我已經在jQuery中創建了一個ajax調用我的服務器,我現在面臨的麻煩是我的響應是打印?即使將正確的整數值寫入輸出流。下面給出了Ajax函數。獲取?當提醒ajax響應或發佈到文本字段
$dntb.on('click', 'button', function(event) {
var i = $(this).closest('tr').index(); //have to get the row where the button is clicked
var sditmId = $("#sditm").val();
var sdhedId = $("#sdhed").val();
$.get('getstock', {
sditmId: sditmId,
sdhedId: sdhedId
}, function(response) {
alert(response);
var stk = ""+response;
$("#stk").val(stk);
});
});
該功能被稱爲下面
顯示在我的表中的問題按鈕單擊該服務器代碼如下
int stk = null;
switch (userPath) {
case "/getstock":
stk = opo.getStockData(request.getParameter("sditmId") request.getParameter("sdhedId")); //value to write into the output stream.
break;
case "/temp":
//er = opo.checkCatUniqueForEdit(request.getParameter("catName"), request.getParameter("catId"));
break;
}
System.out.println(stk); //Printing correctly
response.setContentType("text/html");
response.getWriter().write(stk);
代碼提供給獲得價值
public int getStockData(String sditm, String sdhed) {
int stk = 0;
try {
String query = "Select stk.Stk_instk from tbstk stk inner join tbsditm itm on itm.Sditm_prdid=stk.Stk_prdid where itm.Sditm_sdhed=" + sdhed + " and itm.Sditm_id=" + sditm;
Statement stmt = dcon.con.createStatement();
ResultSet rs = stmt.executeQuery(query);
if (rs.next()) {
stk = rs.getInt("Stk_instk");
}
} catch (SQLException ex) {
Logger.getLogger(Op_OrdConf.class.getName()).log(Level.SEVERE, null, ex);
}
return stk;
}
ajax調用成功發生,但是當我提醒我得到的響應,並且我正在服務器中正確獲取值時,但是在客戶端它是?請幫我解決這個
您正在返回內容類型html中的響應,在$ .get()中提及顯式數據類型。希望這會有所幫助。 – Mahendra 2015-02-23 07:17:57