我想用AJAX显示一个隐藏值,但是当我按下提交按钮并将我重定向到另一个页面时,它不会停留在同一页面上。我的脚本标签如下所示。脚本标签上方的任何内容都可以。我想知道是否有人可以帮我解决这个问题。
<script>
function loadXMLDoc() {
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;
}
}
var hiddenUserID = encodeURIComponent(document.getElementBuId("hiddenID").value);
xmlhttp.open("GET", "displayUserScore.jsp?hUserID="+hiddenUserID,true);
xmlhttp.send(null);
}
</script>
</head>
<body>
<%
String user1 = request.getParameter("compareUser1");
String user2 = request.getParameter("compareUser2");
out.println(user1 + " " + user2);
Class.forName("com.mysql.jdbc.Driver");
Connection conn = DriverManager
.getConnection("jdbc:mysql://localhost/survey?user=root&password=dso123");
String retrieveUserData = "SELECT u.userID, u.fullName, sum(uma.isCorrect) score, u.endTime FROM users u, usermcqanswer uma WHERE u.userID = uma.userID AND u.fullName = ?";
PreparedStatement pstmt = conn.prepareStatement(retrieveUserData);
pstmt.setString(1, user1);
ResultSet rs = pstmt.executeQuery();
%>
<table border=1>
<tr>
<td>User ID</td>
<td>Full name</td>
<td>Score</td>
<td>Other information</td>
</tr>
<%
while (rs.next()) {
out.println("<tr>");
out.println("<td>" + rs.getString("userID") + "</td>");
out.println("<td>" + rs.getString("fullName") + "</td>");
out.println("<td>" + rs.getString("score") + "</td>");
out.println("<td>");
%>
<form action="displayUserScore.jsp" method="get">
<input type="hidden" value=<%=rs.getString("userID")%> id="hiddenID">
<input type="submit" onclick="loadXMLDoc()" value="View Scenario Score">
</form>
<%
out.println("</td>");
}
%>
</table>
<div id="myDiv"></div>
</body>
答案 0 :(得分:0)
添加引号:
value='<%=rs.getString("userID")%>'
并将type="submit"
更改为type="button"
请考虑使用JSTL
。