我正在尝试以一种不涉及任何页面刷新的方式重写表单。换句话说,我不希望浏览器在提交时发出任何GET/POST请求。jQuery应该可以帮助我。这是我的表单(我有几个):
<!-- I guess this action doesn't make much sense anymore -->
<form action="/save-user" th:object="${user}" method="post">
<input type="hidden" name="id" th:value="${user.id}">
<input type="hidden" name="username"
th:value="${user.username}">
<input type="hidden" name="password"
th:value="${user.password}">
<input type="hidden" name="name" th:value="${user.name}">
<input type="hidden" name="lastName"
th:value="${user.lastName}">
<div class="form-group">
<label for="departments">Department: </label>
<select id="departments" class="form-control"
name="department">
<option th:selected="${user.department == 'accounting'}"
th:value="accounting">Accounting
</option>
<option th:selected="${user.department == 'sales'}"
th:value="sales">Sales
</option>
<option th:selected="${user.department == 'information technology'}"
th:value="'information technology'">IT
</option>
<option th:selected="${user.department == 'human resources'}"
th:value="'human resources'">HR
</option>
<option th:selected="${user.department == 'board of directors'}"
th:value="'board of directors'">Board
</option>
</select>
</div>
<div class="form-group">
<label for="salary">Salary: </label>
<input id="salary" class="form-control" name="salary"
th:value="${user.salary}"
min="100000" aria-describedby="au-salary-help-block"
required/>
<small id="au-salary-help-block"
class="form-text text-muted">100,000+
</small>
</div>
<input type="hidden" name="age" th:value="${user.age}">
<input type="hidden" name="email" th:value="${user.email}">
<input type="hidden" name="enabledByte"
th:value="${user.enabledByte}">
<!-- I guess I should JSON it somehow instead of turning into regular strings -->
<input type="hidden" th:name="authorities"
th:value="${#strings.toString(user.authorities)}"/>
<input class="btn btn-primary d-flex ml-auto" type="submit"
value="Submit">
</form>
以下是我的JS:
$(document).ready(function () {
$('form').on('submit', async function (event) {
event.preventDefault();
let user = {
id: $('input[name=id]').val(),
username: $('input[name=username]').val(),
password: $('input[name=password]').val(),
name: $('input[name=name]').val(),
lastName: $('input[name=lastName]').val(),
department: $('input[name=department]').val(),
salary: $('input[name=salary]').val(),
age: $('input[name=age]').val(),
email: $('input[name=email]').val(),
enabledByte: $('input[name=enabledByte]').val(),
authorities: $('input[name=authorities]').val()
/*
↑ i tried replacing it with authorities: JSON.stringify($('input[name=authorities]').val()), same result
*/
};
await fetch(`/users`, {
method: 'PUT',
headers: {
...getCsrfHeaders(),
'Content-Type': 'application/json',
},
body: JSON.stringify(user) // tried body : user too
});
});
});
function getCsrfHeaders() {
let csrfToken = $('meta[name="_csrf"]').attr('content');
let csrfHeaderName = $('meta[name="_csrf_header"]').attr('content');
let headers = {};
headers[csrfHeaderName] = csrfToken;
return headers;
}
下面是我的REST控制器处理程序:
// maybe I'll make it void. i'm not sure i actually want it to return anything
@PutMapping("/users")
public User updateEmployee(@RequestBody User user) {
service.save(user); // it's JPARepository's regular save()
return user;
}
User
实体:
@Entity
@Table(name = "users")
@Data
@EqualsAndHashCode
public class User implements UserDetails {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column
private long id;
@Column(nullable = false, unique = true)
private String username;
@Column(nullable = false)
private String password;
@Column
private String name;
@Column(name = "last_name")
private String lastName;
@Column
private String department;
@Column
private int salary;
@Column
private byte age;
@Column
private String email;
@Column(name = "enabled")
private byte enabledByte;
@ManyToMany
@JoinTable(name = "user_role",
joinColumns = {@JoinColumn(name = "user_id", referencedColumnName = "id"),
@JoinColumn(name = "username", referencedColumnName = "username")},
inverseJoinColumns = {@JoinColumn(name = "role_id", referencedColumnName = "id"),
@JoinColumn(name = "role", referencedColumnName = "role")})
@EqualsAndHashCode.Exclude
private Set<Role> authorities;
Role
实体:
@Entity
@Table(name = "roles")
@Data
@EqualsAndHashCode
public class Role implements GrantedAuthority {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column
private long id;
@Column(name = "role", nullable = false, unique = true)
private String authority;
@ManyToMany(mappedBy = "authorities")
@EqualsAndHashCode.Exclude
private Set<User> userList;
当我按下提交按钮时,我的控制台里会出现这个
WARN 18252 --- [io-8080-exec-10] .w.s.m.s.DefaultHandlerExceptionResolver : Resolved [org.springframework.http.converter.HttpMessageNotReadableException: JSON parse error: Cannot deserialize value of type `java.util.HashSet<pp.spring_bootstrap.models.Role>` from String value (token `JsonToken.VALUE_STRING`)]
似乎我应该以某种方式传递Collection
的JSON表示,而不仅仅是String
。
@Override
public void addFormatters(FormatterRegistry registry) {
registry.addFormatter(new Formatter<Set<Role>>() {
@Override
public Set<Role> parse(String text, Locale locale) {
Set<Role> roleSet = new HashSet<>();
String[] roles = text.split("^\\[|]$|(?<=]),\\s?");
for (String roleString : roles) {
if (roleString.length() == 0) continue;
String authority =
roleString.substring(roleString.lastIndexOf("=") + 2,
roleString.indexOf("]") - 1);
roleSet.add(service.getRoleByName(authority));
}
return roleSet;
}
@Override
public String print(Set<Role> object, Locale locale) {
return null;
}
});
}
我谷歌了一下,发现Thymeleaf没有任何toJson()
方法。我的意思是我可以自己写方法,但我不知道如何在Thymeleaf模板中使用它们。此外,it may not be the most optimal solution
这是一个 Boot 项目,所以我有Jackson数据绑定库
如何将Collection从表单正确传递到JS事件处理程序,然后传递到REST控制器?
我检查了StackOverflow提出的多个类似问题,它们看起来并不相关(例如,它们涉及不同的编程语言,如C#或PHP)
**UPD:**我刚才试过了,可惜也不行!(错误信息相同)
// inside my config
@Bean
public Function<Set<Role>, String> jsonify() {
return s -> {
StringJoiner sj = new StringJoiner(", ", "{", "}");
for (Role role : s) {
sj.add(String.format("{ \"id\" : %d, \"authority\" : \"%s\" }", role.getId(), role.getAuthority()));
}
return sj.toString();
};
}
<input type="hidden" th:name="authorities"
th:value="${@jsonify.apply(user.authorities)}"/>
不过,该方法可以按预期工作
$(document).ready(function () {
$('form').on('submit', async function (event) {
/*
↓ logs:
authorities input: {{ "id" : 1, "authority" : "USER" }}
*/
console.log('authorities input: ' +
$('input[name=authorities]').val());
**UPD 2:**GPT 4建议
authorities: JSON.parse($('input[name=authorities]').val())
现在真的很奇怪。数据库仍然没有改变,但是!IDE控制台现在没有错误,也没有提到PUT请求(以前的尝试中有)!此外,浏览器日志有以下消息
Uncaught (in promise) SyntaxError: Expected property name or '}' in JSON at position 1
at JSON.parse (<anonymous>)
at HTMLFormElement.<anonymous> (script.js:28:31)
at HTMLFormElement.dispatch (jquery.slim.min.js:2:43114)
at v.handle (jquery.slim.min.js:2:41098)
我不知道这是什么意思!
**UPD 3:**GPT 4很聪明。无论如何,比我聪明。这是绝对正确的。它在UPD 2中不起作用的原因是我忽略了它说的另一件事:
authorities字段应该作为对象的数组而不是字符串发送。
这意味着我应该使用方括号,而不是花括号,作为我的StringJoiner
前缀和后缀:
// I also added some line breaks, but I doubt it was necessary
@Bean
public Function<Set<Role>, String> jsonify() {
return s -> {
StringJoiner sj = new StringJoiner(",\n", "[\n", "\n]");
for (Role role : s) {
sj.add(String.format("{\n\"id\" : %d,\n\"authority\" : \"%s\"\n}", role.getId(), role.getAuthority()));
}
return sj.toString();
};
}
我也改了,比如说这个
username: $('input[name=username]').val()
我真傻,没有马上做。
username: $(this).find('input[name=username]').val()
和-中提琴-它现在工作!
GPT 4也注意到我用了
'input[name=department]'
而不是
'select[name=department]'
我也修好了
1条答案
按热度按时间vom3gejh1#
1.它应该是一个对象数组(即使它是一个
Collection
,而不是一个数组),所以new StringJoiner(", ", "{", "}")
→new StringJoiner(", ", "[", "]")
1.它应该以窗体的子级为目标
username: $('input[name=username]').val()
→username: $(this).find('input[name=username]').val()
,依此类推department
由<select>
元素so表示'input[name=department]'
→'select[name=department]'