Question

想从地址表中搜索城市。

实体：酒店

@JsonManagedReference
    @OneToMany(cascade = CascadeType.ALL, mappedBy = "hotel")
    private List<Address> address;

实体：地址

@ManyToOne(fetch = FetchType.LAZY, optional = false)
    @JoinColumn(name = "hotel_id", nullable = false)
    @OnDelete(action = OnDeleteAction.CASCADE)
    @JsonBackReference
    private Hotel hotel;

控制器：

@GetMapping("/forSearch")
    public String searchHotelPage(Model model, Address address){
        model.addAttribute("address", address);
        return "search-hotel";
    }

    @GetMapping("/search-hotel-city/{city}")
    public String searchHotel(@RequestParam("city") String city, Model model){
        List<Address> addresses = addressService.searchAddressByCity(city);
        System.out.println("this is address: "+ addresses);
        model.addAttribute("address",addresses );
        return "search-hotel";
    }

服务：

public List<Address> searchAddressByCity(String city) {
        return addressRepository.findAll();
    }

搜索和列表视图：

<form th:action="@{/search-hotel-city/city?city=${address.city}}" method="get">
            <div class="form-group mb-2">
                <input type="text" class="form-control" name="city" id="city" placeholder="Search "/>
                <input type="submit" value="search" class="btn btn-primary">

            </div>
        </form>


        <hr/>


        <table class="table">
            <thead>
            <tr>
                <th scope="col">Id</th>
                <th scope="col">roadNumber</th>
                <th scope="col">city</th>
                <th scope="col">country</th>
            </tr>
            </thead>
            <tbody>
            <tr th:each="address: ${addresses}">
                <th scope="row" th:text="${address.id}"></th>
                <td th:text="${address.roadNumber}"></td>
                <td th:text="${address.city}"></td>
                <td th:text="${address.country}"></td>
            </tr>

            </tbody>
        </table>

错误：

Hibernate: select address0_.hotel_id as hotel_id5_0_0_, address0_.id as id1_0_0_, address0_.id as id1_0_1_, address0_.city as city2_0_1_, address0_.country as country3_0_1_, address0_.hotel_id as hotel_id5_0_1_, address0_.road_number as road_num4_0_1_ from address address0_ where address0_.hotel_id=?
2019-07-28 12:58:28.184 ERROR 9836 --- [nio-8080-exec-2] o.a.c.c.C.[.[.[/].[dispatcherServlet]    : Servlet.service() for servlet [dispatcherServlet] in context with path [] threw exception [Handler dispatch failed; nested exception is java.lang.StackOverflowError] with root cause

java.lang.StackOverflowError: null
    at java.lang.Exception.<init>(Exception.java:102) ~[na:1.8.0_212]
    at java.lang.ReflectiveOperationException.<init>(ReflectiveOperationException.java:89) ~[na:1.8.0_212]

我想搜索城市。但是当我搜索StackOverflowError时发生。我已经尝试过但无法解决此问题。请帮我。谢谢

Answer 1

从System.out.println("this is address: "+ addresses);中删除searchHotel。它可能导致无限调用，从而导致 StackOverflowError 。

可能的原因：

大多数List实现的toString()方法通过 List 元素进行迭代并在其上调用toString()。 1
结合使用Lombok和自动生成的 hashCode ，以便运行无限调用 2

StackOverflowError

1 个答案: