如何在Spring中将列表用作模型属性?

问题描述

我有这样的Home控制器:

I have my Home controller like this:

    @RequestMapping("/")
    public ModelAndView welcome(@ModelAttribute("myValuesInRows") List<String> myValuesInRows,  ModelMap model) {
        List<Spravochnik> dropDown = spravochnikService.findAll("sprav_of_spravs");
        List<String> justValuesInRows = new ArrayList<>();
        for(Spravochnik sprav : dropDown) {
            for(List<String> vals : sprav.getValuesInRows()) {
                for(String v : vals) {
                    justValuesInRows.add(v);
                }
            }
        }
        for(int i=1; i<justValuesInRows.size(); i+=2) {
            myValuesInRows.add(justValuesInRows.get(i));
        }
        model.addAttribute("myValuesInRows", myValuesInRows);
        return new ModelAndView("home", model);
    }

并且我的主视图具有我正在使用的选择":

and my Home view has this Select I'm using:

<f:form>
<f:select path="myValuesInRows" items="${myValuesInRows}" name="tableName" id="tableName">
</f:select>
</f:form>

当我尝试显示它时，Spring显示此错误:

When I try to show it Spring shows this error:

找不到接口java.util.List 的主要或默认构造函数.

No primary or default constructor found for interface java.util.List.

我想将我的选择链接到我正在传递的列表中，如何在这里完成?

I would like to connect my select to this List I'm passing, how is it done here?

推荐答案

使用实现 List 的类之一，例如 ArrayList 或 LinkedList ，它们具有默认构造函数.

Use one of the classes which implement List, such as ArrayList or LinkedList, which have default constructors.

@ModelAttribute("myValuesInRows") ArrayList<String> myValuesInRows

您可以在此处找到 List 的所有已知实现类的列表:

You can find a list of all known implementing classes of List here:

https://docs.oracle.com/javase/8/docs/api/java/util/List.html

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