见下文。 rs.getString("tags")
是超过1行的子查询。我想迭代那个子查询(rs.getString(“tags”)----喜欢rs.next()
。
while (rs.next()) {
emailDto emaildto = new emailDto();
emaildto.setMid(rs.getInt("id"));
emaildto.setSub(rs.getString("sub"));
emaildto.setMessage(rs.getString("message"));
while(rs.getString("tags").next()){
arrtags[i] = rs.getString(1);
}
emaildto.setTags(arrtags);
rs.getString("tags")
不起作用 - 并且包含多于1行。如何提取它。有没有技术?
答案 0 :(得分:0)
尝试这样的事情:
Array tagsArray = rs.getArray(“tags”); String [] tags =(String [])tagsArray.getArray();
答案 1 :(得分:0)
这不是答案。这是完整的代码
列出emails = new ArrayList();
String listQuery = "select mid, sub, message, "
+ " (select emailid from sub_ids where sub_ids.messageid= sub_mail_list.mid ) // this query fetch more than one row.
作为refid" +"来自sub_mail_list";
PreparedStatement ps = null;
ResultSet rs;
try {
ps = DatabaseConnectionUtil.getConnection().prepareStatement(
listQuery);
rs = ps.executeQuery(listQuery);
while (rs.next()) {
emailDto emaildto = new emailDto();
emaildto.setMid(rs.getInt("mid"));
emaildto.setSub(rs.getString("sub"));
emaildto.setMessage(rs.getString("message"));
Array tagsArray = rs.getArray("refid");
List<vtbDto> vtbdtosvr = new ArrayList<vtbDto>();
int[] tags = (int[])tagsArray.getArray();
for (int i = 0; i < tags.length; i++) {
vtbDto vtbdto = new vtbDto();
vtbdto.setRefid(tags[i]);
vtbdtosvr.add(vtbdto);
}
emaildto.setAr(tagsArray);
emails.add(emaildto);
}
} catch (Exception ex) {
ex.printStackTrace();
} finally {
DatabaseConnectionUtil.closeAll(ps);
}
return emails;
这是打印功能的代码
列出emailDtos = emaildao.getAllemails();
for (emailDto emailDto2 : emailDtos) {
System.out.println( emailDto2.getMid());
System.out.println( emailDto2.getSub());
System.out.println( emailDto2.getMessage());
List<vtbDto> vtbdtos= emailDto2.getVtbdtolst();
for (vtbDto vtbdto2 : vtbdtos) {
System.out.print(vtbdto2.getRefid() + ", ");
}
}
并且控制台打印&#34;子查询返回超过1行&#34 ;;