我正在使用PHP爲JQGrid生成JSON。我已經添加另一個屬性的PHP對象,我JSON編碼:將JSON字符串中的附加參數發送到jqGrid
$sql_array = preg_split('/LIMIT/', $sql);
$pass_sql = $sql_array[0];
$response->sql = $pass_sql;
<~SNIP~>
echo json_encode($response);
這解析客戶端上的精細和用JSON填充jqGrid的是這樣的:
{"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
我需要將該SQL參數文本從JSON回覆中拉出並將其隱藏在DIV中以備後用。這可能嗎?
這裏可能有一些設計問題......但除此之外,您可以在填充網格的處理程序中執行此操作。 – Orbit 2011-03-03 22:34:35