java HTTP 状态 400 - 所需的字符串参数“walletName”不存在

HTTP Status 400 - Required String parameter 'walletName' is not present

提问by Chaklader Asfak Arefe

I work with Java/ Spring MVC RESTfulapp and get 400 HTTP status errorwhile doing a POSTrequest. The @RestControllermethod is provided,

我使用Java/ Spring MVC RESTful应用程序并400 HTTP status error在执行POST请求时获取。@RestController提供的方法，

@RequestMapping(value = "/generateAddress", method = RequestMethod.POST)
    public ResponseEntity<WalletInfoWrapper> generateAddress(@RequestParam("walletName") String walletName,
                                                             @RequestParam("currencyName") String currencyName) {

        logger.info("walletName {} and currencyName {}", walletName, currencyName);

        // return if the wallet name or the currency is null
        if (Objects.isNull(walletName) || Objects.isNull(currencyName)) {
            return new ResponseEntity<WalletInfoWrapper>(HttpStatus.NOT_ACCEPTABLE);
        }

        WalletInfo walletInfo = walletService.generateAddress(walletName, currencyName);

        if (Objects.isNull(walletInfo)) {
            return new ResponseEntity<WalletInfoWrapper>(HttpStatus.NOT_ACCEPTABLE);
        }

        WalletInfoWrapper walletInfoWrapper = new WalletInfoWrapper();
        walletInfoWrapper.setName(walletInfo.getName());

        return new ResponseEntity<WalletInfoWrapper>(walletInfoWrapper, HttpStatus.CREATED);
    }

The POSTrequest in the Postmanprovided below,

下面提供的POST请求Postman，

The error message informs, Required String parameter 'walletName' is not present

错误消息通知， Required String parameter 'walletName' is not present

I can also provide the code for the servicesand the dataaselayers for observing the drop-down operations. What is the issue here?

我还可以提供用于观察下拉操作services的dataase层和层的代码。这里有什么问题？

UPDATE

UPDATE

I updated the Postmanrequest like this and still having the same error,

我Postman像这样更新了请求，但仍然出现相同的错误，

UPDATE 1

UPDATE 1

I still have the same issue,

我还是有同样的问题

I POSTwith the data,

我POST用数据，

{"walletName":"puut","currencyName":"Bitcoin"}

The code is provided below,

下面提供了代码，

@RequestMapping(value = "/generateAddress", method = RequestMethod.POST, produces = MediaType.APPLICATION_JSON_VALUE)
    public ResponseEntity<WalletInfoWrapper> generateAddress(@RequestBody WalletWithMoneyRequest walletWithMoneyRequest) {

        String walletName = walletWithMoneyRequest.getWalletName();

        String currencyName = walletWithMoneyRequest.getCurrencyName();

        logger.info("walletName {} and currencyName {}", walletName, currencyName);

        // return if the wallet name or the currency is null
        if (Objects.isNull(walletName) || Objects.isNull(currencyName)) {
            return new ResponseEntity<WalletInfoWrapper>(HttpStatus.NOT_ACCEPTABLE);
        }

        WalletInfo walletInfo = walletService.generateAddress(walletName, currencyName);

        if (Objects.isNull(walletInfo)) {
            return new ResponseEntity<WalletInfoWrapper>(HttpStatus.NOT_ACCEPTABLE);
        }

        WalletInfoWrapper walletInfoWrapper = new WalletInfoWrapper();
        walletInfoWrapper.setName(walletInfo.getName());

        return new ResponseEntity<WalletInfoWrapper>(walletInfoWrapper, HttpStatus.CREATED);
    }

The POJOis provided,

该POJO提供，

private class WalletWithMoneyRequest {

        String walletName;

        String currencyName;

        public WalletWithMoneyRequest(String walletName, String currencyName) {
            this.walletName = walletName;
            this.currencyName = currencyName;
        }

        public WalletWithMoneyRequest() {
        }

        public String getWalletName() {
            return walletName;
        }

        public String getCurrencyName() {
            return currencyName;
        }

        public void setCurrencyName(String currencyName) {
            this.currencyName = currencyName;
        }

        public void setWalletName(String walletName) {
            this.walletName = walletName;
        }
    }

This is the error message,

这是错误信息，

Here ia the Tomcat server info,

这里是 Tomcat 服务器信息，

org.springframework.web.servlet.mvc.support.DefaultHandlerExceptionResolver: 08/19/2017 19:45:55 - Failed to read HTTP message: org.springframework.http.converter.HttpMessageNotReadableException: JSON parse error: Cannot construct instance of `mobi.puut.controllers.WalletRestController$WalletWithMoneyRequest` (although at least one Creator exists): can only instantiate non-static inner class by using default, no-argument constructor; nested exception is com.fasterxml.Hymanson.databind.exc.MismatchedInputException: Cannot construct instance of `mobi.puut.controllers.WalletRestController$WalletWithMoneyRequest` (although at least one Creator exists): can only instantiate non-static inner class by using default, no-argument constructor
 at [Source: (PushbackInputStream); line: 1, column: 2]

Tomcat Localhost log

Tomcat Localhost log

Tomcat Catalina log

Tomcat Catalina log

采纳答案by zerpsed

Your controller is expecting 2 request parameters that normally look like this: /someurl?walletName=my-wallets-name&currencyName=dollars.

您的控制器需要 2 个通常如下所示的请求参数：/someurl?walletName=my-wallets-name¤cyName=dollars。

You're sending a json string in the post body, but no formal parameters. You need to update either your POST, or your controller to make the two ends agree. I think you probably want to replace the two @RequestParam annotated Strings, with a Java pojo that has two String members: walletName and currencyName, drop that pojo in your request method as an argument and precede it with the annotation @RequestBody. This will match your json post.

您在帖子正文中发送一个 json 字符串，但没有形式参数。您需要更新您的 POST 或您的控制器以使两端一致。我认为您可能想用一个 Java pojo 替换两个带 @RequestParam 注释的字符串，该 Java pojo 具有两个字符串成员：walletName 和 currencyName，将该 pojo 作为参数放入您的请求方法中，并在其前面加上注释 @RequestBody。这将匹配您的 json 帖子。

To have your controller accept the post with JSON in the body edit it like this:

要让您的控制器接受正文中带有 JSON 的帖子，请像这样编辑它：

@RequestMapping(value = "/generateAddress", method = RequestMethod.POST)
public ResponseEntity<WalletInfoWrapper> generateAddress(@RequestBody
    WalletWithMoneyRequest myJsonRequestComingIn) {
    logger.info("walletName {} and currencyName {}", myJsonRequestComingIn.getWalletName(), myJsonRequestComingIn.getCurrencyName());

And your pojo

还有你的 pojo

public class WalletWithMoneyRequest{ 

    private String walletName;
    private String currencyName;

    //getters and setters down here. 

回答by tima

Edit

编辑

In your Postman request, instead of sending JSON, send the values as x-www-form-urlencoded.

在您的 Postman 请求中，不是发送 JSON，而是将值发送为x-www-form-urlencoded.

回答by Kim Rasmussen

To elaborate on zerpsed's answer.

详细说明 zerpsed 的回答。

Change the signature to:

将签名更改为：

public ResponseEntity<WalletInfoWrapper> generateAddress(@ResponseBody WalletPOJO walletCurrency)

Where the WalletPOJO has the two fields walletName and currencyName

WalletPOJO 有两个字段 walletName 和 currencyName

回答by Chaklader Asfak Arefe

I believe the issue is solved for now. I have used a POJOas suggested with the @RequestBodyparameter in the RESTfulmethod. The catch here is I need to make the POJOout of the class (in the same file though) and later, put in the entity directory as an entity.

我相信这个问题现在已经解决了。我已经POJO按照建议使用@RequestBody了RESTful方法中的参数。这里的问题是我需要制作POJO出类（尽管在同一个文件中），然后将其作为实体放入实体目录中。

class WalletWithMoneyRequest {

    String walletName;

    String currencyName;

    public WalletWithMoneyRequest(String walletName, String currencyName) {
        this.walletName = walletName;
        this.currencyName = currencyName;
    }

    public WalletWithMoneyRequest() {
    }

    public String getWalletName() {
        return walletName;
    }

    public String getCurrencyName() {
        return currencyName;
    }

    public void setCurrencyName(String currencyName) {
        this.currencyName = currencyName;
    }

    public void setWalletName(String walletName) {
        this.walletName = walletName;
    }
}

The main issue is believe was an error in the HQL,

主要问题是相信是错误的HQL，

I wrote currency =: currencywhere it should be currency = :currency

我写currency =: currency了它应该在的地方currency = :currency

I still can't have the data in the database as I will need to modify the method in the databaselayer.

我仍然无法在数据库中拥有数据，因为我需要修改database图层中的方法。

回答by Paul Guacan

En POSTMAN set variables in params

En POSTMAN 在 params 中设置变量