我有一個js文件,它將webmethod中的數據附加到gridview。我的問題是每當我再次點擊我的按鈕時,它會添加新的數據並在gridview中添加以前的數據。有沒有辦法刪除最後附加的數據並追加新的數據?另一個問題。在綁定gridview時,我需要在加載頁面時綁定一個datasoure,以便讓它顯示我想要追加的數據。如何刪除綁定的數據源在嘗試附加新數據源時綁定?很抱歉,很長很多問題。謝謝!如何刪除最後附加的數據
$.ajax({
url: "Default.aspx/getDetails",
data: Data,
type: "POST",
dataType: "json",
contentType: "application/json; charset=utf-8",
success: function (mydata) {
for (var i = 0; i < mydata.d.length; i++) {
$(".rpt").html("<table id=\"tbl\"><tr>" +
"<th style='background-color: #75a844' class='tbl_h'>Date Joined</th>" +
"<th style='background-color: #75a844' class='tbl_h'>Memberid</th>" +
"<th style='background-color: #75a844' class='tbl_h'>Member ID</th>" +
"<th style='background-color: #75a844' class='tbl_h'>User ID</th>" +
"<th style='background-color: #75a844' class='tbl_h'>Username</th>" +
"<th style='background-color: #75a844' class='tbl_h'>Password</th>" +
"<th style='background-color: #75a844' class='tbl_h'>Full Name</th>" +
"<th style='background-color: #75a844' class='tbl_h'>First Name</th>" +
"<th style='background-color: #75a844' class='tbl_h'>Last Name</th>" +
"<th style='background-color: #75a844' class='tbl_h'>Birth Date</th>" +
"<th style='background-color: #75a844' class='tbl_h'>Address</th>" +
"<th style='background-color: #75a844' class='tbl_h'>Location</th>" +
"<th style='background-color: #75a844' class='tbl_h'>Sponsor ID</th>" +
"<th style='background-color: #75a844' class='tbl_h'>Parent ID</th>" +
"<th style='background-color: #75a844' class='tbl_h'>Placement</th>" +
"</tr>");
for (var i = 0; i < mydata.d.length; i++) {
$("#tbl").append("<tr><td>" + mydata.d[i].datejoined +
"</td><td>" + mydata.d[i].memid +
"</td><td>" + mydata.d[i].memberid +
"</td><td>" + mydata.d[i].userid +
"</td><td>" + mydata.d[i].usrname +
"</td><td>" + mydata.d[i].pssword +
"</td><td>" + mydata.d[i].fullname +
"</td><td>" + mydata.d[i].fname +
"</td><td>" + mydata.d[i].mname +
"</td><td>" + mydata.d[i].lname +
"</td><td>" + mydata.d[i].bdate +
"</td><td>" + mydata.d[i].address +
"</td><td>" + mydata.d[i].location +
"</td><td>" + mydata.d[i].sponsorid +
"</td><td>" + mydata.d[i].parentid +
"</td><td>" + mydata.d[i].placement + "</td></tr>");
}
$("#tbl").append("</table>");
}
});
您是否嘗試過使用replaceWith? –
你能舉個例子嗎?我沒有遇到replaceWith。謝謝! – ljpv14
查看http://api.jquery.com/replaceWith/或http://api.jquery.com/html/ –