Java JSF、JPA 应用程序中的 SQL 查询错误
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SQL Query error in JSF, JPA application
提问by Jean Carlos Suárez Marranzini
I have a problem with a query in my application. This is the query method doing the query:
我的应用程序中的查询有问题。这是执行查询的查询方法：
public List<Product> obtainProductListByCategory(String category)
{
Query query = em.createQuery("SELECT p FROM PRODUCT p WHERE CATEGORY='" + category + "'");
List<Product> ret = query.getResultList();
if (ret == null)
{
return new ArrayList<Product>();
}
else
{
return ret;
}
}
And this is the error: javax.ejb.EJBException
这是错误： javax.ejb.EJBException
And in the trace I found:
在跟踪中我发现：
Caused by: java.lang.IllegalArgumentException: An exception occurred while creating a query in EntityManager: Exception Description: Syntax error parsing [SELECT * FROM PRODUCT WHERE CATEGORY='Humano']. [22, 22] A select statement must have a FROM clause. [7, 7] The left expression is missing from the arithmetic expression. [9, 22] The right expression is not an arithmetic expression
引起：java.lang.IllegalArgumentException：在EntityManager中创建查询时发生异常：异常描述：语法错误解析[SELECT * FROM PRODUCT WHERE CATEGORY='Humano']。[22, 22] select 语句必须有一个 FROM 子句。[7, 7] 算术表达式中缺少左表达式。[9, 22] 正确的表达式不是算术表达式
Any ideas? My objective is to refresh a JSF datatable in my webpage.
有任何想法吗？我的目标是刷新我网页中的 JSF 数据表。
Edited my code based on @Ilya's answer, and now I got this exception
根据@Ilya 的回答编辑了我的代码，现在我得到了这个异常
Caused by: java.lang.IllegalArgumentException: An exception occurred while creating a query in EntityManager: Exception Description: Problem compiling [SELECT p FROM PRODUCT p WHERE CATEGORY='Humano']. [14, 21] The abstract schema type 'PRODUCT' is unknown. [30, 38] The identification variable 'CATEGORY' is not defined in the FROM clause.
引起：java.lang.IllegalArgumentException：在 EntityManager 中创建查询时发生异常：异常描述：编译问题 [SELECT p FROM PRODUCT p WHERE CATEGORY='Humano']。[14, 21] 抽象模式类型“产品”是未知的。[30, 38] 标识变量“CATEGORY”未在 FROM 子句中定义。
As requested by @Ilya, I post my Productclass:
EDIT: Added @Table to the annotations.
根据@Ilya 的要求，我发布了我的Product课程：编辑：在注释中添加了@Table。
package model;
import java.io.Serializable;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.Table;
@Table
@Entity
public class Product implements Serializable, IProduct
{
private static final long serialVersionUID = 1L;
@GeneratedValue(strategy = GenerationType.AUTO)
@Id
private String name;
private int stock;
private float price;
private String category;
private String description;
@Override
public String getDescription() {
return description;
}
public void setDescription(String description) {
this.description = description;
}
public Product()
{
}
public Product(String name, int stock, float price, String category, String description)
{
this.name = name;
this.stock = stock;
this.price = price;
this.category = category;
this.description = description;
}
@Override
public String getName()
{
return name;
}
public void setName(String name)
{
this.name = name;
}
@Override
public float getPrice()
{
return price;
}
public void setPrice(float price)
{
this.price = price;
}
@Override
public int getStock()
{
return stock;
}
public void setStock(int stock)
{
this.stock = stock;
}
@Override
public int hashCode()
{
return name.hashCode();
}
@Override
public boolean equals(Object object)
{
if (!(object instanceof Product))
{
return false;
}
Product other = (Product) object;
if (name.equals(other.getName()))
{
return true;
}
return false;
}
@Override
public String getCategory()
{
return category;
}
@Override
public String toString()
{
return "Marketv2.model.Product[ name=" + name + " ]";
}
}
Thanks for the help so far. Here I post another query in my application, which is working properly:
感谢你目前的帮助。在这里，我在我的应用程序中发布了另一个查询，该查询正常工作：
public void removeProduct(Product g)
{
Query q = em.createQuery("SELECT x FROM BasketItem x WHERE x.product.name = '" + g.getName() + "'");
List<BasketItem> bItems = q.getResultList();
for (BasketItem i : bItems)
{
em.remove(i);
}
q = em.createQuery("DELETE FROM Product x WHERE x.name = '" + g.getName() + "'");
q.executeUpdate();
}
}
采纳答案by Ilya
1) You should specify alias for tables in FROMclause, and SELECTclause should contains aliasProductshould be an entity
1）你应该在FROM子句中为表指定别名，SELECT子句应该包含别名Product应该是一个实体
em.createQuery("SELECT p FROM Product p WHERE p.category='" + category + "'");
If PRODUCTisn't an entity, you should create nativeQuery
如果PRODUCT不是实体，则应创建nativeQuery
em.createNativeQuery("SELECT p FROM PRODUCT p WHERE p.CATEGORY='" + category + "'");
EntityManager::createQueryis for JPQL (Java Persistence query language)
EntityManager::createNativeQueryis for SQL
EntityManager::createQuery适用于 JPQL（Java 持久性查询语言）
EntityManager::createNativeQuery适用于 SQL
2) JPA throws the "Unknown abstract schema type"error when JPA fails to locate your entity class
Also add entity to persistence.xml
2）JPA"Unknown abstract schema type"无法定位您的实体类时，JPA 抛出错误
并将实体添加到persistence.xml
<persistence-unit ...>
<class>com.package.Product</class>
3) Add @Tableannotation to your @Entity
4) As I see in documentation, JPQL is case-sensitive.
3)@Table向您的@Entity
4 ) 添加注释，正如我在文档中看到的，JPQL 区分大小写。
With the exception of names of Java classes and properties, queries are case-insensitive. So SeLeCT is the same as sELEct is the same as SELECT, but org.hibernate.eg.FOO and org.hibernate.eg.Foo are different, as are foo.barSet and foo.BARSET.
除了 Java 类和属性的名称之外，查询不区分大小写。所以SeLeCT和sELEct一样，SELECT和SELECT一样，但是org.hibernate.eg.FOO和org.hibernate.eg.Foo是不同的，foo.barSet和foo.BARSET也是不同的。
So JPQL query should be
所以JPQL查询应该是
SELECT p FROM Product p WHERE p.category = '...
