[英]Php/ajax return JSON in same form as javascript array
我正在使用Charts.js插件来创建图表。 它使用javascript数组data
生成图表。
var data = {
labels: ["January", "February", "March", "April", "May", "June", "July"],
datasets: [
{
label: "My First dataset",
fillColor: "rgba(220,220,220,0.2)",
strokeColor: "rgba(220,220,220,1)",
pointColor: "rgba(220,220,220,1)",
pointStrokeColor: "#fff",
pointHighlightFill: "#fff",
pointHighlightStroke: "rgba(220,220,220,1)",
data: [65, 59, 80, 81, 56, 55, 40]
},
{
label: "My Second dataset",
fillColor: "rgba(151,187,205,0.2)",
strokeColor: "rgba(151,187,205,1)",
pointColor: "rgba(151,187,205,1)",
pointStrokeColor: "#fff",
pointHighlightFill: "#fff",
pointHighlightStroke: "rgba(151,187,205,1)",
data: [28, 48, 40, 19, 86, 27, 90]
}
]
};
我想使用ajax返回不同时间范围内的结果。 我如何用php生成响应,以便当我以ajax成功读取它时,可以以与使用var data
生成初始图表相同的方式使用它。
$.ajax('/chart.php', {
type: 'POST',
data: {range: range},
dataType: 'json',
success: function(data) {
console.log(data);
}
});
您想要的是创建一个对象数组,然后使用json_encode
,它应该会产生所需的内容。 尝试这个:
$array = array();
$dataset = new stdClass;
$dataset->label = "My first dataset";
// repeat for each field
$array[] = $dataset;
echo json_encode($array);
我们可以做的是缩短时间,而不是每次我们可以使用数组并将其类型转换为这样的对象时都声明一个新的数据集:
$datasets = array();
$datasets[] = (object) array(
'label' => "My First dataset",
'fillColor' => "rgba(220,220,220,0.2)",
'strokeColor' => "rgba(220,220,220,1)",
'pointColor' => "rgba(220,220,220,1)",
'pointStrokeColor' => "#fff",
'pointHighlightFill' => "#fff",
'pointHighlightStroke' => "rgba(220,220,220,1)",
'data' => [65, 59, 80, 81, 56, 55, 40]
);
查看此eval.in中的示例。
更新
对于完整的结构,您需要:
$data = (object) [
'labels' => ["January", "February", "March", "April", "May", "June", "July"],
datasets => []
];
$data->datasets[] = (object) [
'label' => "My First dataset",
'fillColor' => "rgba(220,220,220,0.2)",
'strokeColor' => "rgba(220,220,220,1)",
'pointColor' => "rgba(220,220,220,1)",
'pointStrokeColor' => "#fff",
'pointHighlightFill' => "#fff",
'pointHighlightStroke' => "rgba(220,220,220,1)",
'data' => [65, 59, 80, 81, 56, 55, 40]
];
echo json_encode($data);
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