JPA：如何使用 SHA1 加密保留列？

发布于 2024-12-11 15:13:11 字数 930 浏览 0 评论 0原文

我对 JPA（以及 JPQL）相当陌生。希望有人能够启发我解决这个问题。

我试图执行的查询...

String query = "select u from user u where u.email = '" + userEmail + "' and u.password = sha1('"+ userPassword + "')";
List resultList = emf.createEntityManager().createQuery(query).getResultList();

并且我收到以下异常...

Caused by: Exception [EclipseLink-8025] (Eclipse Persistence Services - 2.2.0.v20110202-r8913): org.eclipse.persistence.exceptions.JPQLException
Exception Description: Syntax error parsing the query [select u from Utilisateur u where u.email = 'myuser' and u.mot_Passe = sha1('mypassword')], line 1, column 73: unexpected token [(].
Internal Exception: NoViableAltException(83@[()* loopback of 383:9: (d= DOT right= attribute )*])
    at org.eclipse.persistence.exceptions.JPQLException.unexpectedToken(JPQLException.java:372)
...

显然，我在这里遗漏了一些东西，我应该以某种方式转义该字符吗？以另一种方式指定它？我将非常感谢对此的任何帮助。

原文

I'm fairly new to JPA (and JPQL by extension). Hopefully, someone will be able to enlighten me with this problem.

The query I'm trying to execute ...

String query = "select u from user u where u.email = '" + userEmail + "' and u.password = sha1('"+ userPassword + "')";
List resultList = emf.createEntityManager().createQuery(query).getResultList();

And I'm getting the follwowing exception ...

Caused by: Exception [EclipseLink-8025] (Eclipse Persistence Services - 2.2.0.v20110202-r8913): org.eclipse.persistence.exceptions.JPQLException
Exception Description: Syntax error parsing the query [select u from Utilisateur u where u.email = 'myuser' and u.mot_Passe = sha1('mypassword')], line 1, column 73: unexpected token [(].
Internal Exception: NoViableAltException(83@[()* loopback of 383:9: (d= DOT right= attribute )*])
    at org.eclipse.persistence.exceptions.JPQLException.unexpectedToken(JPQLException.java:372)
...

Obviously, I'm missing something here, should I escape the character somehow? specify it in another way? I'll appreciate very much any help on this.

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感受沵的脚步 2024-12-18 15:13:11

我认为JPA没有sha1功能。因此，您有两个选择：

您可以在 Java 代码上实现 sha1 函数，并在加载查询中的参数之前使用它们。

字符串查询 = "从用户 u 中选择 u，其中 u.email = :userEmail" +
   " 和 u.password = :userPassword";
查询 jpqlQuery = em.createQuery(query)
    .setParameter("用户邮箱", 用户邮箱)
    .setParameter("用户密码",sha1(用户密码));

如果您的数据库具有 sha1 函数，您可以编写本机查询。检查此链接：http://www.oracle.com/technetwork/articles /vasiliev-jpql-087123.html

列表<客户>;客户 = (List<客户>)em.createNativeQuery
        （“从客户中选择*”，jpqlexample.entities.Customer.class）
                      .getResultList(); 
迭代器 i =customers.iterator();
客户定制；
while (i.hasNext()) {
    cust = (客户) i.next();
    //做某事
}

I think that JPA doesn't have a sha1 function. So you have two options:

You can implementa a sha1 function on you Java code and use them before load the parameters in the query.

String query = "select u from user u where u.email = :userEmail" +
   " and u.password = :userPassword";
Query jpqlQuery = em.createQuery(query)
    .setParameter("userEmail", userEmail)
    .setParameter("userPassword",sha1(userPassword));

If your database has the sha1 function you can write a native query. Check this link: http://www.oracle.com/technetwork/articles/vasiliev-jpql-087123.html

List<Customer> customers = (List<Customer>)em.createNativeQuery
        ("SELECT * FROM customers", jpqlexample.entities.Customer.class)
                      .getResultList(); 
Iterator i = customers.iterator();
Customer cust;
while (i.hasNext()) {
    cust = (Customer) i.next();
    //do something
}

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要走就滚别墨迹 2024-12-18 15:13:11

通过查询接口设置参数，而不是将它们逐字添加到 JPQL 字符串中。这样你也可以保护自己免受 SQL 注入：

String query =
   "select u from user u where u.email = :userEmail"
   + " and u.password = sha1(:userPassword)";
Query jpqlQuery = em.createQuery(query)
    .setParameter("userEmail", userEmail)
    .setParameter("userPassword",userPassword)

Set the parameters through the Query interface instead of adding them literally to the JPQL String. That way you are also protecting yourself from SQL injection:

String query =
   "select u from user u where u.email = :userEmail"
   + " and u.password = sha1(:userPassword)";
Query jpqlQuery = em.createQuery(query)
    .setParameter("userEmail", userEmail)
    .setParameter("userPassword",userPassword)

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