Question

我有两个在Hibernate中映射的类 1.班级实行帐户
2.该类实现了帐户类型重要！一个帐户可以有多种类型。

 Account.class
    @Entity
    @Table(name = "dp_account",
            uniqueConstraints={@UniqueConstraint(columnNames={"id"})})
        public class Account {

            @Id
            @Column(name="id", nullable = false, unique = true, length = 11)
            @GeneratedValue(strategy = GenerationType.AUTO)
            private long id;

            @Column(name = "price", nullable = false)
            private int price;

            @Column(name = "customer_id", nullable = false, unique = true, length = 11)
            private long customerId;

            @Column(name = "customer", length = 100, nullable = false, unique = true)
            private String customer;

            @Column(name = "comment", length = 1000)
            private String comment;

            @Column(name = "date", columnDefinition="TIMESTAMP")
            @Temporal(TemporalType.TIMESTAMP)
            private Date date;

            @Column(name = "is_deleted", nullable = false)
            private boolean deleted = false;

            @ManyToMany(fetch = FetchType.EAGER, targetEntity = AccountType.class, cascade = { CascadeType.ALL })
            @JoinTable(name = "account_accountType",
                    joinColumns = { @JoinColumn(name = "account_id") },
                    inverseJoinColumns = { @JoinColumn(name = "type_id") })
            private Set<AccountType> accountTypes;
        // дальше конструкторы и геттеры/сеттеры
        }

具有名称，描述，ID和所有者名称的帐户类型的本质。

    AccountType.class
    @Entity
    @Table(name = "account_type")
    public class AccountType {
        private static final Long serialVersionUID = -4727727495060874301L;

        @Id
        @Column(name = "id")
        @GeneratedValue(strategy = GenerationType.AUTO)
        private long id;

        @Column(name = "name", length = 25, nullable = false, unique = true)
        private String name;

        @Column(name = "description", length = 255)
        private String description;

        @Column(name = "user_login", length = 45, nullable = false, unique = true)
        private String login;
   //дальше геттеры/сеттеры и конструкторы
}

问题是 - 有一个页面，我填写表格来创建帐户（Account.class），我multiple select，以填写帐户类型（AccountType.class）。 Account.class具有字段Set<AccountType> accountTypes，其中每个帐户都包含类型列表。（这是我试图填写的列表，然后在MVC控制器中传递Account.class）。但是在控制器中我甚至不会得到400错误。经过多次测试后，我意识到Spring MVC没有绑定我的列表。我想知道怎么做？

MVC控制器：

@RequestMapping(value = "/account/addAccount", method = RequestMethod.POST)
    public void addAccount(@ModelAttribute("accountAttribute") Account account) {
        System.out.println("Account on the top");
        System.out.println(account);
        ServiceFactory.getInstance().getAccountService().addAccount(account);
    }

JSP：

<div style="margin-top: 40px">
            <form:form action="${pageContext.request.contextPath}/account/addAccount" method="post" modelAttribute="accountAttribute">

                <label>
                    Сумма: <input type="number" name="price">
                </label>
                <br> <br>

                <label>
                    Категория:
                    <select multiple name="accountTypes">
                        <c:forEach items="${accountTypes}" var="accountType">
                        <option value="${accountType.name}">${accountType.name}</option>
                        </c:forEach>
                    </select>
                </label>
                <br><br>

                <label>Комментарий:
                    <br> <br>
                    <textarea name="comment" rows="5" cols="30"></textarea>
                </label>
                <br><br>

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

            </form:form>
        </div>

屏幕截图示例：

更新：

@RequestMapping(value = "/welcome", method = RequestMethod.GET)
    public ModelAndView welcomePage() {
        ModelAndView model = new ModelAndView();
        List<GrantedAuthority> grantedAuthorityList = (List<GrantedAuthority>) SecurityContextHolder.getContext().getAuthentication().getAuthorities();
        Authentication authentication = SecurityContextHolder.getContext().getAuthentication();
        String login = authentication.getName();
        if (grantedAuthorityList.contains(new SimpleGrantedAuthority("ROLE_ADMIN"))) {
            model.setViewName("admin");
        } else {
            List<AccountType> accountTypes = ServiceFactory.getInstance().getAccountService().getAllAccountTypeByLogin(login);
            model.addObject("accountTypes", accountTypes);
            model.setViewName("user");
        }
        return model;
    }

Answer 1

您正在使用html select元素，但您需要做的是使用spring tag library＆select;

<form:select path="accountTypes" multiple="true">
  <form:options items="${accountTypes}"/>
</form:select>

这样spring会绑定控制器中的值。

尝试使用此

@RequestMapping(value = "/welcome", method = RequestMethod.GET)
public ModelAndView welcomePage() {
    ModelAndView model = new ModelAndView();
    List<GrantedAuthority> grantedAuthorityList = (List<GrantedAuthority>) SecurityContextHolder.getContext().getAuthentication().getAuthorities();
    Authentication authentication = SecurityContextHolder.getContext().getAuthentication();
    String login = authentication.getName();
    if (grantedAuthorityList.contains(new SimpleGrantedAuthority("ROLE_ADMIN"))) {
        model.setViewName("admin");
    } else {
        List<AccountType> accountTypes = ServiceFactory.getInstance().getAccountService().getAllAccountTypeByLogin(login);
        model.addObject("accountTypes", accountTypes);
        model.setViewName("user");
//Add here
model.addObject("accountAttribute", new Account());
    }
    return model;
}

Answer 2

问题是您需要注册一个属性编辑器或转换器，它能够将您的帐户类型名称转换为AccountType对象。您的视图发送accountTypes元素的字符串值。 Spring需要知道如何将String值绑定到Object。

public class AccountTypeEditor extends PropertyEditorSupport{

   public void setAsText(String text){
        AccountType accountType = // some logic to find account type by name
        setValue(accountType);
   }
}

在你的控制器中，你需要一个@InitBinder方法来注册你的编辑器

@InitBinder
public void initBinder(WebDataBinder binder){
   binder.registerCustomEditor(AccountType.class , new AccountTypeEditor());
}

}

Answer 3

我能够解决这个问题，我开始分别通过设置和分别对象本身的帐号。问题是我从jsp帐户传递，我想要设置，这是不可能的，因为我只选择名称类型帐户。这是解决我的问题：

JSP：

<form:form action="${pageContext.request.contextPath}/account/addAccount" method="post" modelAttribute="accountAttribute">

                <label>
                    Сумма: <input type="number" name="price">
                </label>
                <br> <br>

                <select multiple name="accountTypeNameSet" size="4s">
                    <c:forEach items="${accountTypes}" var="accountType">
                        <option >${accountType}</option>
                    </c:forEach>
                </select>

                <label>Комментарий:
                    <br> <br>
                    <textarea name="comment" rows="5" cols="30"></textarea>
                </label>
                <br><br>

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

            </form:form>

控制器：

@RequestMapping(value = "/account/addAccount", method = RequestMethod.POST)
    public void addAccount(@ModelAttribute("accountAttribute") Account account, @RequestParam(value="accountTypeNameSet") Set<String> accountTypeNameSet) {
    // implementation here
// accountTypeNameSet - This is the list of names account type, which user selected. On the basis of these names we can pull your invoice from database.

带有集合的MVC和对象＆lt;＆gt;

3 个答案: