用于创建Jqwidgets动态列网格的JSON格式

时间:2016-01-31 14:04:16

标签: php json jqwidget

我无法获得用于创建动态列网格的有效JSON格式。 这是我有的: PHP

$query="select * from mytable";
$rs = mysql_query($query);
$rows = array();
while($r = mysql_fetch_object($rs)) {
    $rows[] = $r;
}
echo json_encode($rows);

//Results:
[{"EmpID":"12345","Name":"James","Account":"New Account1","Group1":"Library","Food_Bill_Date":null,"Food_Bill_Amount":null,"Food_Bill_Number":null},
{"EmpID":"12346","Name":"David","Account":"New Account2","Group2":"Lab","Food_Bill_Date":null,"Food_Bill_Amount":null,"Food_Bill_Number":null},
{"EmpID":"12347","Name":"Ramsy","Account":"New Account2","Group2":"Lab","Food_Bill_Date":null,"Food_Bill_Amount":null,"Food_Bill_Number":null},
............
.............
............
]

Jqwidgets Dynamic Column

需要从结果数组中分隔行和列。

中需要
[
    {
        "columns": [
            {
                "text": "SampleK",
                "datafield": "sample1",
                "width": "30"
            },
            {
                "text": "ID",
                "datafield": "id",
                "width": "30"
            }
        ]
    }
    {
        "rows": [
            {
                "id": "1",
                "name": "Hot Chocolate",
                "type": "Chocolate Beverage",
                "calories": "370",
                "totalfat": "16g",
                "protein": "14g"
            },
            {
                "id": 2,
                "name": "Peppermint Hot Chocolate",
                "type": "Chocolate Beverage",
                "calories": "440",
                "totalfat": "16g",
                "protein": "13g"
            },
        ]
    }
]

请帮忙。

0 个答案:

没有答案