将json数据获取到Highcharts散点图中


Get json data into Highcharts scatter plot

我无法让Highcharts绘制2个系列的散点图。事实上,这张图根本没有显示出来。请帮我确定我做错了什么。我找不到一个散点图的例子,我对此很陌生。我从一个php文件中得到了json数据,它看起来像:

[[65,44],[66,37],[67,42],[68,55],[65,50],[65,41],[65,41],[68,41],[69,42],[70,47],[69,55],[67,45],[67,49],[67,53],[67,49],[68,51],[68,55],[68,57],[70,53],[69,66],[68,54],[69,52],[68,48]][[75,36],[72,42],[75,44],[69,56],[72,40],[73,37],[77,34],[77,40],[74,50],[77,45],[77,43],[75,47],[73,52],[73,50],[75,44],[72,54],[71,57],[72,57],[74,55],[74,54],[75,47],[78,45],[75,43]]

这应该是(x,y)格式的两个系列。我想把这些放在HighCharts的散点图上。我的HighCharts代码是:

  <script type="text/javascript">
$(function () {
    var chart;
    $(document).ready(function() {
        $.getJSON("comfortgb1b.php", function(json) {
               chart = new Highcharts.Chart({
                    chart: {
                    renderTo: 'container4',
                    type: 'scatter',
                    marginRight: 175,
                    marginBottom: 50
                },
                title: {
                    text: 'Comfort Level',
                    x: -20 //center
                },
                subtitle: {
                    text: '',
                    x: -20
                },
                xAxis: {
                    title: {
                        enabled: true,
                        text: 'Temp (F)'
                    },
                    min: 60,
                    max: 85,
                    startOnTick: true,
                    endOnTick: true,
                    showLastLabel: true
                },
                yAxis:  {
                    title: {
                        text: 'Humidity (%RH)'
                    },                  
                    min: 30,
                    max: 100
                },                
                plotOptions: {
                scatter: {
                    marker: {
                        radius: 5,
                        states: {
                            hover: {
                                enabled: true,
                                lineColor: 'rgb(100,100,100)'
                            }
                        }
                    },
                    states: {
                        hover: {
                            marker: {
                                enabled: false
                            }
                        }
                    },
                     tooltip: {
                        headerFormat: '<b>{series.name}</b><br>',
                        pointFormat: '{point.x} F, {point.y} %RH'
                    }
                },
                 series: [{
                    name: 'Night',
                    data: json(1)
                               }, {
                                    name: 'Night',
                    data: json(2)
                     });
        });
    });
});
        </script>

提前感谢!

下面是创建json数据的php文件。如何用逗号分隔这些数组?

  $result1 = mysql_query("SELECT round(AVG(d_internal_duct_return),0) AS 'avg_return', round(AVG(d_evap_pre_humidity),0) AS 'avg_hum' FROM pheom.pheom_gb WHERE timestamp between subdate(curdate(), interval 2 month) and curdate()  AND HOUR(Timestamp) NOT BETWEEN 9 AND 22 GROUP BY DAY(Timestamp) ORDER BY Timestamp");
$ret1 = array();
while($item = mysql_fetch_array($result1)) {
    $avg_return1 = $item['avg_return'];
    $avg_hum1 = $item['avg_hum'];
    $ret1[] = array($avg_return1,$avg_hum1);
}
$result2 = mysql_query("SELECT round(AVG(d_internal_duct_return),0) AS 'avg_return', round(AVG(d_evap_pre_humidity),0) AS 'avg_hum' FROM pheom.pheom_gb WHERE timestamp between subdate(curdate(), interval 2 month) and curdate()  AND HOUR(Timestamp) BETWEEN 9 AND 22 GROUP BY DAY(Timestamp) ORDER BY Timestamp");
$ret2 = array();
while($item = mysql_fetch_array($result2)) {
    $avg_return2 = $item['avg_return'];
    $avg_hum2 = $item['avg_hum'];
    $ret2[] = array($avg_return2,$avg_hum2);
}
echo json_encode($ret1, JSON_NUMERIC_CHECK);
echo json_encode($ret2, JSON_NUMERIC_CHECK);

对此不确定,但乍一看,我认为从php文件返回的数组需要在其外部添加一个方括号,才能被解析为正确的json。目前是:

[[65,44],[66,37]..][[75,36],[72,42]..]

据我所知,这只是两个数组。您想要的是将这些数组封装在一个数组中。尝试将其更改为:

[[[65,44],[66,37]..],[[75,36],[72,42]..]]

也就是说,在外部添加一个额外的方括号,并使用逗号分隔两个数组。

此外,此处:

series: [{
        name: 'Night',
        data: json(1)
    }, {
        name: 'Night',
        data: json(2)
});

json(1)和json(2)被解释为函数调用。您应该使用:

series: [{
        name: 'Night',
        data: json[0]
    }, {
        name: 'Night',
        data: json[1]
});

编辑----根据OP编辑进行编辑

同样根据要求,为了添加逗号和正确的格式,可以在最后两行更改php文件,如下所示:

echo "[".json_encode($ret1, JSON_NUMERIC_CHECK).",";
echo json_encode($ret2, JSON_NUMERIC_CHECK)."]";

您遇到了一些问题。看看这把小提琴:http://jsfiddle.net/nbwN9/1/

首先,我认为json的格式不适合2系列图。您有两个数据数组,它们只是相互对接,而不是在另一个数组中。链接的fiddle纠正了这一点(并将其填充到var中,因为getJSON()调用将在jsFiddle中失败)。每个点都是(x,y)坐标的数组。每个系列数据都是一组点。json需要是series.data数组的数组。因此,我们正在深入研究嵌套数组3。

其次,您似乎有一组格式错误的图表选项。最值得注意的是,您的系列节点(带有名称和数据)本应位于plotOptions节点的外部,但却位于该节点的内部。而且该系列节点没有正确终止。

第三,一旦您正确格式化了json数据和图表选项,访问json数组的操作如下:

    series: [{
        name: 'Night',
        data: json[0]
    }, {
        name: 'Day',
        data: json[1]
    }]

数组是基于0的(因此第一条记录将用0索引,第二条记录用1索引,等等),并且使用括号[]而不是括号()进行访问

对不起,我把你的一个系列改名为"Day",这样我就可以在图表中看到区别了。

至于PHP脚本。。。试试这个:

  $result1 = mysql_query("SELECT round(AVG(d_internal_duct_return),0) AS 'avg_return', round(AVG(d_evap_pre_humidity),0) AS 'avg_hum' FROM pheom.pheom_gb WHERE timestamp between subdate(curdate(), interval 2 month) and curdate()  AND HOUR(Timestamp) NOT BETWEEN 9 AND 22 GROUP BY DAY(Timestamp) ORDER BY Timestamp");
  $ret1 = array();
  while($item = mysql_fetch_array($result1)) {
      $avg_return1 = $item['avg_return'];
      $avg_hum1 = $item['avg_hum'];
      $ret1[] = array($avg_return1,$avg_hum1);
  }
  $result2 = mysql_query("SELECT round(AVG(d_internal_duct_return),0) AS 'avg_return', round(AVG(d_evap_pre_humidity),0) AS 'avg_hum' FROM pheom.pheom_gb WHERE timestamp between subdate(curdate(), interval 2 month) and curdate()  AND HOUR(Timestamp) BETWEEN 9 AND 22 GROUP BY DAY(Timestamp) ORDER BY Timestamp");
  $ret2 = array();
  while($item = mysql_fetch_array($result2)) {
      $avg_return2 = $item['avg_return'];
      $avg_hum2 = $item['avg_hum'];
      $ret2[] = array($avg_return2,$avg_hum2);
  }
  echo json_encode(array($ret1,$ret2), JSON_NUMERIC_CHECK);