我一直在研究这个具体问题几天,找不到任何解决方案。因此,我在这里。
情况:
修改
我打算如何处理查询:我的servlet正在尝试显示连接到给定jboss的所有信息。用户从jsp页面的jboss名称列表中选择一个jboss。然后执行查询并发生错误。
EDIT2:
将查询更改为内部子选择为const bool = "101011";
const bool2 = "111011";
const and = (parseInt(bool,2) & parseInt(bool2,2)).toString(2);
console.log("bool: "+bool+", bool2: "+bool2+", &: "+and);
会在测试小程序时导致相同的错误
EDIT4:
尝试使用查询作为脚本创建视图,然后尝试使用where in
从创建的视图中获取所有条目,但没有成功。
问题:
当我尝试使用servlet运行以下select * from vtest
时,会产生ORA-01427。如果我在select
中运行此语句,我会得到所需的结果。
Toad for Oracle
问题:
查询中是否存在我遗漏的内容?即使查询本身是正确的,select * from table1 where number in (
select jboss2 from table2 where jboss1 = (
select number from table1 where name = 'nam1'))
是否有可能无法处理多行的子查询?
表1:
jdbc
表2:
+--------+------+-----------+
| Number | Name | values... |
+--------+------+-----------+
| 000001 | nam1 | vals1 |
| 000002 | nam2 | vals2 |
| 000003 | nam3 | vals3 |
+--------+------+-----------+
结果Toad / Desired result:
+--------+--------+
| JBoss1 | JBoss2 |
+--------+--------+
| 000001 | 000002 |
| 000001 | 000003 |
| 000002 | 000003 |
+--------+--------+
EDIT3:
相关的Java类。省略了不相关的查询。
+--------+------+-----------+
| Number | Name | values... |
+--------+------+-----------+
| 000002 | nam2 | vals2 |
| 000003 | nam3 | vals3 |
+--------+------+-----------+
错误消息
class QuickInfoAction implements Action{
@Override
public String execute(HttpServletRequest request, HttpServletResponse response) throws ActionException {
Connection conn = null;
PreparedStatement prep = null;
ResultSet rs = null;
Map<String,String> queries = Queries.getInfoQueries(request);
try {
conn = DatabaseConnector.getConnection();
Map<String, Result> res = new HashMap<String, Result>();
for (Map.Entry<String, String> entry: queries.entrySet()) {
prep = conn.prepareStatement(entry.getValue());
rs = prep.executeQuery();
while(rs.next()) {
res.put(entry.getKey(), ResultSupport.toResult(rs));
}
}
request.setAttribute("results", res);
} catch (Exception e) {
throw new ActionException(e.getStackTrace().toString());
} finally {
try {
conn.close();
prep.close();
rs.close();
} catch (Exception e) {
throw new ActionException(e.getStackTrace().toString());
}
}
return "results";
}
}
public static Map<String, String> getInfoQueries(HttpServletRequest request) {
String jboss_res = "select jboss.name, jboss.port, jboss.apache_nummer, jboss.bere_mandant_id, "
+ "maschine.name as maschine, maschine.ip_adresse "
+ "from jboss "
+ "inner join maschine on jboss.maschine_nummer = maschine.nummer "
+ "where jboss.name = '" + request.getParameter("jboss") + "'";
String jboss_db = "select datenbank.nummer, datenbank.name, db_schema.name as schema "
+ "from datenbank "
+ "inner join db_schema on datenbank.db_schema_nummer = db_schema.nummer "
+ "where datenbank.nummer = ("
+ "select datenbank_nummer "
+ "from jboss_datenbank "
+ "where jboss_nummer = ("
+ "select nummer "
+ "from jboss "
+ "where name = '" + request.getParameter("jboss") + "'))";
String jboss_tux = "select tuxedo.*, datenbank.name as datenbank, db_schema.name as schema "
+ "from tuxedo, datenbank,db_schema "
+ "where tuxedo.nummer = ("
+ "select tuxedo_nummer "
+ "from jboss "
+ "where name = '" + request.getParameter("jboss") + "') "
+ "and datenbank.nummer = ("
+ "select datenbank_nummer "
+ "from tuxedo_datenbank "
+ "where tuxedo_nummer = tuxedo.nummer) "
+ "and db_schema.nummer = ("
+ "select db_schema_nummer "
+ "from datenbank "
+ "where nummer = ("
+ "select datenbank_nummer "
+ "from tuxedo_datenbank "
+ "where tuxedo_nummer = tuxedo.nummer))";
String jboss_corr = "select * from jboss where nummer in ("
+ "select jboss_nummer_2 from jboss_corr where jboss_nummer_1 in ("
+ "select nummer from jboss where name = '" + request.getParameter("jboss") + "'))";
Map<String, String> queries = new HashMap<String,String>();
queries.put("jboss", jboss_res);
queries.put("datenbank", jboss_db);
queries.put("tuxedo", jboss_tux);
queries.put("corr", jboss_corr);
return queries;
答案 0 :(得分:0)
似乎你也需要内部子选择
select * from table1 where number in (
select jboss2 from table2 where jboss1 in (
select number from table1 where name = 'nam1'))
答案 1 :(得分:0)
不确定为什么带有where-in的查询不适合您,但这里是一个查询,其中第二个子查询被连接替换:
select * from table1 where number in (
select jboss2 from table2
join table1 on table2.jboss1 = table1.number
and table1.name = 'nam1'
)
答案 2 :(得分:0)
我找到了解决方案。另一个查询也返回了多个结果。此查询不是为此而设计的,因此失败了。这是我的疏忽。