如何从 thymeleaf 的两部分创建链接? [英] How to create a link from two parts in thymeleaf?
问题描述
我是 thymeleaf 的新手.我需要创建由控制器路径(PersonController 类)和对象 id 组成的链接,我从列表中获取.我希望列表中的每个人对象都有自己的链接,如下所示:href="/personData/person.id'.这是我的代码,它给了我一个错误 org.thymeleaf.exceptions.TemplateInputException: An error occurred during template parsing:
<div th:each="person : ${persons}" th:with="hrefToPerson=${/personData/+ ${person.id}}"><a th:href=hrefToPerson>编辑</a><p th:text="${person.id} + ' ' + ${person.getLastName()} + ' ' + ${person.getFirstName()} + ' ' + ${person.getPatronymic()}"/><p>电话:</p><ol><li th:each="phone: ${person.getPhoneNumbers()}" th:text="${phone.getType()} + ' ' + ${phone.getNumber()}"></里></ol><p>地址:</p><ol><li th:each="address:${person.addresses}" th:text="${address.getZipCode()} + ', ' + ${address.getCountry()} +', ' + ${address.getRegion()} + ', ' + ${address.getCity()} + ', ' + ${address.getAddressLine2()} + ' ' + ${address.getAddressLine1()}"></li></ol>
控制器:
@Controller@RequestMapping("/personData")公共类 PersonController {@自动连线私有 PersonRepository personRepo;@GetMapping("{person}")public String personEditForm(@PathVariable Person person, Model model) {model.addAttribute("person", person);return "personEdit";}@PostMappingpublic String personSave(@RequestParam Person person, Model model) {return "personEdit";}}
<小时>
类人:
@Entity@NamedQuery(name = "Person.findByPhone",query = "select p from Person as p join p.phoneNumbers as pn where pn.number = :number")@Table(唯一约束={@UniqueConstraint(columnNames = {"lastName", "firstName", "patronymic"})})公共类人{@ID@GeneratedValue(strategy=GenerationType.AUTO)私有整数 ID;私人字符串姓氏;@NotNull私人字符串名字;private String 父名;@ElementCollection(fetch = FetchType.EAGER)@CollectionTable(name = "person_phone_numbers", joinColumns = @JoinColumn(name = "person_id"))私人设置<电话号码>phoneNumbers = new HashSet<>();@ElementCollection(fetch = FetchType.EAGER)@CollectionTable(name = "person_addresses", joinColumns = @JoinColumn(name = "person_id"))@AttributeOverrides({@AttributeOverride(name = "addressLine1", column = @Column(name = "house_number")),@AttributeOverride(name = "addressLine2", column = @Column(name = "street"))})私有集<地址>地址 = 新的 HashSet<>();公共人(){}公共人(字符串姓氏,字符串名字,字符串父名,设置<电话号码>电话号码,设置<地址>地址){this.lastName = 姓氏;this.firstName = firstName;this.patronymic = 父名;this.phoneNumbers = phoneNumbers;this.addresses = 地址;}
getter 和 setter...}
它的结构应该是这样的:
<a th:href="@{/personData/{id}(id=${person.id})}">编辑</a>请参阅标准网址语法.你可以用 th:with
做同样的表达,但我不认为你有任何理由想要这样做,除非你重复使用 url.
<div th:each="person: ${persons}" th:with="hrefToPerson=@{/personData/{id}(id=${person.id})}"><a th:href="${hrefToPerson}">编辑</a>
I am newbie in thymeleaf.
I need to create links that consists of the path to controller(class PersonController) and id of object, which I take from list. I want that eachPerson object from list have its own link like this: href=" /personData/person.id'.
It's my code, which give me an error org.thymeleaf.exceptions.TemplateInputException: An error happened during template parsing:
<div th:each="person : ${persons}" th:with="hrefToPerson=${/personData/ + ${person.id}}">
<a th:href=hrefToPerson>Edit</a>
<p th:text="${person.id} + ' ' + ${person.getLastName()} + ' ' + ${person.getFirstName()} + ' ' + ${person.getPatronymic()}"/>
<p>Phones:</p>
<ol>
<li th:each="phone: ${person.getPhoneNumbers()}" th:text="${phone.getType()} + ' ' + ${phone.getNumber()}"></li>
</ol>
<p>Address: </p>
<ol>
<li th:each="address:${person.addresses}" th:text="${address.getZipCode()} + ', ' + ${address.getCountry()} +
', ' + ${address.getRegion()} + ', ' + ${address.getCity()} + ', ' + ${address.getAddressLine2()} + ' ' + ${address.getAddressLine1()}"></li>
</ol>
</div>
Controller:
@Controller
@RequestMapping("/personData")
public class PersonController {
@Autowired
private PersonRepository personRepo;
@GetMapping("{person}")
public String personEditForm(@PathVariable Person person, Model model) {
model.addAttribute("person", person);
return "personEdit";
}
@PostMapping
public String personSave(@RequestParam Person person, Model model) {
return "personEdit";
}
}
class Person:
@Entity
@NamedQuery(name = "Person.findByPhone",
query = "select p from Person as p join p.phoneNumbers as pn where pn.number = :number")
@Table(uniqueConstraints={
@UniqueConstraint(columnNames = {"lastName", "firstName", "patronymic"})
})
public class Person {
@Id
@GeneratedValue(strategy=GenerationType.AUTO)
private Integer id;
private String lastName;
@NotNull
private String firstName;
private String patronymic;
@ElementCollection(fetch = FetchType.EAGER)
@CollectionTable(name = "person_phone_numbers", joinColumns = @JoinColumn(name = "person_id"))
private Set<PhoneNumber> phoneNumbers = new HashSet<>();
@ElementCollection(fetch = FetchType.EAGER)
@CollectionTable(name = "person_addresses", joinColumns = @JoinColumn(name = "person_id"))
@AttributeOverrides({
@AttributeOverride(name = "addressLine1", column = @Column(name = "house_number")),
@AttributeOverride(name = "addressLine2", column = @Column(name = "street"))
})
private Set<Address> addresses = new HashSet<>();
public Person() {
}
public Person(String lastName, String firstName, String patronymic,
Set<PhoneNumber> phoneNumbers, Set<Address> addresses) {
this.lastName = lastName;
this.firstName = firstName;
this.patronymic = patronymic;
this.phoneNumbers = phoneNumbers;
this.addresses = addresses;
}
getters and setters...
}
解决方案 It should be structured like this:
<div th:each="person: ${persons}">
<a th:href="@{/personData/{id}(id=${person.id})}">Edit</a>
See the standard url syntax. You could do the same expression with th:with
, but I don't see any reason you'd want to do that unless you are reusing the url.
<div th:each="person: ${persons}" th:with="hrefToPerson=@{/personData/{id}(id=${person.id})}">
<a th:href="${hrefToPerson}">Edit</a>
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