我有一个在网站上注册的用户列表，当管理员点击批准时，它必须获取id，然后将该行移动到另一个名为login的表...现在问题来自jsp如何获取id在servlet中，它可以传递给服务和DAO层的方法。
以下是我的jsp：
<tr bgcolor="white">
<td><b>id</b></td>
<td><b>FirstName</b></td>
<td><b>LastName</b></td>
<td><b>Gender</b></td>
<td><b>Category</b></td>
<td><b>DateOfBirth</b></td>
<td><b>Age</b></td>
<td><b>Address</b></td>
<td><b>Country</b></td>
<td><b>State</b></td>
<td><b>City</b></td>
<td><b>PinCode</b></td>
<td><b>EmailId</b></td>
<td><b>ContactNo</b></td>
<td><b>MobileNo</b></td>
</tr>
<c:forEach items="${clients}" var="client">
<tr>
<td><c:out value="${client.id}"/></td>
<td><c:out value="${client.firstName}"/></td>
<td><c:out value="${client.lastName}"/></td>
<td><c:out value="${client.gender}"/></td>
<td><c:out value="${client.category}"/></td>
<td><c:out value="${client.date}"/></td>
<td><c:out value="${client.age}"/></td>
<td><c:out value="${client.address}"/></td>
<td><c:out value="${client.country}"/></td>
<td><c:out value="${client.state}"/></td>
<td><c:out value="${client.city}"/></td>
<td><c:out value="${client.pinCode}"/></td>
<td><c:out value="${client.emailId}"/></td>
<td><c:out value="${client.contactNo}"/></td>
<td><c:out value="${client.mobileNo}"/></td>
<td><a href="ClientApproveServlet?id=${client.id}">APPROVE</a></td> // I have fetched the id here
<td><a href="Reject.jsp">REJECT</a></td>
<td><a href="OnHold.jsp">ONHOLD</a></td>
</tr>
</c:forEach>
Stack TRace：
java.lang.RuntimeException: com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '?' at line 1
at com.affiliate.DAO.ClientApproveDAO.insertClient(ClientApproveDAO.java:28)
at com.affiliate.service.ClientApproveService.clientApprove(ClientApproveService.java:12)
at com.affiliate.servlet.ClientApproveServlet.doGet(ClientApproveServlet.java:35)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:621)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:728)
at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:305)
at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:210)
at org.apache.tomcat.websocket.server.WsFilter.doFilter(WsFilter.java:51)
at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:243)
at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:210)
at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:222)
at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:123)
at org.apache.catalina.authenticator.AuthenticatorBase.invoke(AuthenticatorBase.java:502)
at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:171)
at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:100)
at org.apache.catalina.valves.AccessLogValve.invoke(AccessLogValve.java:953)
at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:118)
at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:408)
at org.apache.coyote.http11.AbstractHttp11Processor.process(AbstractHttp11Processor.java:1041)
at org.apache.coyote.AbstractProtocol$AbstractConnectionHandler.process(AbstractProtocol.java:603)
at org.apache.tomcat.util.net.JIoEndpoint$SocketProcessor.run(JIoEndpoint.java:312)
at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1145)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:615)
at java.lang.Thread.run(Thread.java:724)
Caused by: com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '?' at line 1
at sun.reflect.NativeConstructorAccessorImpl.newInstance0(Native Method)
at sun.reflect.NativeConstructorAccessorImpl.newInstance(NativeConstructorAccessorImpl.java:57)
at sun.reflect.DelegatingConstructorAccessorImpl.newInstance(DelegatingConstructorAccessorImpl.java:45)
at java.lang.reflect.Constructor.newInstance(Constructor.java:526)
at com.mysql.jdbc.Util.handleNewInstance(Util.java:411)
at com.mysql.jdbc.Util.getInstance(Util.java:386)
at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:1054)
at com.mysql.jdbc.MysqlIO.checkErrorPacket(MysqlIO.java:4190)
at com.mysql.jdbc.MysqlIO.checkErrorPacket(MysqlIO.java:4122)
at com.mysql.jdbc.MysqlIO.sendCommand(MysqlIO.java:2570)
at com.mysql.jdbc.MysqlIO.sqlQueryDirect(MysqlIO.java:2731)
at com.mysql.jdbc.ConnectionImpl.execSQL(ConnectionImpl.java:2812)
at com.mysql.jdbc.StatementImpl.executeUpdate(StatementImpl.java:1811)
at com.mysql.jdbc.StatementImpl.executeUpdate(StatementImpl.java:1725)
at org.apache.tomcat.dbcp.dbcp.DelegatingStatement.executeUpdate(DelegatingStatement.java:228)
at org.apache.tomcat.dbcp.dbcp.DelegatingStatement.executeUpdate(DelegatingStatement.java:228)
at com.affiliate.DAO.ClientApproveDAO.insertClient(ClientApproveDAO.java:23)
... 23 more
答案 0 :(得分：1)
嘿，我决定改变我的DAO：
public class ClientApproveDAO {
private DataSource dataSource;
public void setDataSource(DataSource dataSource) {
this.dataSource = dataSource;
}
Connection conn=null;
PreparedStatement statement=null;
ResultSet rs=null;
public void insertClient(int id) {
try{
conn=dataSource.createConnection();
PreparedStatement ps=conn.prepareStatement("insert into login(?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?)");
ps.executeUpdate("insert into login(id,FirstName,LastName,Gender,Category,Dateofbirth,Age,Address,Country,State,city,PinCode,EmailId,ContactNo,MobileNo)select * from register where id="+id+"");
ps.setInt(1, id);
}
catch (SQLException e) {
throw new RuntimeException(e);
} finally {
if (rs != null) try { rs.close(); } catch (SQLException ignore) {}
if (statement != null) try { statement.close(); } catch (SQLException ignore) {}
if (conn != null) try { conn.close(); } catch (SQLException ignore) {}
}
}
}
感谢所有人的帮助，这是我的最终代码，它的工作非常精细，并且符合要求。
答案 1 :(得分：0)
删除$和{client.id}
<a href="ClientApproveServlet?id= $ {client.id}">
↑
应该是
<a href="ClientApproveServlet?id=${client.id}">
这意味着我们克服了NumberFormat。现在，错误在insertClient（）方法中。
我可以从异常中说插入语句的语法错误。
只需添加条件where id=?，因为select * from register会获取所有记录。
PreparedStatement ps=conn.prepareStatement("insert into login(id,FirstName,
LastName,Gender,Category,Dateofbirth,Age,Address,Country,State,city,
PinCode,EmailId,ContactNo,MobileNo)select * from register where id=?");
ps.setInt(1, id);
答案 2 :(得分：0)
只需检查您的SQL查询。我的观点是你的列数不匹配。尽量给予
select id,firstname... etc
而不是
select * from tablename
答案 3 :(得分：0)
我认为你在$ {client.id}中有一个空格，只需删除它就可以了。
<a href="ClientApproveServlet?id= ${client.id}">
我认为你查询的格式也不太好试试：
String query ="insert into login(id,FirstName,LastName,Gender,Category,Dateofbirth,Age,"
" Address,Country,State,city,PinCode,EmailId,ContactNo,MobileNo)"+
" select id,..,..,.. FROM register; ";
答案 4 :(得分：0)
您必须在以下位置更改插入代码：
insert into login (FirstName,LastName,Gender,Category,
Dateofbirth,Age,Address,Country,State,city,PinCode,
EmailId,ContactNo,MobileNo)values(?,?,?,?,?,?,?,?,?,?);
您应该像以下一样检查此用户模型：
private int userid;
public int getUserid()
{
return userid;
}
public void setUserid(int userid)
{
this.userid=userid;
}
您应该为插入类型更改此类型：
public void addUser(User user)
{
try
{
PreparedStatement preparedStatement=connection.prepareStatement("insert into users (firstname,lastname,dob,email)values(?,?,?,?)");
preparedStatement.setString(1, user.getFirstName());
preparedStatement.setString(2, user.getLastName());
preparedStatement.setDate(3, new java.sql.Date(user.getDob().getTime()));
preparedStatement.setString(4, user.getEmail());
preparedStatement.executeUpdate();
}
catch(SQLException e)
{
e.printStackTrace();
}
}
答案 5 :(得分：0)
好的，你可以试试这个
执行此行之前
public void insertClient(int id) {
try{
conn=dataSource.createConnection();
PreparedStatement ps=conn.prepareStatement("insert into login(id,FirstName,LastName,Gender,Category,Dateofbirth,Age,Address,Country,State,city,PinCode,EmailId,ContactNo,MobileNo)select * from register");
ps.setInt(1, id);
ps.executeUpdate();
}
catch (SQLException e) {
throw new RuntimeException(e);
将此添加到其中
String x=null;
Resultset rs=conn.prepareStatement("select id from old_table where FirstName='request.getParameter('firstname')'and LastName='request.getParameter('lastname')'.......and so on.
while(rs.next){
x=rs.getString('id');
}
然后使用此x并将其传递给insert语句..
答案 6 :(得分：0)
最简单的方法是将它们放入会话或cookie中，如果只有一个或两个值。
session.setAttribute("id",id);
cookie.setAttribute("name",name);