从Mysql修改JSON&；PHP与人力车图形兼容

我想通过PHP显示MySQL的最后一个数据

它必须看起来像这样：

[
    {
        "data": [ { "x": 1350995278, "y": 26.12 }, 
                  { "x": 1350995276, "y": 26.19 }, 
                  { "x": 1350995273, "y": 26.12 }, 
                  { "x": 1350995271, "y": 26.19 }, 
                  { "x": 1350995268, "y": 26.19 } ]
    }
]

但是我的PHP和

json\u encode

函数返回的结果有点不同。我该怎么修

PHP代码：

    while($row = mysql_fetch_assoc($select)) {
        $r[] = $row;
        $items[] = array('x' => $row['unixtime'], 'y' => $row['sensor']);
    }
    print json_encode($items);

PHP输出：

[{"x":"1350996886","y":"26.06"},
 {"x":"1350996884","y":"26.06"},
 {"x":"1350996881","y":"26.06"},
 {"x":"1350996879","y":"26.06"},
 {"x":"1350996876","y":"26.06"},
 {"x":"1350996874","y":"26.06"},
 {"x":"1350996871","y":"26.06"},
 {"x":"1350996869","y":"26.06"},
 {"x":"1350996866","y":"26.06"},
 {"x":"1350996864","y":"26.06"}]

---

在

x

：和

Y

之后我不需要

”

：我是

PHP

新手，不知道如何计算，请给出建议。

如果不想引用，请将值转换为数字：

$items[] = array('x' => (float) $row['unixtime'], 'y' => (float) $row['sensor']);