Question

我的代码中存在此错误：

"Request processing failed; nested exception is javax.persistence.PersistenceException: 
org.hibernate.exception.SQLGrammarException: could not extract ResultSet] con causa raíz
com.ibm.db2.jcc.am.SqlSyntaxErrorException: DB2 SQL Error: SQLCODE=-206, 
SQLSTATE=42703, SQLERRMC=INVOICESEL2_.CODE_K, DRIVER=4.14.88"

我找不到错误，我知道SQL代码-206引用的对象是在无效的上下文中指定的。

我的代码是这样的：

public InvoiceSeller getSellerByEmail(String email, String debtorCode) {

    String qString = "SELECT d FROM  " + InvoiceSeller.class.getSimpleName() + " d "
            + " INNER JOIN "+ InvoiceSellerContact.class.getSimpleName() + " sc "
            + " ON d.code = sc.invoiceSeller.code AND d.invoiceDebtor.code = sc.invoiceSeller.invoiceDebtor.code "
            + " INNER JOIN "+ Contact.class.getSimpleName() + " c ON sc.contact.id = c.id "
            + " INNER JOIN " + ActiveRegisterMaster.class.getSimpleName() + " a "
            + " ON d.code = a.code AND d.invoiceDebtor.code = a.invoiceDebtor "
            + " WHERE a.tableName = :tablename AND d.invoiceDebtor.code = :debtorCode "
            + " AND c.email = :email AND d.status = :status";


    TypedQuery<InvoiceSeller> query = this.em.createQuery(qString, InvoiceSeller.class);
    System.out.println(qString);

    System.out.println(query.getFirstResult());

    query.setParameter("debtorCode", debtorCode);
    query.setParameter("email", email);
    query.setParameter("tablename", InvoiceSeller.TABLENAME);
    query.setParameter("status", "A");

    InvoiceSeller seller = query.getSingleResult();

    System.out.println(seller.toString());

    return seller;
}

谢谢！

Answer 1

您的HQL是错误的，请记住，HQL语法不同于SQL，并且当您调用=VALUE("red")时，您正在尝试创建HQL。

因此，您的HQL应该是这样的：

select * from SOME_VIEW
union all select a0,b0,c0,d0... from dual
union all select a1,b1,c1,d1... from dual
...

请注意，我在INNER JOIN之后删除了 ON 关键字，因为HQL不需要它。

休眠错误：无法提取ResultSet

1 个答案: