Question

我正在尝试将多个系列绘制到flotchart中。我成功获得了1个标签，但我不得不用PHP获得更多标签并编码为json。

$connection = mysql_connect($server,$user,$password);
    $db = mysql_select_db($database,$connection);

    $query = "SELECT SC,SR FROM high";
    $result = mysql_query($query);        

    while($row = mysql_fetch_assoc($result))
    {   
        $int = $row['SR'];
        $join = intval($int);
        $int2 = $row['SC'];
        $join2 = intval($int2);
        $dataset1[] = array($join2,$join);
    }

    $final = json_encode($dataset1);
echo $final;

RESULT

[[1,3],[2,20],[3,30],[4,10],[5,4],[6,40],[7,67],[8,100],[9,5],[10,11]]

将JSON解析为Javascript

$.ajax({
    dataType:'json',  /*to avoid calling JSON.parse(data) in your callback function*/
    url: 'chart-data.php',
    success: function (data) {
        console.log(data);//as mentioned in comments
        //1.either call plot again 
        /*calling plot as seen in your code - start*/

        var d1 = data;/* JSON.parse(data) */
        $.plot($("#placeholder"), 
            [{
                label: "SCR",
                data: d1,
            },
            /*{
                label: "SCR",
                data: d1,
                },*/
            ],

FOKUS ON {label：“SCR”，数据：d1，}，

问题：如何将JSON数据解析为JS，如：

[ { label: "Foo", data: [ [10, 1], [17, -14], [30, 5] ] },
  { label: "Bar", data: [ [11, 13], [19, 11], [30, -7] ] }
]

我只是想使用数组：

$s= array('label'=> "aaaa",
    'data'=> $final);
print_r($s);

并获得结果

Array ( [label] => aaaa [data] => [[1,3],[2,20],[3,30],[4,10],[5,4],[6,40],[7,67],[8,100],[9,5],[10,11]] )

Answer 1

在使用PHP编码之前创建数组：

$dataset = array("label" => "FOO/BAR", "data" => array(/* Your `data` array */));

并将所有label/data值附加到dataset变量，最后对其进行编码和回显

如何用PHP获得更多标签

1 个答案: