Question

怎么通过ajax的回调函数返回一个数据库表格 后台怎么写

Answer 1

//C#后台
[WebMethod]
public static DataSet GetServer(int PartID)
{
DataTable dt = GameBLL.SelectDataTableByID(CommandType.StoredProcedure, "SelectServerByPartID", SelectBy.PartID, PartID);
DataSet ds = new DataSet();
ds.Tables.Add(dt.Copy());
return ds;
}
//jquery

$.ajax({
//要用post方式
type: "Post",
//方法所在页面和方法名
url: "SendGoods.aspx/GetServer",
contentType: "application/json; charset=utf-8",
dataType: "json",
data: json_str,
success: function (data) {
var rows = data.d.tables[0].rows;
for (var rowIndex = 0; rowIndex < rows.length; ++rowIndex) {

$("<a>", {
id: "s" + rows[rowIndex]["GameServerID"],
class: 'gamea',
value: rows[rowIndex]["GameServerID"],
text: rows[rowIndex]["GameServerName"],
}).appendTo("#ServerDiv .info");
}
},
error: function (err) {
alert(err);
}
});