当UserType表示单行对象集时。如何用类似条件查询Hibernate自定义的UserType？ [英] When UserType represents Set of Objects in single line. How to query Hibernate custom UserType with like criteria?

问题描述

当试图用查询这个集合就像一样，条件，使用 JPA CriteriaBuilder Hibernate抛出 IllegalArgumentException
$ b

参数值字符串与预期类型不匹配java.util.Set

是否有解决方法？

以下是我们使用的UserType：

  public class SetStringType implements UserType，LiteralType< Set< String>> {
 
 final private static String SEPARATOR =|; 
 final private static String SEPARATOR_REGEXP =\\ |; 
 
 @Override 
 public int [] sqlTypes（）{
 return new int [] {Types.VARCHAR}; 
 
 $ b @Override 
 public Object nullSafeGet（ResultSet rs，String [] names，SessionImplementor session，Object owner）throws HibernateException，
 SQLException {
的HashSet<字符串> resultValues = new HashSet< String>（）; 
 
字符串值= rs.getString（names [0]）; 
 if（value == null）
 return resultValues; 
 
 String [] values = value.split（SEPARATOR_REGEXP）; 
 resultValues.addAll（Arrays.asList（values））; 
返回resultValues; 
 
 $ b @Override 
 @SuppressWarnings（unchecked）
 public void nullSafeSet（PreparedStatement st，Object value，int index，SessionImplementor session）throws HibernateException，
 SQLException {
 st.setString（index，StringUtils.collectionToDelimitedString（（Collection< String>）value，SEPARATOR））; 
 
 $ b @Override 
 @SuppressWarnings（rawtypes）
 public Class returnedClass（）{
 return Set.class; 
 
 $ b @Override 
 public boolean equals（Object x，Object y）throws HibernateException {
 return ObjectUtils.equals（x，y）; 
} 
 
 @Override 
 public int hashCode（Object x）throws HibernateException {
 assert x！= null; 
 return x.hashCode（）; 
 
 $ b @覆盖
 @SuppressWarnings（unchecked）
 public Object deepCopy（Object value）抛出HibernateException {
 if（value == null ）
返回null; $（bash）
 if（value of instance of HashSet）
 return（（HashSet< String>）value）.clone（）; 
 
返回新的HashSet< String>（（Collection< String>）value）; 
} 
 
 @Override 
 public boolean isMutable（）{
 return true; 
 
 $ b @Override 
 public可串行化反汇编（Object value）抛出HibernateException {
 return（Serializable）deepCopy（value）; 
 
 $ b @Override 
 public Object assemble（Serializable cached，Object owner）throws HibernateException {
 return deepCopy（cached）; 
} 
 
 @Override 
 public Object replace（Object original，Object target，Object owner）throws HibernateException {
 return original; 
 
 $ b @Override 
 public String objectToSQLString（Set< String> value，Dialect dialect）throws Exception {
 return StringUtils.collectionToDelimitedString（value，SEPARATOR）; 
 
 
 
 $ div $解析方案

spring javadoc



  collectionToDelimitedString 
 
 public static String collectionToDelimitedString（Collection coll，
 String delim） 
 
便捷方法将Collection作为分隔符（例如CSV）字符串返回。例如。用于toString（）实现。 
 
参数：
 coll  - 要显示的集合
 delim  - 要使用的分隔符（可能是，）
返回：
分隔字符串
  

所以你的代码行是不正确的，因为你设置的字符串不是Set，因为它是预期的

  st.setString（index，StringUtils.collectionToDelimitedString（（Collection< String>）value，SEPARATOR））; 
  




We are using custom Hibernate UserType to store a Set of Strings in a single line.

When trying to Query this set with like criteria, using JPA CriteriaBuilder Hibernate throws IllegalArgumentException

Parameter value String did not match expected type java.util.Set

Is there a workaround for this?

Here is a UserType we are using:

public class SetStringType implements UserType, LiteralType<Set<String>> {

final private static String SEPARATOR = "|";
final private static String SEPARATOR_REGEXP = "\\|";

@Override
public int[] sqlTypes() {
    return new int[] { Types.VARCHAR };
}

@Override
public Object nullSafeGet(ResultSet rs, String[] names, SessionImplementor session, Object owner) throws HibernateException,
        SQLException {
    HashSet<String> resultValues = new HashSet<String>();

    String value = rs.getString(names[0]);
    if (value == null)
        return resultValues;

    String[] values = value.split(SEPARATOR_REGEXP);
    resultValues.addAll(Arrays.asList(values));
    return resultValues;
}

@Override
@SuppressWarnings("unchecked")
public void nullSafeSet(PreparedStatement st, Object value, int index, SessionImplementor session) throws HibernateException,
        SQLException {
    st.setString(index, StringUtils.collectionToDelimitedString((Collection<String>) value, SEPARATOR));
}

@Override
@SuppressWarnings("rawtypes")
public Class returnedClass() {
    return Set.class;
}

@Override
public boolean equals(Object x, Object y) throws HibernateException {
    return ObjectUtils.equals(x, y);
}

@Override
public int hashCode(Object x) throws HibernateException {
    assert x != null;
    return x.hashCode();
}

@Override
@SuppressWarnings("unchecked")
public Object deepCopy(Object value) throws HibernateException {
    if (value == null)
        return null;

    if (value instanceof HashSet)
        return ((HashSet<String>) value).clone();

    return new HashSet<String>((Collection<String>) value);
}

@Override
public boolean isMutable() {
    return true;
}

@Override
public Serializable disassemble(Object value) throws HibernateException {
    return (Serializable) deepCopy(value);
}

@Override
public Object assemble(Serializable cached, Object owner) throws HibernateException {
    return deepCopy(cached);
}

@Override
public Object replace(Object original, Object target, Object owner) throws HibernateException {
    return original;
}

@Override
public String objectToSQLString(Set<String> value, Dialect dialect) throws Exception {
    return StringUtils.collectionToDelimitedString(value, SEPARATOR);
}
}

解决方案

From spring javadoc

 collectionToDelimitedString

public static String collectionToDelimitedString(Collection coll,
                                             String delim)

 Convenience method to return a Collection as a delimited (e.g. CSV) String. E.g. useful for toString() implementations.

Parameters:
    coll - the Collection to display
    delim - the delimiter to use (probably a ",") 
Returns:
    the delimited String

So your line is not correct because you are setting a String not a Set as it's expected

st.setString(index, StringUtils.collectionToDelimitedString((Collection<String>) value, SEPARATOR));

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