使用 Spring 框架发送 POST 请求答案

【问题标题】：Sending POST request using Spring framework使用 Spring 框架发送 POST 请求
【发布时间】：2018-04-04 13:38:54
【问题描述】：

我有一个 MySql 记录表，其中有一列名为 ConsignmentCode。我想发送一个 HTTP 请求以通过 ConsignmentCode 搜索表。

当我使用 Postman 向 http://localhost:8081/records/consignmentCode 发出 POST 请求时，正文中包含 {consignmentCode: "123456789"}，我收到 404 Not Found 错误。有谁知道为什么？

控制器类：

  /**
     * POST request to search by ConsignmentCode
     *
     * /records/consignmentCode
     *
     * Input ex: {consignmentCode: "123456789"}
     * @param params
     * @return
     */

    @CrossOrigin
    @ResponseBody
    @RequestMapping(
            value = "/consignmentCode",
            method = RequestMethod.POST,
            consumes= MediaType.APPLICATION_JSON_VALUE)
    public java.lang.String SearchRecordsByConsignmentCode(@RequestBody String params) {
        System.out.print("this SearchRecordsByConsignmentCode respond happened");
        JSONObject obj = new JSONObject();
        JSONObject jsonObj = new JSONObject(params);
        String likeConsignment= jsonObj.getString("consignmentCode");

        List<record> results=RecordDao.SearchRecordsByConsignmentCode(likeConsignment);
        obj.put("results", results);
        return obj.toString();
    }

道类：

 /**
     * Search by ConsignmentCode.
     *
     * @param consignmentCode
     * @return
     */

    public List<record> SearchRecordsByConsignmentCode(String consignmentCode) {
        final String sql = "SELECT * FROM records WHERE ConsignmentCode = ?";
        List<record> Record = jdbcTemplate.query(sql, new RowMapper<record>() {
            public record mapRow(ResultSet resultSet, int Id) throws SQLException {
                record records = new record();
                records.setId(resultSet.getInt("Id"));
                records.setNumber(resultSet.getString("Number"));
                records.setTitle(resultSet.getString("Title"));
                records.setScheduleId(resultSet.getInt("ScheduleId"));
                records.setTypeId(resultSet.getInt("TypeId"));
                records.setConsignmentCode(resultSet.getString("ConsignmentCode"));
                records.setStateId(resultSet.getInt("StateId"));
                records.setContainerId(resultSet.getInt("ContainerId"));
                records.setLocationId(resultSet.getInt("LocationId"));
                records.setCreatedAt(resultSet.getDate("CreatedAt"));
                records.setUpdatedAt(resultSet.getDate("UpdatedAt"));
                records.setClosedAt(resultSet.getDate("ClosedAt"));
                System.out.print(records);
                return records;
            }
        }, consignmentCode);
        return Record;
    }

【问题讨论】：

好吧，你的 DAO 类可能与错误无关。你应该在Spring的DispatcherServlet中下一个断点，看看错误在哪里调试doService方法。

标签： mysql spring rest spring-mvc

【解决方案1】：

使用：

@RequestBody String consignmentCode

代替：

@RequestBody String params

【讨论】：

【解决方案2】：

您不必期望收到一个字符串然后解析为 JSon。它是 Spring MVC 中的内置行为。

只需导入杰克逊依赖项。如果您使用 Maven，请添加您的 pom.xml 依赖项：

<dependency>
    <groupId>org.codehaus.jackson</groupId>
    <artifactId>jackson-mapper-asl</artifactId>
    <version>1.9.9</version>
</dependency>
<dependency>
    <groupId>org.codehaus.jackson</groupId>
    <artifactId>jackson-core-asl</artifactId>
    <version>1.9.9</version>
</dependency>

然后以这种方式更改您的控制器：

@Controller
public class RestController {

    @ResponseBody
    @RequestMapping(
            value = "/consignmentCode",
            method = RequestMethod.POST,
            consumes= MediaType.APPLICATION_JSON_VALUE)
    public List<Record> SearchRecordsByConsignmentCode(@RequestBody ConsignmentCode code) {
        System.out.print("this SearchRecordsByConsignmentCode respond happened");

        List<Record> results=RecordDao.SearchRecordsByConsignmentCode(code.getConsignmentCode());
        return results;
    }

您需要创建一个封装消息正文的类：

import java.io.Serializable;

public class ConsignmentCode implements Serializable {

    private String consignmentCode;

    public String getConsignmentCode() {
        return consignmentCode;
    }

    public void setConsignmentCode(String consignmentCode) {
        this.consignmentCode = consignmentCode;
    }   


}

然后，启动它并尝试将这条消息发布到您的控制器 url（请记住将 Accept/Content-Type 标头设置为 application/json）。属性名称也必须放在分号之间，就是这样：

{"consignmentCode":"123456789"}

你会得到正确的json格式的结果

【讨论】：