Question

<%
RESOURCES res = new RESOURCES();
ResultSet rs = res.getAll();
while(rs.next()){

if( res.getSecId(rs.getInt("res_id")).equals(request.getParameter("sec")) ){
    out.println(res.getTitle(rs.getInt("res_id")));
}
}
%>

对于此代码，我有以下错误。

java.sql.SQLException: Before start of result set
at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:1094)
at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:997)
at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:983)
at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:928)
at com.mysql.jdbc.ResultSetImpl.checkRowPos(ResultSetImpl.java:850)
at com.mysql.jdbc.ResultSetImpl.getStringInternal(ResultSetImpl.java:5768)
at com.mysql.jdbc.ResultSetImpl.getString(ResultSetImpl.java:5688)
at com.mysql.jdbc.ResultSetImpl.getString(ResultSetImpl.java:5728)
.
.
.

注意：getSecId，getTitle和getAll是RESOURCES java类中的方法，而我的db连接是在另一个java类中。

public String getSecId(int id) throws SQLException {
    String SQL = "select section_sec_id from resource where res_id = " + id;
    ResultSet res = db.getStmt().executeQuery(SQL);
    return  res.getString("section_sec_id");

}

 public String getTitle(int id) throws SQLException {
    String SQl = "select title from resource where res_id = " + id;
    ResultSet res = db.getStmt().executeQuery(SQl);
    res.next();
    return res.getString("title");
}

Answer 1

Before start of result set

这意味着您在调用next（）之前尝试从ResultSet获取值，以便在结果集的开头进行定位。事实上：

ResultSet res = db.getStmt().executeQuery(SQL);
return  res.getString("section_sec_id");

应该是

ResultSet res = db.getStmt().executeQuery(SQL);
if (res.next()) {
    return res.getString("section_sec_id");
}
else {
    throw new IllegalStateException("I expect to have at least one row, but there is none");
}

那就是说，请不要把Java代码，甚至更少的数据库访问代码放在JSP中。这不应该是它应该的地方。 JSP是一种视图技术。它的职责是生成HTML，而不是访问数据库。

使用jsp页面从db中提取数据

1 个答案: