我正在尝试从servlet页面获取响应并在成功时显示警报。但它总是向我显示错误。我无法弄清楚。
我的ajax代码:
$(document).ready(function() {
$("#srch").click(function() {
var txt1 = $("#store-qsearch").val();
alert(txt1)
$.ajax({
url : 'http://localhost:8080/searchengine/SearchDataServlet',
data : 'val='+txt1,
type : 'GET',
success : function(response) {
alert("Success");
// create an empty div in your page with some id
},
error: function(){ alert("error");
}
});
});
});
我的servlet代码:
public class SerachDataServlet extends HttpServlet {
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
// TODO Auto-generated method stub
String searchkey = request.getParameter("val").toString();
SearchInput searchinput = new SearchInput();
searchinput.searchkeys = searchkey;
System.out.println(searchkey);
SearchParser searchparser = new SearchParser();
searchparser.searchData(searchkey);
PrintWriter output = response.getWriter();
output.println("successs");
}
}
最佳答案
将此行data : 'val='+txt1,
更改为data: { val: txt1},
参见example