jQuery $ .post（）JSON对象

我有一个JSON对象

{ "widgetSettings":[{"maxDisplay": 6, "maxPerRow": 2}], "widgets": [ {"wigetID": 1, "show": false, "weight": 0, "widgetTitle": "Widget 1", "widgetColor": "defualt"}, {"wigetID": 2, "show": false, "weight": 0, "widgetTitle": "Widget 2", "widgetColor": "defualt"}, {"wigetID": 3, "show": false, "weight": 0, "widgetTitle": "Widget 3", "widgetColor": "defualt"}, {"wigetID": 4, "show": false, "weight": 0, "widgetTitle": "Widget 4", "widgetColor": "defualt"}, {"wigetID": 5, "show": false, "weight": 0, "widgetTitle": "Widget 5", "widgetColor": "defualt"}, {"wigetID": 6, "show": false, "weight": 0, "widgetTitle": "Widget 6", "widgetColor": "defualt"}, {"wigetID": 7, "show": false, "weight": 0, "widgetTitle": "Widget 7", "widgetColor": "defualt"}, {"wigetID": 8, "show": false, "weight": 0, "widgetTitle": "Widget 8", "widgetColor": "defualt"}, {"wigetID": 9, "show": false, "weight": 0, "widgetTitle": "Widget 9", "widgetColor": "defualt"}, {"wigetID": 10, "show": false, "weight": 0, "widgetTitle": "Widget 10", "widgetColor": "defualt"}, {"wigetID": 11, "show": false, "weight": 0, "widgetTitle": "Widget 11", "widgetColor": "defualt"}, {"wigetID": 12, "show": false, "weight": 0, "widgetTitle": "Widget 12", "widgetColor": "defualt"}, {"wigetID": 13, "show": false, "weight": 0, "widgetTitle": "Widget 13", "widgetColor": "defualt"}, {"wigetID": 14, "show": false, "weight": 0, "widgetTitle": "Widget 14", "widgetColor": "defualt"}, {"wigetID": 15, "show": false, "weight": 0, "widgetTitle": "Widget 15", "widgetColor": "defualt"}, {"wigetID": 16, "show": false, "weight": 0, "widgetTitle": "Widget 16", "widgetColor": "defualt"} ]} 

我想用jQuery将它发布到服务器端脚本，这样我就可以将它保存在数据库中。 当我说保存时，我的意思是作为JSON对象。 然而，当你发布post/获得其他方式时，我发布的JSON格式是我发布的JSON格式，因此我可以保存在数据库中，看起来很迷茫，数据库会留下空值。 我可能做错的任何想法..inheritance人jQuery部分。

 $.post('ui-DashboardWidgetsPost.php', {"dashWidgets":dashboardJSON}, function(msg) { if(msg.error == "yes"){console.log('Error Found: '+ msg.errorMsg);} else { } }); 

编辑 PHP

Answer 1

如果要将JSON（作为字符串，而不是实际值）发送到DB，或许您应该将其视为一个？

 $.post('ui-DashboardWidgetsPost.php', { json: dashboardJSON }, function(msg) { msg=jQuery.parseJSON(msg); if (msg.error == "yes") { console.log('Error Found: ' + msg.errorMsg); } else { ... } }); 

Answer 2

您确定您的服务器正在正确解析它吗？ 它到目前为止的事实意味着问题出在你的PHP中。

您最好也确保数据正确地通过网络传输，您可以通过Chrome / Firebug的网络选项卡进行操作。 话虽这么说，我更喜欢使用外部数据包嗅探器，如Fiddler（或Mac上的HTTPScoop）。

Answer 3

你可以这样做：

1. 使用Javascript对象方法JSON.parse
2. 将其设置为某个POST值，我们会说“json”
3. 在服务器上读取并解码。 使用PHP，这就像json_decode（$ _ POST [‘json’]）。

因此，客户端上的代码可能是：

 $.post('ui-DashboardWidgetsPost.php', 'json=' + JSON.parse(dashboardJSON), function(msg) { if(msg.error == "yes"){console.log('Error Found: '+ msg.errorMsg);} else { } }); 

在PHP中：

 $jsonDecoded = json_decode($_POST['json']) 

Answer 4

使用json作为第四个参数。

 $.post('ui-DashboardWidgetsPost.php', dashboardJSON, function(msg) { if(msg.error == "yes"){console.log('Error Found: '+ msg.errorMsg);} else { } }, 'json');