下面小编就为大家带来一篇javaweb中ajax请求后台servlet(实例)。小编觉得挺不错的,现在就分享给大家,也给大家做个参考。一起跟随小编过来看看吧
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>
以上是javaweb中ajax如何请求后台servlet的示例代码分享的详细内容。更多信息请关注PHP中文网其他相关文章!