java - Spring MVC HttpServletRequest请求: which button clicked on

Question

我有一个带有三个按钮的表单，一个用于保存、更新和查询，但是我想知道在服务器端用户单击了哪个按钮。我尝试使用 request.getParameter("action") 和 request.getAttribute("action") 但它们都返回 null。有什么方法可以在服务器上获取此内容，也许我可以在 session 中存储“操作”值？如果是这样，我如何创建和存储 session 变量？

代码

<button class="btn" value="save"  id="action"></button>

<button class="btn" value="update"  id="action"></button>

<button class="btn" value="query"  id="action"></button>

基本上我试图避免重新发布到服务器。我知道我应该使用 POST/redirect/GET 模式，但是我的方法不支持重定向，即使更改并且表单有错误，我也将无法返回服务器验证。

Controller

@RequestMapping(value = "crime_registration_save.htm", method = RequestMethod.POST)
    public ModelAndView handleSave(@Valid @ModelAttribute Crime crime,HttpServletRequest request,
            HttpServletResponse response,BindingResult result, ModelMap m, Model model) throws Exception {


        String action = request.getParameter("action");

    logger.info("The requested action is "+ action);
        if (result.hasErrors()) {

            logger.debug("Has Errors In crime_registration_save");
            model.addAttribute("dbcriminals", myCriminalList);
            model.addAttribute("dbvictims", myVictimList);
            model.addAttribute("status", myStatusList);
            model.addAttribute("crimeCategory", myCrimeCategoryList);
            model.addAttribute("crimeLevel", myCrimeLevelList);
            model.addAttribute("officers", myOfficerList);

            model.addAttribute("victimList", crime.getVictims());
            model.addAttribute("criminalList", crime.getCriminals());

            model.addAttribute("crimeTypeList",
                    crimeTypeManager.getCrimeTypeList(crime.getOffenceCatId()));
            model.addAttribute("icon", "ui-icon ui-icon-circle-close");
            model.addAttribute("results", "Error: Unable to Save Record!");

            return new ModelAndView("crime_registration");
        }
        logger.debug("No errors going to preform save");

        int crimeRecNo;

        crimeRecNo = crimeManager.saveCrime(crime);

        model.addAttribute("dbcriminals", myCriminalList);
        model.addAttribute("dbvictims", myVictimList);
        model.addAttribute("status", myStatusList);
        model.addAttribute("crimeCategory", myCrimeCategoryList);
        model.addAttribute("crimeLevel", myCrimeLevelList);
        model.addAttribute("officers", myOfficerList);
        model.addAttribute("save", "disabled");
        model.addAttribute("victimList", crime.getVictims());
        model.addAttribute("criminalList", crime.getCriminals());

        model.addAttribute("crimeTypeList",
                crimeTypeManager.getCrimeTypeList(crime.getOffenceCatId()));
        model.addAttribute("crimeRecNo", crimeRecNo);
        model.addAttribute("crimeRecordNoStatus", "true");
        model.addAttribute("icon", "ui-icon ui-icon-circle-check");
        model.addAttribute("results", "Record Was Saved");



        return new ModelAndView("crime_registration");
    }

jquery

function submitPage(urlMapping,method,action) {
        alert(urlMapping);
        document.getElementById("crime_registration").action = urlMapping;
        document.getElementById("crime_registration").target = "_self";
        document.getElementById("crime_registration").method = method;
        document.getElementById("crime_registration").submit();
        $('#action').val(action);

        alert($('#action').val());
        return false;
    }

Answer 1

表单内容不是请求属性，而是参数。

使用提交，为其命名，然后通过该名称检索参数。

<input type="submit" class="btn" value="save" name="action" />

String button = request.getParameter("action");

如果它不在表单中，您将需要一些 JavaScript，但您的问题暗示它是。