java - 数据截断 : Data to long for column

标签 java mysql hibernate encryption

我正在尝试将加密的密码插入到 UserProfile 表中。

我创建表的sql文件是这样的:

-- ---------- 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;
    }

}

用户配置文件.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 + "]";
    }

}

消息错误:

Data truncation: Data too long for column 'enPassword' at row 1; SQL [n/a]; nested exception is org.hibernate.exception.DataException: Data truncation: Data too long for column 'enPassword' at row 1

我该如何解决这个问题?

(doEncryptedPassword方法返回的字符串长度为129个字符)

最佳答案

当为 enPassword 列传递的值大于 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 中返回的字符串长度是多少

关于java - 数据截断 : Data to long for column,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/27883324/

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