Question

我试图将加密的密码插入UserProfile表。

我的sql文件用于创建表格，如：

-- ---------- Table for validation queries from the connection pool. ----------
DROP TABLE PingTable;
CREATE TABLE PingTable (foo CHAR(1));
-- ------------------------------ UserProfile ----------------------------------
DROP TABLE UserProfile;
CREATE TABLE UserProfile (
    usrId BIGINT NOT NULL AUTO_INCREMENT,
    loginName VARCHAR(30) COLLATE latin1_bin NOT NULL,
    enPassword VARCHAR(150) NOT NULL, 
    firstName VARCHAR(30) NOT NULL,
    lastName VARCHAR(40) NOT NULL,
    email VARCHAR(60) NOT NULL,
    CONSTRAINT UserProfile_PK PRIMARY KEY (usrId),
    CONSTRAINT LoginNameUniqueKey UNIQUE (loginName))
    ENGINE = InnoDB;

CREATE INDEX UserProfileIndexByLoginName ON UserProfile (loginName);

我注册用户的实现：

private String doEncryptedPassword(String clearPassword) {
    HashFunction hf = Hashing.sha512();
    HashCode hc = hf.newHasher(clearPassword.length()).putString(clearPassword, StandardCharsets.UTF_8).hash();
    return hc.toString();
}

public UserProfile registerUser(String loginName, String clearPassword,
        UserProfileDetails userProfileDetails)
        throws DuplicateInstanceException {

    try {
        userProfileDao.findByLoginName(loginName);
        throw new DuplicateInstanceException(loginName,
                UserProfile.class.getName());
    } catch (InstanceNotFoundException e) {
        String encryptedPassword = doEncryptedPassword(clearPassword);

        UserProfile userProfile = new UserProfile(loginName,
                encryptedPassword, userProfileDetails.getFirstName(),
                userProfileDetails.getLastName(), userProfileDetails
                    .getEmail());


        userProfileDao.save(userProfile);
        return userProfile;
    }

}

UserProfile.java：

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.SequenceGenerator;

@Entity
public class UserProfile {

    private Long userProfileId;
    private String loginName;
    private String encryptedPassword;
    private String firstName;
    private String lastName;
    private String email;

    public UserProfile() {
    }

    public UserProfile(String loginName, String encryptedPassword,
            String firstName, String lastName, String email) {

        /**
         * NOTE: "userProfileId" *must* be left as "null" since its value is
         * automatically generated.
         */

        this.loginName = loginName;
        this.encryptedPassword = encryptedPassword;
        this.firstName = firstName;
        this.lastName = lastName;
        this.email = email;
    }

    @Column(name = "usrId")
    @SequenceGenerator( // It only takes effect for
    name = "UserProfileIdGenerator", // databases providing identifier
    sequenceName = "UserProfileSeq")
    // generators.
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO, generator = "UserProfileIdGenerator")
    public Long getUserProfileId() {
        return userProfileId;
    }
    public void setUserProfileId(Long userProfileId) {
        this.userProfileId = userProfileId;
    }

    public String getLoginName() {
        return loginName;
    }
    public void setLoginName(String loginName) {
        this.loginName = loginName;
    }

    @Column(name = "enPassword")
    public String getEncryptedPassword() {
        return encryptedPassword;
    }
    public void setEncryptedPassword(String encryptedPassword) {
        this.encryptedPassword = encryptedPassword;
    }

    public String getFirstName() {
        return firstName;
    }
    public void setFirstName(String firstName) {
        this.firstName = firstName;
    }

    public String getLastName() {
        return lastName;
    }
    public void setLastName(String lastName) {
        this.lastName = lastName;
    }

    public String getEmail() {
        return email;
    }
    public void setEmail(String email) {
        this.email = email;
    }

    @Override
    public String toString() {
        return "UserProfile [userProfileId=" + userProfileId + ", loginName="
                + loginName + ", encryptedPassword=" + encryptedPassword
                + ", firstName=" + firstName + ", lastName=" + lastName
                    + ", email=" + email + "]";
    }

}

邮件错误：

数据截断：对于列＆＃39; enPassword＆＃39;在第1行; SQL [不适用];嵌套异常是org.hibernate.exception.DataException：数据截断：对于列＆＃39; enPassword＆＃39;在第1行

如何解决这个问题？

（方法 doEncryptedPassword 中字符串返回的长度为129个字符）

Answer 1

当列enPassword传递的值大于150个字符时，将发生此异常。尝试增加enPassword列大小

同时检查方法doEncryptedPassword

中字符串返回的长度

Answer 2

我发现了问题。

我有两个数据库Data和DataTest。在＆＃39;数据＆＃39;一切正常。

In＆＃39; DataTest＆＃39; （执行测试的数据库）有这个＆＃39; UserProfile＆＃39;之前有多个外键（之所以不删除）。

旧表的 enPassword列定义为

enPassword VARCHAR(30) NOT NULL

出现错误的原因（129个字符的字符串无法存储到30个大小的列中。）

要解决使用外键删除表的问题，请执行此操作

SET FOREIGN_KEY_CHECKS=0;
DROP TABLE IF EXISTS UserProfile;
SET FOREIGN_KEY_CHECKS=1;

数据截断：数据为列的长度

2 个答案: