Question

在我的 java 项目中，我正在使用 jsp 页面。这里我只传递一个参数，我想从数据库中检索多个（3）值。我知道如何只传递和获取单个参数。但我不知道如何通过传递单个参数来获取多个值。我的ajax代码如下：

var request;
function sendInfo1() {
    var v = document.Example.wayid.value;
    // var url = 'Showroutelink.jsp?val='+v+'&start='+v1+'&end='+v2;
    var url = 'Showroutelink.jsp?val='+v;

    if (window.XMLHttpRequest) {

        request = new XMLHttpRequest();
    } else if (window.ActiveXObject) {
        request = new ActiveXObject("Microsoft.XMLHTTP");
    }

    try {

        request.onreadystatechange = getInfo1;
        request.open("GET", url, true);
        request.send();

    } catch (e) {
        alert("Unable to connect to server");
    }
}

function getInfo1() {

    if (request.readyState == 4) {
          var val = request.responseText.split('|');



        document.getElementById("start").innerHTML = val[0];
        document.getElementById("end").innerHTML = val[1];
        document.getElementById("waypoints").innerHTML = val[2];
    }
}




This is My Setver side Java code

<%
    String R_Name=request.getParameter("val");
    int w_id=0;


            String wplist = "";
            // String R_Name = request.getParameter("way");

            try {

                String buffer = "<select name='state'>";
                Class.forName("com.mysql.jdbc.Driver");
                java.sql.Connection con = DriverManager.getConnection("jdbc:mysql://localhost/googlemap", "root",
                        "root");
                Statement st = con.createStatement();

                ResultSet rs = st.executeQuery("select start from route where Route_Name='" + R_Name + "'");

                while (rs.next()) {
                    String Startval = rs.getString("start");

                    buffer = buffer + "<option value='" + Startval + "'>" + Startval + "</option>";


                }
                buffer = buffer + "</select>";
                response.getWriter().println(buffer);

            } catch (Exception e) {

            }




             try {

                String buffer1 = "<select name='state'>";
                Class.forName("com.mysql.jdbc.Driver");
                java.sql.Connection con = DriverManager.getConnection("jdbc:mysql://localhost/googlemap", "root",
                        "root");
                Statement st = con.createStatement();

                ResultSet rs = st.executeQuery("select end from route where Route_Name='" + R_Name + "'");

                while (rs.next()) {

                    String Endval = rs.getString("end");

                    buffer1 = buffer1 + "<option value='" + Endval + "'>" + Endval + "</option>";

                }
                buffer1 = buffer1 + "</select>";
                response.getWriter().println(buffer1);
            } catch (Exception e) {

            }




            try {

                String buffer2 = "<select name='state'>";
                Class.forName("com.mysql.jdbc.Driver");
                java.sql.Connection con1 = DriverManager.getConnection("jdbc:mysql://localhost/googlemap", "root",
                        "root");
                Statement st1 = con1.createStatement();

                ResultSet rs1 = st1.executeQuery("select wplist from route where Route_Name='" + R_Name + "'");
                while (rs1.next()) {
                    wplist = rs1.getString("wplist");

                    String wplist1[] = wplist.split("/");
                    buffer2 = buffer2 + "<option value='" + wplist1 + "'>" + wplist1 + "</option>";

                }
                buffer2 = buffer2 + "</select>";
                response.getWriter().println(buffer2);
            } catch (Exception e) {

            } 


    %>

Answer 1

如果我的问题是正确的，那么您就是在询问如何在请求中添加多个查询参数。

查询参数结构如下：

Showroutelink.jsp?key_1=value_1&key_2=value_2&key_n=value_n

所以你将遵循这种模式。

如果您询问如何从服务器调用返回多个值，请尝试使用JSON。

Answer 2

如果您的ajax调用返回的数据是JSON字符串，即（responseText = '{ "name":"John", "age":30, "city":"New York"}'），那么您可以使用以下内容：

function getInfo1() {

    if (request.readyState == 4) {
        var val = request.responseText;
        var obj = JSON.parse(request.responseText);

        document.getElementById("start").innerHTML = obj.start;

        document.getElementById("end").innerHTML = obj.end;
        document.getElementById("waypoints").innerHTML = obj.waypoints;
    }
}

如果您能提供ajax调用返回的响应，我可以更好地回答您的问题。

Answer 3

我没有完全理解这个问题，因为你还没有分享你的java代码。但在JS中你总是可以创建一个数组并将其分配给url。 var v = [para1，para2，para3];

我认为这应该有效。创建一个数组。

var v= [];

然后添加所有参数

v.push(document.Example.wayid1.value);
v.push(document.Example.wayid2.value);
v.push(document.Example.wayid3.value);

然后像这样添加到URL。

var url = 'Showroutelink.jsp?val=' + v.toString();

在java端你可以得到。

String[] routeNames=request.getParameter("val");

\\in case error occurs use 
String[] routeNames=request.getParameter("val").split(",");

还有一件事，不是使用带有“=”的where子句，而是使用“in”，只是循环遍历结果集。

如何使用ajax传递单个参数并获取多个值？

3 个答案: