将JSON字符串中的附加参数发送到jqGrid

Question

我正在使用PHP为JQGrid生成JSON。我已经添加另一个属性的PHP对象，我JSON编码：

$sql_array = preg_split('/LIMIT/', $sql); 
$pass_sql = $sql_array[0]; 
$response->sql = $pass_sql; 
<~SNIP~> 
echo json_encode($response); 

这解析客户端上的精细和用JSON填充jqGrid的是这样的：

{"page":"1","total":28,"records":"685","sql":"SELECT * FROM fires ORDER BY id desc ","rows":[{"id":"3065","cell":["Southern","Lost Fire","National Forests in Mississippi","492","100","0000-00-00",null,null,null,null,null,null,null,null,null,null,null,"3065","2011-03-03 00:00:00"]},{"id":"3064","cell":["Southern","PineTree","East Central Area Dispatch Office","420","80","2011-03-02",null,null,null,null,null,null,null,null,null,null,null,"3064","2011-03-03 00:00:00"]},{"id":"3063","cell":["Southern","LILAC ROAD","Georgia Forestry Commission","100","100","2011-03-01",null,null,null,null,null,null,null,null,null //etc 

我需要将该SQL参数文本从JSON回复中拉出并将其隐藏在DIV中以备后用。这可能吗？

Answer 1

从服务器到jqGrid的发送附加信息的最简单方法是userdata（见this answer。如果你的JSON数据就会像

{ 
    "page":"1", 
    "total":28, 
    "records":"685", 
    "userdata":"SELECT * FROM fires ORDER BY id desc ", 
    "rows":[ 
     ... 
    ] 
} 

或

{ 
    "page":"1", 
    "total":28, 
    "records":"685", 
    "userdata": { 
     sql: "SELECT * FROM fires ORDER BY id desc " 
    }, 
    "rows":[ 
     ... 
    ] 
} 

附加信息会保存在jqGrid里面，你可以使用$("#grid_id").jqGrid('getGridParam','userData')

在的情况下要小心。在JSON数据中，它必须是userdata和getGridParam：'userData'。