javaweb中ajax请求后台servlet(实例)
时间:2020-12-02 13:05:34|栏目:JAVA代码|点击: 次
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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栏 目:JAVA代码
本文标题:javaweb中ajax请求后台servlet(实例)
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