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javaweb中ajax请求后台servlet(实例)

2024-07-13 10:08:29
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废话不多说,直接上代码

public class DZFP_jdbc extends HttpServlet{  private static final long serialVersionUID = 1L;  public static Connection conn;   public static ResultSet rs = null ;    public static PreparedStatement ps = null ;  private static String url = "jdbc:oracle:thin:@192.168.100.11:1111:CRM";  private static String name = "name";  private static String pwd = "pwd";  protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {    /*PreparedStatement ps;    ResultSet rs = null;*/    response.setCharacterEncoding("utf-8");    request.setCharacterEncoding("utf-8");    response.setHeader("content-type", "text/html;charset=UTF-8");    PrintWriter out = response.getWriter();    ***********    out.print("{/"errorno/":[{/"list/":error}]}");  }}$.ajax({   type: "post",   url: "DZFP_jdbc",   dataType: "text",     data : {      taxcode : taxcode,      mobilenum : mobilenum  },  timeout : 50000,    success: function (data) {   var jsonobjs = eval("(" + data + ")");  var list = jsonobjs.errorno[0].list;    },   error: function() {      alert("网络异常,请稍后重试");  } });<servlet>  <servlet-name>DZFP_jdbc</servlet-name>  <servlet-class>  weishijiestudio.hangxinwx.servlet.DZFP_jdbc  </servlet-class></servlet><servlet-mapping>  <servlet-name>DZFP_jdbc</servlet-name>  <url-pattern>/DZFP_jdbc</url-pattern></servlet-mapping>

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