我的PHP側生成服務器上一個JSON對象,加上迭代它在客戶端
$i = 1;
while ($row = mysql_fetch_array($query)) {
$productarray[$i][] = $row['product_id'];
$productarray[$i][] = $row['product_model'];
$productarray[$i][] = $row['product_type'];
$productarray[$i][] = $row['product_return'];
$i++;
}
$jsonstring = json_encode($productarray);
這就是即時得到這是一個有效的JSON字符串
{
"1":["1","HFJ5G1.5","plat","graviteits"],
"2":["2","HHJ5S2.5","holle plunjer","veer"],
"3":["3","HTJ5S7.5","inbouw","veer"]
}
我無法通過這個有人迭代請告訴我如何迭代此json
從json站點我認爲這是格式,如果是的話我怎麼能改變上述json字符串到這種格式?
{
"1":[{"1","HFJ5G1.5","plat","graviteits"}],
"2":[{"2","HHJ5S2.5","holle plunjer","veer"}],
"3":[{"3","HTJ5S7.5","inbouw","veer"}]
}
var j = 1; (var i in o){ alert(o [i] [j]);/*它需要提醒每一個值的權利*/ j ++ } – coolguy 2012-02-16 08:46:51
但即時通訊獲取警報[,O,B,J,E,C,T,O,B,J ......等奇怪:( – coolguy 2012-02-16 08:48:02
我修改了我的答案。 – 2012-02-16 09:03:53