在我的页面jsp中，我有一个表单，我可以将用户添加到我的数据库，并且我使用验证器在字段为空时显示错误，但我想要做的是，当我插入重复的条目我的表的主键，验证器向我显示一条消息，例如已经采用此登录而不是Apache的错误！
这是我的用户POJO：
package gestion.delegation.domaine;
import java.io.Serializable;
public class User implements Serializable{
/**
*
*/
private static final long serialVersionUID = 1L;
int id;
String nom;
String prenom;
String login;
String password;
String role;
boolean enable;
public int getId() {
return id;
}
public void setId(int id) {
this.id = id;
}
public String getNom() {
return nom;
}
public void setNom(String nom) {
this.nom = nom;
}
public String getPrenom() {
return prenom;
}
public void setPrenom(String prenom) {
this.prenom = prenom;
}
public String getLogin() {
return login;
}
public void setLogin(String login) {
this.login = login;
}
public String getPassword() {
return password;
}
public void setPassword(String password) {
this.password = password;
}
public boolean getEnable() {
return this.enable;
}
public void setEnable(boolean enable) {
this.enable = enable;
}
public User(int id, String nom, String prenom, String login,
String password, String role, boolean enable) {
super();
this.id = id;
this.nom = nom;
this.prenom = prenom;
this.login = login;
this.password = password;
this.role = role;
this.enable = enable;
}
public String getRole() {
return role;
}
public void setRole(String role) {
this.role = role;
}
public User() {
super();
}
}
这是我的验证器：
package gestion.delegation.validator;
import gestion.delegation.domaine.User;
import org.springframework.validation.Errors;
import org.springframework.validation.ValidationUtils;
import org.springframework.validation.Validator;
public class AddUserValidator implements Validator{
@Override
public boolean supports(Class<?> clazz) {
return User.class.isAssignableFrom(clazz);
}
@Override
public void validate(Object obj, Errors err) {
ValidationUtils.rejectIfEmptyOrWhitespace(err, "nom", "name.required","Choisissez un nom");
ValidationUtils.rejectIfEmptyOrWhitespace(err, "prenom", "prenom.required", "Choisissez un prenom");
ValidationUtils.rejectIfEmptyOrWhitespace(err, "login", "login.required", "Choisissez un login");
ValidationUtils.rejectIfEmptyOrWhitespace(err, "password", "password.required", "Choisissez un password");
ValidationUtils.rejectIfEmpty(err, "role", "role.required", "Choisissez un role");
}
}
形式：
<c:if test="${not empty msg_success}">
<div class="success">Vous avez ajouter un utilisateur avec
succès !</div>
</c:if>
<form:form name="ajf"
action="${pageContext.request.contextPath}/ajouter_user"
method="post" commandName="user">
<table id="tabmenu">
<tr>
<td id="idtab">Nom :</td>
<td><form:input type="text" path="nom"
class="round default-width-input" name="name_" /></td>
<td><form:errors path="nom" Class="errorbox" /></td>
</tr>
<tr>
<td id="idtab">Prénom :</td>
<td><form:input type="text" path="prenom" name="prenom_"
class="round default-width-input" /></td>
<td><form:errors path="prenom" cssClass="errorbox" />
</tr>
<tr>
<td id="idtab">Login :</td>
<td><form:input type="text" path="login" name="login_"
cssClass="round default-width-input" /></td>
<td><form:errors path="login" cssClass="errorbox" /></td>
</tr>
<tr>
<td id="idtab">Password :</td>
<td><form:input type="password" path="password" name="pass_"
class="round default-width-input" /></td>
<td><form:errors path="password" cssClass="errorbox" /></td>
</tr>
<tr>
<td id="idtab">Séléctionner un rôle :</td>
<td><form:select path="role">
<form:option value="" label="" />
<form:option value="ROLE_ADMIN">Administrateur</form:option>
<form:option value="ROLE_USER">Simple utilisateur</form:option>
</form:select></td>
<td><form:errors path="role" cssClass="errorbox" /></td>
</tr>
<tr>
<td id="idtab">Activé :</td>
<td><form:input type="checkbox" value="true" path="enable" />
Oui</td>
</tr>
<tr></tr>
<tr></tr>
<tr>
<td><input
class="button round blue image-right ic-right-arrow"
type="submit" value="Créer" /></td>
<td><input
class="button round blue image-right ic-right-arrow"
type="reset" value="Initialiser" /></td>
</tr>
</table>
</form:form>
任何想法？
答案 0 :(得分：3)
在这种情况下，我担心验证器是不够的。虽然您可以扩展AddUserValidator类以检查给定的用户名是否是免费的，但它不会
在两个用户同时尝试使用相同的用户名注册的情况下工作 - 验证将通过，但是其中一个用户将从数据库中收到错误。
为了保护自己免受这种情况的影响，我会将注册逻辑放在try catch块中，并在出现错误时向用户显示正确的消息。这将是一种application-level validation。
答案 1 :(得分：1)
Spring验证器只需在将对象引入数据库之前按照规定的规则检查对象。它对数据库一无所知。要显示使用数据库时发生的错误，您需要手动捕获异常。
答案 2 :(得分：1)
在您的控制器中只需通过查询模型库来检查重复。
@Model Entity
@Table(name = "people")
public class People {
@Column(name = "nic")
@NotEmpty(message = "*Please provide your nic")
private String nic;
@Repository
public interface PeopleRepository extends JpaRepository<People, Integer>{
People findByNic(String nic);
}
@Controller
@RequestMapping(value = "/welcome", method = RequestMethod.POST)
public ModelAndView createNewPeople(Model model,, BindingResult bindingResult) {
ModelAndView modelAndView = new ModelAndView();
People peopleExistsEmail = peopleService.findUserByNic(people.getNic());
if (peopleExistsEmail != null) {
bindingResult
.rejectValue("nic", "error.people",
"There is already a person registered with the nic provided");
}
if (bindingResult.hasErrors()) {
modelAndView.setViewName("welcome");
} else {
peopleService.savePeople(people);
modelAndView.addObject("successMessage", "People has been registered successfully");
modelAndView.addObject("people", new People());
} catch (Exception e) {
e.printStackTrace();
}
}
return modelAndView;
}
答案 3 :(得分：0)
所以这是正确答案：
DAO类实现中的方法如下：
public boolean AddUser(User user) {
boolean t=true;
final String User_INSERT1 = "insert into utilisateurs (login, password, nom, prenom,enable) "
+ "values (?,?,?,?,?)";
final String User_INSERT2="insert into roles (login,role) values(?,?)";
/*
* On récupère et on utilisera directement le jdbcTemplate
*/
MessageDigestPasswordEncoder encoder = new MessageDigestPasswordEncoder("SHA");
String hash = encoder.encodePassword(user.getPassword(), "");
final String check ="select count(*) from utilisateurs where login = ?";
int result= getJdbcTemplate().queryForInt(check, new Object[]{String.valueOf(user.getLogin())});
if (result==0) {
getJdbcTemplate()
.update(User_INSERT1,
new Object[] {user.getLogin(),
hash, user.getNom(),
user.getPrenom(), user.getEnable(),
});
getJdbcTemplate().update(User_INSERT2, new Object[]{user.getLogin(),user.getRole()});
return t;
}
else { t = false ; return t;}
}
控制器：
@RequestMapping(value ="/ajouter_user", method = RequestMethod.POST)
public String add(@ModelAttribute User us,BindingResult result,ModelMap model) {
AddUserValidator uservalid=new AddUserValidator();
uservalid.validate(us, result);
if (result.hasErrors()) {
model.addAttribute("usersystem", userservice.getAllUsers());
return "gestionUser";
}else {
boolean e = userservice.AddUser(us);
if (e==false){
model.addAttribute("msg_failed","true");
}
else {
model.addAttribute("msg_success","true");}
model.addAttribute("usersystem", userservice.getAllUsers()); /*verifier*/
return "gestionUser";
}
}
并显示jsp文件中的错误：
<c:if test="${not empty msg_failed}">
<div class="errorblock">Il existe déjà un utilisateur avec cet login </div>
</c:if>
答案 4 :(得分：0)
DAO类实现中的方法是这样的：
public final long insert(final User user) throws MyException {
String sql = "INSERT INTO users ....";
String chkSql = "SELECT count(*) FROM users WHERE username=:username";
Map namedParams = new HashMap();
namedParams.put("username", user.getUsername());
long newId = namedTemplate.queryForInt(chkSql, namedParams);
if (newId > 0) {
try {
throw new MyException(-1);
} catch (MyException e) {
throw e;
}
}
newId = ...;// could be generated if not inc field or triggered...
namedParams.put("username", user.getUserName());
...
...
namedParams.put("id", newId);
namedTemplate.update(sql, namedParams);
return newId;
}
像这样的MyException：
public class MyException extends Exception{
private static final long serialVersionUID = 1L;
int a;
public MyException(int b) {
a=b;
}
}
控制器：
@RequestMapping(value = "/user", method = RequestMethod.POST)
public final ModelAndView actionUser(
final ModelMap model,
@ModelAttribute("myitemuser") @Valid final User user,
final BindingResult bindResult,
...);
...
try {
userService.addUser(user); // or dao... ;)
} catch (Exception e) {
UserValidator userValidator = new UserValidator();
userValidator.custError(bindResult);
}
...
@Override
public final void validate(final Object object, final Errors errors) {
User user = (User) object;
...
if (user.getPassword().isEmpty()) {
errors.rejectValue("password", "error.users.emptypass");
}
}
public final void custError(final Errors errors){
errors.rejectValue("username"/* or login */, "error.users.uniqname");
}
...
Нукактотак））））