我在jsp中创建了一个表单,并尝试使用Spring JDBC将表单详细信息发送到数据库。我创建了一个JSP表单,一个servlet文件db_Servlet.java,用于从form和db_Service.java中获取数据,并将数据发送到数据库。无法使用Spring JDBC将jsp表单详细信息发送到数据库
当我使用tomcat运行项目时,它在创建db_Service类的对象时卡住了。 以下代码是db_Servlet.java的servlet doPost方法。如果你使用Spring,然后
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
System.out.println("entered the servlet");
response.setContentType("text/html");
PrintWriter out = response.getWriter();
String name = request.getParameter("name");
String ph = request.getParameter("phone");
long phone = Long.parseLong(ph);
String emailid = request.getParameter("emailid");
customer c= new customer(name,phone,emailid);
try {
System.out.println("entered the try block");
db_Service service = new db_Service();
int result = service.addtodb(c);
System.out.println(result);
String title = "Thank you";
String doctype = "<!doctype html public \"-//w3c//dtd html 4.0 " +
"transitional//en\">\n";
out.println(doctype +
"<html>\n" +
"<head><title>" + title + "</title></head>\n" +
"<body>" +
"Thank you for wasting your precious time"
+ " </body></html>"
);
}
finally {
out.close();
}
}
及以下是db_Service类
public class db_Service {
public void addtodb(customer c){
System.out.println("Entered the service");
ApplicationContext context =
new ClassPathXmlApplicationContext("Beans.xml");
JDBCtemplate customerJDBCtemplate =
(JDBCtemplate)context.getBean("JDBCtemplate");
customerJDBCtemplate.create(c.getName(),c.getPhone(),c.getEmailid());
}
}
注释“中输入服务”的评论后不显示“进入try块”
尝试捕获异常并打印出堆栈跟踪。 – Jens
我试图打印stacktrace,但它不打印任何东西 –
比使用调试器找出发生了什么 – Jens