Javascript 使用ajax如何将参数从一个jsp页面发送到另一个servlet
我正在使用azax动态更改jsp页面,我希望将该jsp页面中的数据发送到servlet。jsp代码:Javascript 使用ajax如何将参数从一个jsp页面发送到另一个servlet,javascript,jsp,servlets,Javascript,Jsp,Servlets,我正在使用azax动态更改jsp页面,我希望将该jsp页面中的数据发送到servlet。jsp代码: <input type="submit" oninput="loadXMLDoc(this.value)" value="ok" name="ok"> <div id="myDiv"> Insert Id:<input id="p1" type="text" name="edit1" value=""style="visibility:h
<input type="submit" oninput="loadXMLDoc(this.value)" value="ok" name="ok">
<div id="myDiv">
Insert Id:<input id="p1" type="text" name="edit1" value=""style="visibility:hidden" size="30"/>
</div>
function loadXMLDoc(str){
var xmlhttp;
if (window.XMLHttpRequest){// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}else{// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function(){
if (xmlhttp.readyState==4 && xmlhttp.status==200){
document.getElementById("myDiv").innerHTML=xmlhttp.responseText;
}
}
xmlhttp.open("POST","edit?q="+str,true);
xmlhttp.send();
}
插入Id:
函数loadXMLDoc(str){
var-xmlhttp;
if(window.XMLHttpRequest){//IE7+、Firefox、Chrome、Opera、Safari的代码
xmlhttp=新的XMLHttpRequest();
}else{//IE6、IE5的代码
xmlhttp=新的ActiveXObject(“Microsoft.xmlhttp”);
}
xmlhttp.onreadystatechange=函数(){
if(xmlhttp.readyState==4&&xmlhttp.status==200){
document.getElementById(“myDiv”).innerHTML=xmlhttp.responseText;
}
}
open(“POST”,“edit?q=“+str,true”);
xmlhttp.send();
}
servlet代码:
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
response.setContentType("text/html;charset=UTF-8");
PrintWriter out = response.getWriter(); Connection conn=null;
String url = "jdbc:mysql://localhost:3306/";
String dbName = "studentdatabase";
String driver = "com.mysql.jdbc.Driver";
String userName = "root";
String password = "1234";
String student=request.getParameter("str");
Statement stmt;out.println(student);
try {
Class.forName(driver).newInstance();
conn = DriverManager.getConnection(url+dbName,userName,password);
String query = "select name1,telephone,email,department from studentinfo where studentid='"+student+"";
stmt = conn.createStatement();
ResultSet rs = stmt.executeQuery(query);
while(rs.next()){
String s = rs.getObject(1).toString();
out.println("<p> " +s+ "</p>");
}
conn.close;
//System.out.println("Disconnected from database");
} catch (Exception e) {
e.printStackTrace();
}
}
protectedvoiddopost(HttpServletRequest请求,HttpServletResponse响应)抛出ServletException,IOException{
setContentType(“text/html;charset=UTF-8”);
PrintWriter out=response.getWriter();连接连接=null;
String url=“jdbc:mysql://localhost:3306/";
字符串dbName=“studentdatabase”;
String driver=“com.mysql.jdbc.driver”;
字符串userName=“root”;
字符串密码=“1234”;
字符串student=request.getParameter(“str”);
报表stmt;out.println(学生);
试一试{
Class.forName(driver.newInstance();
conn=DriverManager.getConnection(url+dbName、用户名、密码);
String query=“从studentinfo中选择名称1、电话、电子邮件、部门,其中studentid=”+student+”;
stmt=conn.createStatement();
ResultSet rs=stmt.executeQuery(查询);
while(rs.next()){
字符串s=rs.getObject(1.toString();
out.println(“”+s+“”);
}
康涅狄格州关闭;
//System.out.println(“从数据库断开”);
}捕获(例外e){
e、 printStackTrace();
}
}
字符串student显示为
null
,即使数据库中有studentid=student的值。由于使用HTTP-POST,您必须将参数放入send方法中
...
var params = "q="+str;
xmlhttp.open("POST", url, true);
xmlHttp.setRequestHeader("Content-type","application/x-www-form-urlencoded");
xmlHttp.setRequestHeader("Content-length", params.length);
xmlHttp.setRequestHeader("Connection", "close");
xmlHttp.send(params);
...
您需要在服务器端代码中进行一些调试(或者至少告诉我们aobut it)——您看到了什么?+s+能生产吗?如果您看到的是null,那么您有一行,但其中的值为null……我尝试过这个方法,但它不起作用。它显示为null。+s+。servlet代码应该是什么:String student=request.getParameter(“str”);可以吗?