int i = 0;
while(rs.next()) {
    i++;
}

我也有同样的问题。以这种方式使用ResultSet.first()在执行后解决它:

if(rs.first()){
    // Do your job
} else {
    // No rows take some actions
}

文档(链接):

布尔第一次() 抛出SQLException异常 将光标移动到此ResultSet对象的第一行。 返回: 如果光标位于有效的 行;如果结果集中没有行，则为False 抛出: SQLException—如果发生数据库访问错误;在封闭的结果集或结果集类型为TYPE_FORWARD_ONLY时调用此方法 SQLFeatureNotSupportedException—如果JDBC驱动程序不支持 这个方法 自: 1．2

最简单的方法，运行Count(*)查询，执行resultSet.next()指向第一行，然后执行resultSet.getString(1)获得计数。代码:

ResultSet rs = statement.executeQuery("Select Count(*) from your_db");
if(rs.next()) {
   int count = rs.getString(1).toInt()
}

如果你有一个ResultSet类型的ResultSet。TYPE_FORWARD_ONLY您希望保持这种方式(而不是切换到ResultSet)。TYPE_SCROLL_INSENSITIVE或ResultSet。TYPE_SCROLL_INSENSITIVE，以便能够使用.last())。

我建议一种非常有效的破解方法，即在顶部添加第一个包含行数的伪/伪行。

例子

假设您的查询如下

select MYBOOL,MYINT,MYCHAR,MYSMALLINT,MYVARCHAR
from MYTABLE
where ...blahblah...

输出是这样的

true    65537 "Hey" -32768 "The quick brown fox"
false  123456 "Sup"    300 "The lazy dog"
false -123123 "Yo"       0 "Go ahead and jump"
false       3 "EVH"    456 "Might as well jump"
...
[1000 total rows]

简单地重构你的代码，就像这样:

Statement s=myConnection.createStatement(ResultSet.TYPE_FORWARD_ONLY,
                                         ResultSet.CONCUR_READ_ONLY);
String from_where="FROM myTable WHERE ...blahblah... ";
//h4x
ResultSet rs=s.executeQuery("select count(*)as RECORDCOUNT,"
                           +       "cast(null as boolean)as MYBOOL,"
                           +       "cast(null as int)as MYINT,"
                           +       "cast(null as char(1))as MYCHAR,"
                           +       "cast(null as smallint)as MYSMALLINT,"
                           +       "cast(null as varchar(1))as MYVARCHAR "
                           +from_where
                           +"UNION ALL "//the "ALL" part prevents internal re-sorting to prevent duplicates (and we do not want that)
                           +"select cast(null as int)as RECORDCOUNT,"
                           +       "MYBOOL,MYINT,MYCHAR,MYSMALLINT,MYVARCHAR "
                           +from_where);

您的查询输出将如下所示

1000 null     null null    null null
null true    65537 "Hey" -32768 "The quick brown fox"
null false  123456 "Sup"    300 "The lazy dog"
null false -123123 "Yo"       0 "Go ahead and jump"
null false       3 "EVH"    456 "Might as well jump"
...
[1001 total rows]

所以你必须

if(rs.next())
    System.out.println("Recordcount: "+rs.getInt("RECORDCOUNT"));//hack: first record contains the record count
while(rs.next())
    //do your stuff

这是一种简单的行计数方法。

ResultSet rs = job.getSearchedResult(stmt);
int rsCount = 0;

//but notice that you'll only get correct ResultSet size after end of the while loop
while(rs.next())
{
    //do your other per row stuff 
    rsCount = rsCount + 1;
}//end while

当使用rs.last()时，我得到了一个异常

if(rs.last()){
    rowCount = rs.getRow(); 
    rs.beforeFirst();
}

:

java.sql.SQLException: Invalid operation for forward only resultset

因为默认情况下它是ResultSet。TYPE_FORWARD_ONLY，这意味着只能使用rs.next()

解决方案是:

stmt=conn.createStatement(ResultSet.TYPE_SCROLL_INSENSITIVE,
    ResultSet.CONCUR_READ_ONLY);