java - Spring 形式什么也不做

Question

我正在尝试更新用户设置信息。当我点击提交时，没有任何反应。

这是我的设置 Controller

@Secured({ "ROLE_USER", "ROLE_ADMIN" })
@RequestMapping(value = "/settings**", method = RequestMethod.GET)
public ModelAndView settings() {
    Authentication auth = SecurityContextHolder.getContext().getAuthentication();
    String userName = auth.getName();
    User userInfo = userDAO.getUserInfo(userName);
    Settings settings = userDAO.getSettingsInfo(userInfo.getDetails().getMacAddress());
    List<Plugs> plugs = userInfo.getDetails().getPlugs();
    ModelAndView model = new ModelAndView();
    model.addObject("settings", settings);
    model.addObject("plugs", plugs.get(plugs.size() - 1));
    return model;
}

@Secured({ "ROLE_USER", "ROLE_ADMIN" })
@RequestMapping(value = "/settings", method = RequestMethod.POST)
public ModelAndView settings(@Valid @ModelAttribute("settings") Settings settings, BindingResult result) {
    ModelAndView model = new ModelAndView();
    Authentication auth = SecurityContextHolder.getContext().getAuthentication();
    String userName = auth.getName();
    User userInfo = userDAO.getUserInfo(userName);
    if (result.hasErrors()) {
        model.setViewName("welcome");
    } else {
        model.setViewName("settings");
        userDAO.updateSettings(userInfo);
    }

    return model;
}

JSP 表单

<form:form method="POST" modelAttribute="settings" commandName="settings">

这是我的提交按钮

<input type="submit" value="submit" name ="submit">

如何让我的表单执行任何操作？它应该更新 MySQL 数据库中的用户详细信息。如果您想发布更多代码，我会添加它。

Answer 1

您可以尝试在表单标签中添加action属性吗？

<form:form method="POST" modelAttribute="settings" commandName="settings" action="/settings">

如果您的 Springframework DispatcherServlet servlet 映射的 url 模式不是 web.xml 中的“/”，请将该模式附加在“/settings”之前，如下所示:

<form:form method="POST" modelAttribute="settings" commandName="settings" action="{url pattern}/settings">