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我創建了InformationServlet,只要我需要一些細節,我可以發送它我想要的(使用AJAX),它會返回給我信息。如何發送和捕獲參數從JavaScript(AJAX請求)到Servlet
我搜索如何做到這一點Ajax和根據: How to send parameter to a servlet using Ajax Call
我用:url: "InformationServlet?param=numberOfPlayers"
但在servlet的請求的屬性不包含我發這樣的參數我想是的不正確的練習:
你可以看到屬性大小爲零
的Servlet:
protected void processRequest(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
response.setContentType("text/html;charset=UTF-8");
PrintWriter out = response.getWriter();
try {
Gson gson = new Gson();
Engine engine = (Engine)getServletContext().getAttribute("engine");
String responseJson = "";
if(request.getAttribute("numberOfPlayers") != null)
{
String numberOfPlayers = "";
numberOfPlayers = gson.toJson(String.valueOf(engine.GetNumOfPlayers()));
responseJson = numberOfPlayers;
}
out.print(responseJson);
} finally {
out.close();
}
}
的JavaScript(AJAX請求):
function getNumberOfPlayersAndPrintSoldiers()
{
$.ajax({
url: "InformationServlet?param=numberOfPlayers",
timeout: 2000,
error: function() {
console.log("Failed to send ajax");
},
success: function(numberOfPlayers) {
var r = numberOfPlayers;
}
});
}
嘗試過,但屬性是STIL爲零 – E235 2014-09-27 10:18:49
大小我糾正我的答案。 – eitank 2014-09-27 10:32:18