jQuery的$。员额（）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来张贴到一个服务器端脚本，所以我可以将其保存在数据库中。当我说保存它时，我的意思是作为JSON对象。然而，当你发布帖子/获取其他发布的方式是JSON格式时，我发布的JSON使我可以保存在数据库中，看起来好像丢了，而数据库却留下了一个空值。任何想法，我可能做错了..继承人的jQuery部分。

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

编辑的PHP

<?php 
$validJSON = $_POST['dashWidgets']; 

mysql_connect("127.0.0.1", "", "") or die('{"error": "yes", "errormsg": "'.mysql_error().'"}'); 
mysql_select_db("xxxx") or die('{"error": "yes", "errormsg": "'.mysql_error().'"}'); 

$result = mysql_query("UPDATE dashboardPrefs SET widgetSettings='".$validJSON."' WHERE userID=100") 
or die('{"error": "yes", "errormsg": "'.mysql_error().'"}'); 

echo '{"error": "none"}'; 
?>

来源

2011-08-21 chris

你能告诉我们在服务器端的PHP代码，以及服务器接收的确切数据吗？ – Bojangles

'dashboardJSON'是一个**字符串**（它必须是）？也许你必须将该值赋给一个键：'{data：dashboardJSON}'，然后访问POST'data'值。 –

编辑过的帖子来显示PHP和@Robus，我注意到它刚刚编辑之前结束它:) – chris

如果你想（而不是实际值作为字符串）发送JSON到数据库，也许你应该把它作为一个？

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

来源

2011-08-21 09:38:09 Robus

你确定你的服务器正在解析它吗？事实证明这个问题在你的PHP中。

您最好也确保数据正确移动，您可以通过Chrome/Firebug的网络选项卡执行此操作。话虽如此，我更喜欢使用像Fiddler这样的外部数据包嗅探器（或Mac上的HTTPScoop）。

来源

2011-08-21 09:40:36

你可以做这样的事情：

使用JavaScript对象方法JSON.parse
设置一定的岗位价值，我们会说“JSON”
阅读并解码在服务器上。使用PHP，这会像json_decode（$ _ POST ['json']）。

因此，在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 
     { 
     } 
    });

而在PHP：

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

来源

2011-08-21 09:41:38 FreezeWarp

使用json作为第四参数。

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

来源

2011-08-21 09:43:24

jQuery的$。员额（）JSON对象

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