问题描述
我正在创建一个 springboot web 应用程序,我想为 OnetoOne 关系表插入值,我有员工和地址实体,我可以像下面这样在数据库中插入值,我无法使用 web 表单插入值,
I am creating a springboot web app, I want to insert values for OnetoOne relation table,I have Employee and Address Entities, I am able to insert values in DB like bellow, I am unable to insertvalues using web form,
Employee employee= new Employee();
employee.setEmployeeName("Barry Bingel");
employee.setEmail("[email protected]");
employee.setSalary(50000.00);
Address address1 = new Address();
address1.setCity("Chennai");
address1.setPincode(9087727L);
address1.setState("Tamilnadu");
address1.setStreet("Park Street");
employee.setAddress(address1);
empServices.save(employee);
但我想通过网络表单插入相同的数据,如下所述
But I want to insert the same data through web form, mentioned bellow
<%@ page language="java" contentType="text/html; charset=ISO-8859-1"
pageEncoding="ISO-8859-1"%>
<%@ taglib prefix="form" uri="http://www.springframework.org/tags/form" %>
<!DOCTYPE html>
<html>
<head>
<title>New Student Form</title>
</head>
<body>
<form:form action="save" method="post" modelAttribute="employee">
<table >
<tr>
<td>employeeName</td>
<td><form:input path="employeeName" placeholder="Name" /></td>
</tr>
<tr>
<td>Email</td>
<td><form:input path="email" placeholder="Email"/></td>
</tr>
<tr>
<td>Salary</td>
<td><form:input path="salary" placeholder="salary"/></td>
</tr>
<tr>
<td>Address</td>
<td><form:input path="address" placeholder="address" /></td>
</tr>
</table>
<input type="submit" value="Save">
</form:form>
</body>
</html>
但是我收到了下面提到的错误.
But I am getting error mentioned bellow.
Field error in object 'employee' on field 'address': rejected value [Address [street=Park Street, city=Chennai, state=Tamilnadu, pincode=9087727]]; codes
[typeMismatch.employee.address,typeMismatch.address,typeMismatch.com.tcdc.Entity.Address,typeMismatch]; arguments [org.springframework.context.support.DefaultMessageSourceResolvable: codes
[employee.address,address]; arguments []; default message [address]]; default message [Failed to convert property value of type 'java.lang.String' to required type 'com.tcdc.Entity.Address' for property
'address'; nested exception is java.lang.IllegalStateException: Cannot convert value of type
'java.lang.String' to required type 'com.tcdc.Entity.Address' for property 'address': no matching
editors or conversion strategy found]]
下面提到的我的员工和地址实体类
My Employee and Address Entity Class mentioned bellow
地址.JAVA
@Entity
@Table(name="address_table")
public class Address {
@Id
@Column(name="adress_id")
@GeneratedValue(strategy=GenerationType.IDENTITY)
private Integer addressId;
@Column(name = "street_name",length=50)
private String street;
@Column(name = "city_name",length=50)
private String city;
@Column(name = "state_name")
private String state;
@Column(name = "pin_code")
private Long pincode;
public String getStreet() {
return street;
}
public void setStreet(String street) {
this.street = street;
}
public String getCity() {
return city;
}
public void setCity(String city) {
this.city = city;
}
public String getState() {
return state;
}
public void setState(String state) {
this.state = state;
}
public Long getPincode() {
return pincode;
}
public void setPincode(Long pincode) {
this.pincode = pincode;
}
@Override
public String toString() {
return "Address [street=" + street + ", city=" + city + ", state=" + state + ", pincode=" + pincode + "]";
}
}
员工.JAVA
@Entity
@Table(name="employee_table")
@DynamicUpdate
public class Employee {
@Id
@Column(name="employee_id")
@GeneratedValue(strategy=GenerationType.IDENTITY)
private Integer employeeId;
@Column(name="employee_name",length=200,nullable=false)
private String employeeName;
@Column(name="email")
private String email;
@Column(name="salary")
private Double salary;
@OneToOne(cascade=CascadeType.ALL)
@JoinColumn(name="address_id")
private Address address;
public Integer getEmployeeId() {
return employeeId;
}
public void setEmployeeId(Integer employeeId) {
this.employeeId = employeeId;
}
public String getEmployeeName() {
return employeeName;
}
public void setEmployeeName(String employeeName) {
this.employeeName = employeeName;
}
public String getEmail() {
return email;
}
public void setEmail(String email) {
this.email = email;
}
public Double getSalary() {
return salary;
}
public void setSalary(Double salary) {
this.salary = salary;
}
public void setAddress(Address address) {
this.address = address;
}
public Address getAddress() {
return address;
}
@Override
public String toString() {
return "Employee [employeeId=" + employeeId + ", employeeName=" + employeeName + ", email=" + email
+ ", salary=" + salary + ", address=" + address + "]";
}
}
MyController 类我的控制器.JAVA
MyController ClassMyController.JAVA
@Controller
public class MyController {
@Autowired
EmpServices empServices;
@GetMapping("/home3")
public String methodOne(Model model) {
Employee employee= new Employee();
// employee.setEmployeeName("Barry Bingel");
// employee.setEmail("[email protected]");
// employee.setSalary(50000.00);
Address address1 = new Address();
// address1.setCity("Chennai");
// address1.setPincode(9087727L);
// address1.setState("Tamilnadu");
// address1.setStreet("Park Street");
// employee.setAddress(address1);
// empServices.save(employee);
model.addAttribute("employee", employee);
return "form";
}
@PostMapping("/save")
public String saveMethod(@ModelAttribute("employee") Employee themployee) {
//themployee.setAddress(themployee.getAddress());
System.out.println(themployee.getAddress());
return null;
}
}
推荐答案
看起来你的表单只有一个用于 Address 的文本字段,它是一个具有多个属性的复合对象
It looks like your form has single text field for the Address, which is a composite object with multiple properties
<td><form:input path="address" placeholder="address" /></td>
Spring 无法将 String 转换为预期的对象
Spring cannot convert a String to the expected object
Cannot convert value of type 'java.lang.String' to required type 'com.tcdc.Entity.Address' for property 'address': no matching
您应该为地址提供多个字段(address.street、address.city)
You should provide multiple fields for the address (address.street, address.city)
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