java - spring web应用程序-html按钮-表单外的post方法-404错误

Question

我尝试通过按钮“onClick”方法将 POST 请求从 JSP View 发送到 Controller ，但收到 404 错误，表明 RequestMapping 没有签名，这是为什么？

家庭 Controller :

@RequestMapping(value = "/showSelectedRequest/{id}", method = RequestMethod.POST)
public String loadRequestProducts(@PathVariable("id") int id, Model model) {

    logger.debug("HomeController.RequestIdSelected() - Start");
    logger.debug("HomeController.RequestIdSelected: id: " + id);
    model.addAttribute("RequestIdSelected", id);

    logger.debug("HomeController.RequestIdSelected() - Done");
    return "/home";

}

首页.jsp:

<form action="${contextPath}/requestlist" method="post">    
    <table class="table table-sm">
        <thead class="thead-inverse">
            <tr>
                <th>
                    Id
                </th>
                <th>
                    Name
                </th>
                <th>
                    Show request
                </th>
            </tr>
        </thead>

        <c:forEach items="${requestDTOList}" var="requestDTO">
            <tr>
                <td>
                    ${requestDTO.getId()}
                </td>
                <td>
                    ${requestDTO.getName()}
                </td>
                <td>
                    <button class="btn btn-info"  onclick="post(/showSelectedRequest/${requestDTO.getId()})">Query</button>
                    <input type="hidden"  name="${_csrf.parameterName}"   value="${_csrf.token}"/>
                </td>
            </tr>
        </c:forEach>
    </table>
</form>

Answer 1

当您有表单时，操作字段是当您单击“提交”类型的输入时将执行的操作字段。

作为解决方案，您可以按如下方式修改代码:

<form action="${contextPath}/showSelectedRequest/${requestDTO.getId()}" method="post">  
    // Form elements ...
    <input type="submit" value="Query" />
</form>