我有一个连接到Web服务到数据库的网站
我有这个不一致的错误,上面写着
{"消息":"收集已修改;枚举操作可能无法执行。"," StackTrace":" at System.ThrowHelper.ThrowInvalidOperationException(ExceptionResource resource)\ r \ n在System.Collections.Generic.List
1.Enumerator.MoveNextRare()\r\n at System.Collections.Generic.List
1.Enumerator.MoveNext()\ r \ n在System.Web.Script.Serialization.JavaScriptSerializer.SerializeEnumerable (IEnumerable可枚举,StringBuilder sb,Int32深度,Hashtable objectsInUse,SerializationFormat serializationFormat)\ r \ n在System.Web.Script.Serialization.JavaScriptSerializer.SerializeValueInternal(Object o,StringBuilder sb,Int32 depth,Hashtable objectsInUse,SerializationFormat serializationFormat,MemberInfo currentMember) )\\ n \ n \ n \ n \ n \ n \ n \ n \ n \ n \ n \ n \ nSystem.Web.Script.Serial。 .Serialize(Object obj,Str
我在c#web服务中有这些代码,此函数可能返回0行或更多行。
public class Liked
{
public bool liked;
}
static List<Liked> _get_liked = new List<Liked> { };
[ScriptMethod(ResponseFormat = ResponseFormat.Json)]
[WebMethod]
public List<Liked> get_liked(string userID,string postID)
{
DataTable table = null;
SqlCommand cmd = new SqlCommand();
cmd.CommandText = "Like_Retrieve";
cmd.Parameters.AddWithValue("@UserId", userID);
cmd.Parameters.AddWithValue("@PostId", postID);
cmd.CommandType = System.Data.CommandType.StoredProcedure;
table = this.dbcon.ExecuteDataTable(cmd);
_get_liked.Clear();
Liked _list = new Liked();
if (table.Rows.Count > 0)
{
_list.liked = true;
}
else
{
_list.liked = false;
}
_get_liked.Add(_list);
return _get_liked;
}
错误来源不一致。我在我的javascript中有这些
function checkLike(userId, postId) {
$.ajax({
type: "POST",
url: "../Main.asmx/get_liked",
data: "{'userID':'" + userId + "', 'postID':'" + postId + "'}",
contentType: "application/json; charset=utf-8",
dataType: "json",
success: function (response) {
var result = response.d;
$.each(result, function (index,data) {
var liked = data.liked;
if (liked == true) {
$("#" + postId).prev().parent("button").css("background", "#ccc");
}
else
{
$("#" + postId).prev().parent("button").css("background", "white");
}
});
},
error: function (xhr, ajaxOptions, thrownError) {
alert('error');
console.log(postId);
console.log(userId);
console.log(xhr.responseText);
console.log(thrownError);
}
});
}
我希望有人会帮助我。我现在已经敲了好几个小时