java - Hibernate sql错误:java. lang.String无法转换为java.lang.Integer

Question

我更改了 Hibernate sql，遇到了问题。 我认为这是因为 ids 是字符串。
我该如何解决这个问题？

java.lang.ClassCastException: java.lang.String cannot be cast to java.lang.Integer
    at org.hibernate.type.IntegerType.set(IntegerType.java:64)
    at org.hibernate.type.NullableType.nullSafeSet(NullableType.java:156)
    at org.hibernate.type.NullableType.nullSafeSet(NullableType.java:138)
    at org.hibernate.param.PositionalParameterSpecification.bind(PositionalParameterSpecification.java:68)
    at org.hibernate.loader.hql.QueryLoader.bindParameterValues(QueryLoader.java:567)
    at org.hibernate.loader.Loader.prepareQueryStatement(Loader.java:1612)
    at org.hibernate.loader.Loader.doQuery(Loader.java:717)
    at org.hibernate.loader.Loader.doQueryAndInitializeNonLazyCollections(Loader.java:270)
    at org.hibernate.loader.Loader.doList(Loader.java:2294)
    at org.hibernate.loader.Loader.listIgnoreQueryCache(Loader.java:2172)
    at org.hibernate.loader.Loader.list(Loader.java:2167)
    at org.hibernate.loader.hql.QueryLoader.list(QueryLoader.java:448)...

这是我的代码:

public List getUser(int leader, List<Member> members) {
    String ids = String.valueOf(leader);
    if (GeneralUtil.hasValue(members)) {
        for (Member member : members) {
            ids += "," + member.getUserId(); //1,2,3,4  
        }
    }
    try {
        // Old method works well:
        // StringBuffer qryStr = new StringBuffer(
        //      "from User u where u.userId in (" + ids + ")"); 
        // return this.getHibernateTemplate().find(qryStr.toString());  

        StringBuffer qryStr = new StringBuffer("from User u where u.userId in (?)");    
        return this.getHibernateTemplate().find(qryStr.toString(), new Object[] { ids });   
    } catch (DataAccessException e) {
        e.printStackTrace();
    }
    return null;
}

Answer 1

整数数组怎么样？

Integer[] ids = new Integer[members.size()+1];
ids[0] = leader;
for (int i = 0; i < members.size(); i++) {
   ids[i+1] = member.getUserId();
}