Question

如何将dateOfJoining值（字符串类型）绑定到成员变量＆＃34; dateOfJoining＆＃34;（日期类型）＆＃34; DateInput＆＃34;通过邮递员发送到JSON以下的课程。如何使用相同的格式dd / MM / yyyy将String中的String转换为Date对象。转换日期应该在Date对象中，而不是String。

Json如下所示

{
 "dateOfJoining" : "03/04/2017"
}

服务URL在postman中命中 - localhost / Rest / hello

java中的RestService类 - HandleRestRequest.java

@RestController
public class HandleRestRequest
{

  @RequestMapping("\hello");
  public List setRequestParams(@RequestBody DateInput dateInput)
 {
   .......
  }
 }

 Pojo Class DateInput.java

  public class DateInput
 {
  private  Date dateOfJoining;
   .......
  }

如果我以下面的格式从json发送日期，则其投掷错误为输入不满意

 {
  "dateOfJoining" : 04/04/2017
 }

所以我将它作为字符串格式发送，并将dateOfJoining更改为DateInput.java文件中的字符串，稍后我尝试将其转换为日期对象，如下所示

从Date到String修改DateInput.java文件

 public class DateInput
 {
  private  String dateOfJoining;
   .......
 }

修改了JSON

{
 "dateOfJoining" : "04/04/2017"
}

将用户输入从String转换为Date

的代码

 DateFormat sdf = new SimpleDateFormat("dd/MM/yyyy");
  String convertedDate = sdf.format(dateInput.getDateOfJoining());

它转换为所需的格式但返回类型是String但是期望是发送DAO层的Date对象。所以我尝试了sdf.parse，它返回Date对象但不是必需的格式

Date date = sdf.parse(sdf.format(dateInput.getDateOfJoining()));
output is like -  Tue Apr 04 00:00:00 IST 2017
expected is 04/04/2017 (with return type Date object).

所以请帮助我如何将字符串转换为具有所需格式的日期对象，因为DAO层期望输入为日期对象，格式为dd / MM / yyyy

Answer 1

编辑：根据更新的问题更新答案。

使用Jackson Databind中的注释@JsonFormat指定日期模式。

public class DateInput
 {
  @JsonFormat(shape=JsonFormat.Shape.STRING, pattern="dd-MM-yyyy")
  private  Date dateOfJoining;
   .......
  }

Answer 2

将您的代码更改为以下代码段

public List setRequestParams(@RequestParam("dateOfJoining")@DateTimeFormat(pattern="dd-MM-yyyy")  DateInput dateInput)
{
   ...
}

Answer 3

使用JSON-B（包含在Java EE 8中），您可以执行以下操作：

    $client = new Client(array( 'curl' => array( CURLOPT_SSL_VERIFYPEER => false, CURLOPT_SSL_VERIFYHOST =>0), ));
    $res = $client->request('GET', 'https://assignment-api.uspto.gov/patent/basicSearch?query='.$request->input('query'));
    $respons = $res->getBody()->getContents();
    // dd($respons);
    $xml = XmlParser::extract($respons);
    $doc = $xml->parse([
        'publNum' => ['uses' => 'result.publNum'],
        'inventionTitle' => ['uses' => 'result.inventionTitle'],
        'inventors' => ['uses' => 'result.inventors'],
        'corrName' => ['uses' => 'result.corrName'],
        'corrAddress1' => ['uses' => 'result.corrAddress1'],
        'corrAddress2' => ['uses' => 'result.corrAddress2'],
        'corrAddress3' => ['uses' => 'result.corrAddress3'],
        'patAssigneeName' => ['uses' => 'result.patAssigneeName'],
        'patAssigneeAddress1' => ['uses' => 'result.patAssigneeAddress1'],
        'patAssigneeAddress2' => ['uses' => 'result.patAssigneeAddress2'],
        'patAssigneeCity' => ['uses' => 'result.patAssigneeCity'],
        'patAssigneeState' => ['uses' => 'result.patAssigneeState'],
        'patAssigneeCountryName' => ['uses' => 'result.patAssigneeCountryName'],
    ]);

    dd($doc);

在使用Thorntail 2.4进行的测试中，使用class DateInput { @JsonbDateFormat("dd/MM/yyyy") public Date dateOfJoining; }时不需要ISO格式的任何注释：

java.util.Date

从post man（休息服务）如何发送json日期（字符串格式）到接受日期对象的java

3 个答案: