String sess_user = session.getAttribute("userid").toString();
out.println(sess_user);
try{
String connectionURL = "jdbc:mysql://localhost:3306/books";// books is the database
Connection connection = null;
Class.forName("com.mysql.jdbc.Driver");
connection = DriverManager.getConnection(connectionURL, "root", "bonnysingh");
Statement st = connection.createStatement();
ResultSet rs=st.executeQuery("select email from signup where UserName = '" + sess_user +"'");
while(rs.next()){
userEmail = rs.getString(3);
}
}
catch(Exception ee){
out.println(ee);
}
out.println("Hello");
out.println(userEmail);
userEmail的输出始终为null.我不知道这里有什么问题.但我认为while(rs.next())循环没有正确执行.谁能告诉我这是错误的？
这段代码
SELECT * FROM MBR_MST WHERE MBR_JOIN_DT=MIN(MBR_JOIN_DT)
显示以下错误:
消息147,级别15,状态1,行1聚合可能不会出现在WHERE子句中,除非它位于HAVING子句或选择列表中包含的子查询中,并且聚合的列是外部引用.
它给我一个错误,说没有找到列
import java.sql.*;
class jdbc_demo
{
public static void main(String args[])
{
try
{
Class.forName("sun.jdbc.odbc.JdbcOdbcDriver");
Connection con=DriverManager.getConnection("jdbc:odbc:testc");
//Class.forName("com.mysql.jdbc.Driver");
//Connection con=DriverManager.getConnection("jdbc:mysql://localhost/db","root","abc");
Statement st=con.createStatement();
//st.execute("create table student(rollno int,name varchar(20),class varchar(10))");
//st.execute("insert into student values(1,'Raj','Fy')");
//st.execute("insert into student values(2,'Karan','Sy')");
//st.execute("insert into student values(3,'sameer','Ty')");
//st.execute("delete from student where name='xyza'");
//st.executeUpdate("update student set name='xyza',class='Fy' where rollno=3");
ResultSet rs=st.executeQuery("select*from student");
ResultSetMetaData rsmd=rs.getMetaData();
String c1=rsmd.getColumnName(1);
String c2=rsmd.getColumnName(2);
String c3=rsmd.getColumnName(3);
System.out.println(c1+"\t"+c2+"\t"+c3);
while(rs.next())
{
int r=rs.getInt("roll no");
String n=rs.getString("name");
String c=rs.getString(3);
System.out.println(r+"\t"+n+"\t"+c);
}
con.close();
System.out.println("Connection successfull");
}
catch(Exception e) …在sql server中,当我选择1/2时,它返回0而不是0.5
为什么？
不应该所有分区都返回小数值吗？我可以设置一个设置让它正常划分吗？
我在c#中注意到了同样的
事情背后的逻辑是什么？
当运行以下SQL查询时,我收到消息:
ORA-00907: missing right parenthesis
00907. 00000 - "missing right parenthesis"
*Cause:
*Action:
Error at Line: 32 Column: 18
我最近在查询中添加了dv6.long_name toll_system_type,所以我知道它与之有关,因为之前工作正常.
这是SQL:
with t as (SELECT DISTINCT dv6.long_name toll_system_type,
ti.tollsystem_id,
ti.toll_id,
pp.type,
dv.published_value,
nvl(dv.long_name, 'N/A') paypoint_type,
pp.map_id,
pp.map_id_entry,
nvl(dv2.long_name, 'N/A') tollpaymentmethod,
c.currency,
c.accuracy,
c.per_distance,
nvl(dv3.long_name, 'N/A') currencytype ,
COUNT (DISTINCT ti.toll_id) count_Toll_ID,
(COUNT (DISTINCT c.cost_id)) count_costid,
MIN(asr.subregion) AS region
FROM
WEU_bw_.TC_TOLL_ID ti,
WEU_bw_.TC_PAYPOINT pp,
WEU_bw_.nt_domain_value dv,
WEU_bw_.tc_cost_paymethod cp,
WEU_bw_.nt_domain_value dv2,
WEU_bw_.tc_cost c,
WEU_bw_.nt_domain_value dv3,
WEU_bw_.nt_condition con,
WEU_bw_.nt_nav_strand ns,
WEU_bw_.nt_nav_link nl, …我正在使用此代码，但收到上面的错误消息。为什么会这样？我还为第二台 PC 创建了链接服务器，链接名称 = 192.168.1.101，用户名=xxxx 和密码=xxxx
WITH Source AS
( SELECT id, name FROM Tamio.dbo.memberform WHERE active = 1 or active =0)
MERGE INTO [192.168.1.101].Imatismos.dbo.customermaster AS target
USING Source ON target.id = source.id
WHEN MATCHED THEN UPDATE SET target.name = source.name
WHEN NOT MATCHED THEN INSERT (id, name) VALUES (source.id, source.name)
我有一个复选框,当选中字段时,字段将返回到html表.我似乎无法删除最后一个逗号.
$faultTypesStr = '';
if(!empty($faultTypeIds)){
foreach($faultTypeIds as $faultTypeId){
$faultTypeResult = FaultType::find()->where(['FaultTypeId' => $faultTypeId])->one();
$faultTypesStr = $faultTypesStr.$faultTypeResult->FaultTypeName.', ';
}}
这是我在html表中的显示方式.SMTP
<tr>
<td class="fields"> <b>Type Of Fault</b></td>
<td>'.$faultTypesStr.'</td>
</tr>
我在我的数据库表中添加了一列.它是一个简单的Char列,其值为"Y"或"N".
是否可以将列默认为"N"？如果是这样,怎么样？
当前添加列的脚本:
ALTER TABLE PERSON
ADD IS_ACTIVE VARCHAR2(1);
我有一个名为 empl 的表，还有另一个表 new_empl 与 empl 具有相同的列和定义。是否可以使用以下代码在 new_empl 中从 v_record 插入记录？
DECLARE
v_record empl%rowtype;
BEGIN
Insert into new_empl values v_record;
END;
empl 表中有太多列，我想避免列出它们。
import java.util.ArrayList;
import java.util.Scanner;
public class ToisenJaKolmannenSumma {
public static void main(String[] args) {
Scanner lukija = new Scanner(System.in);
ArrayList<Integer> numbers = new ArrayList<>();
while (true) {
int number = Integer.valueOf(lukija.nextLine());
if (number == 0) {
break;
}
numbers.add(number);
}
System.out.println(numbers.get(1+2));
}
}