Question

我正在尝试使用从JSON设置的id和value属性创建输入。我有一个获取JSON的ajax调用，返回的数据很好，对于JSON中的每个对象，我想创建一个带有ID和JSON值的按钮。

Ajax电话：

$.ajax({
            type: "GET",
            url: '@Url.Action("GetSubjects", "CommAPI")',
            dataType: "json",
            success: function (data) {
                $.each(data, function (index, element) {
                    $('#subjects').append('<input type ="button" id=" ' + element.Id + ' " value=' + element.Title + '  class=k-button  />');
                });
            },

        });

然而，JSON中的对象还有5个附加属性，而不仅仅是Id和Title。将调试器放置在创建输入的行时，Id和Title未定义。

如何从此JSON创建这些输入？

从控制台复制的返回JSON：

14:41:57.928 {"Data":[{"Id":1,"IdCSite":1,"IdDEvent":1,"Title":"Test","CaseName":null,"Description":"dsadasdasda","InsertedDate":"/Date(-62135596800000)/","SelectedUsers":null,"ViewSelectedUsers":null,"IsCurrentUserIncluded":false},{"Id":2,"IdCSite":1,"IdDEvent":1,"Title":"Test2","CaseName":null,"Description":"sdadasdas","InsertedDate":"/Date(-62135596800000)/","SelectedUsers":null,"ViewSelectedUsers":null,"IsCurrentUserIncluded":false},{"Id":3,"IdCSite":1,"IdDEvent":1,"Title":"test 3","CaseName":null,"Description":"sdadasdasda","InsertedDate":"/Date(-62135596800000)/","SelectedUsers":null,"ViewSelectedUsers":null,"IsCurrentUserIncluded":false}],"Total":3,"AggregateResults":null,"Errors":null}1 messageboard:128:25

Answer 1

根据您的JSON，您需要使用Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''id', 'name', 'email', 'website', `comment`, `timestamp`, `articleid`) VALUE' at line 1对象的<? if( $_POST ) { $con = mysql_connect("localhost","iluxcoke_myuser","qwert2012"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("iluxcoke_inmoti6_mysite", $con); $users_name = $_POST['name']; $users_email = $_POST['email']; $users_website = $_POST['website']; $users_comment = $_POST['comment']; $users_name = mysql_real_escape_string($users_name); $users_email = mysql_real_escape_string($users_email); $users_website = mysql_real_escape_string($users_website); $users_comment = mysql_real_escape_string($users_comment); $sql = "SELECT * FROM `comments` LIMIT 0, 30 "; $sql= "INSERT INTO `comments`('id', 'name', 'email', 'website', `comment`, `timestamp`, `articleid`) VALUES ( id, '$users_name', '$users_email', '$users_website', '$users_comment', 'CURRENT_TIMESTAMP ');"; if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } //mysql_query($sql) echo "<h2>Thank you for your Comment!</h2>"; mysql_close($con); } ?>属性。因此将Data更改为data

$.each(data,

Answer 2

您可以直接使用jQuery属性来实现：

A1 =[1     0     0     1     1     0     0     0     0     0     0     0     0     0;...
     1     1     0     0     0     1     0     0     0     0     0     0     0     0;...
     1     1     1     0     0     0     1     0     0     0     0     0     0     0;...
     0     1     1     1     0     0     0     1     0     0     0     0     0     0;...
     0     0     1     1     0     0     0     0     1     0     0     0     0     0;...
     1     1     0     1     1     0     0     1     0     1     0     0     0     0;...
     1     0     1     1     0     1     0     0     1     0     1     0     0     0;...
     1     1     1     0     0     0     1     1     1     0     0     1     0     0;...
     0     1     1     1     1     1     1     0     0     0     0     0     1     0;...
     0     0     0     0     1     1     1     1     1     0     0     0     0     1;...
     0     1     1     1     1     0     1     1     1     0     1     1     1     0;...
     0     0     0     1     0     0     0     1     0     0     0     0     0     0;...
     0     0     1     0     0     0     0     1     0     0     0     0     0     0;...
     1     1     1     1     0     0     0     0     0     0     0     0     0     0;...
     0     0     1     1     0     0     0     0     1     1     0     0     0     0;...
     0     0     1     0     0     0     0     0     0     0     0     0     0     1 ];


D1=[0; 0; 0;  0 ; 0;   0 ;  0;   0 ;  0 ;  0 ;  103 ;  198 ;  105 ;  115;   175  ;  14];





for ii=1:14

   % Find ii-th pivot index between row ii and last row
   PivIndex=find(A1(ii:end,ii),1)+ii-1;

   % Switch ii-th row with Pivot row
   A1([ii PivIndex],:)=A1([PivIndex ii],:);
   D1([ii PivIndex])=D1([PivIndex ii]);

   % Find all rows other than row ii containing a 1 in column ii
   RowIndexes=find(A1(:,ii));
   RowIndexes(RowIndexes==ii)==[];

   % Add row ii to all rows in RowIndexes, do the same in D
   A1(RowIndexes,:)=mod(A1(RowIndexes,:)+repmat(A1(ii,:),numel(RowIndexes),1),2);

%% Problem with my answer was here, as the sum in GF(256) doesn t work like that. (GF(256),+) behaves like ((Z/2Z)^8,+)... See prequisite for summing in GF(256)

   % D1(RowIndexes)=mod(D1(RowIndexes)+repmat(D1(ii),numel(RowIndexes),1),256);

   D1(RowIndexes)=SumInGF256(D1(RowIndexes),repmat(D1(ii),numel(RowIndexes),1));

end

% Now A1 is diagonal, with both last rows being zero. Problem is D
% has to be 0 aswell on the 2 last positions to get 0=0..
% Check if D(15:16)==[0;0] if not the system has no solution

if isequal(D1(15:16),[0;0])

A2=A1(1:14,:);
C=D1(1:14)


else

    disp('No solution')

end

如何命名您的变量，例如＆＃39; Id＆＃39;是Javascript的关键词。如果它是一个用户，你应该像idu一样命名。

享受：）

JQuery - 使用JSON中的属性创建输入

2 个答案: