换为Java日期/时间类.

TL;DR - you can capture it as a string with just @RequestParam, or you can have Spring additionally parse the string into a java date / time class via @DateTimeFormat on the parameter as well.

@RequestParam足以在=符号后获取您提供的日期，但是，它作为String进入方法.这就是为什么它引发强制转换异常的原因.

the @RequestParam is enough to grab the date you supply after the = sign, however, it comes into the method as a String. That is why it is throwing the cast exception.

有几种方法可以做到这一点:

There are a few ways to achieve this:

1. 自己解析日期，以字符串的形式获取值.

@GetMapping("/test")
public Page<User> get(@RequestParam(value="start", required = false) String start){

    //Create a DateTimeFormatter with your required format:
    DateTimeFormatter dateTimeFormat = 
            new DateTimeFormatter(DateTimeFormatter.BASIC_ISO_DATE);

    //Next parse the date from the @RequestParam, specifying the TO type as 
a TemporalQuery:
   LocalDateTime date = dateTimeFormat.parse(start, LocalDateTime::from);

    //Do the rest of your code...
}

1. 利用Spring的能力来自动解析和期望日期格式:

@GetMapping("/test")
public void processDateTime(@RequestParam("start") 
                            @DateTimeFormat(iso = DateTimeFormat.ISO.DATE_TIME) 
                            LocalDateTime date) {
        // The rest of your code (Spring already parsed the date).
}

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