Question

我遇到了一些我现在已经工作了几天的问题，我真的无法解决这个问题。问题是我正在制作一个折线图，其中显示的是PH，氯和温度值，其中包含日期和时间。我不认为我已经将jsoncode格式化为正确，如下所示：https://developers.google.com/chart/interactive/docs/reference#dataparam

当我回显我的jsoncode时，它看起来像这样：

  {"cols":[{"label":"Time","type":"date"},{"label":"Date","type":"date"},     {"label":"PH","type":"number"},{"label":"temperature","type":"number"},   {"label":"Chlorine","type":"number"}],"rows":[{"c":[{"v":"Date(2014, 2, 17, 15, 03, 14)"},   {"v":"7.00"},{"v":"34.00"},{"v":"3.40"}]}]}

我的数据库看起来像这样：

enter image description here

代码：

<?php

    $con=mysql_connect("localhost","root","") or die("Failed to connect with database!!!!");
    mysql_select_db("chart", $con);

    $sth = mysql_query("SELECT * FROM googlechart");

    $rows = array();
    //flag is not needed
    $flag = true;
    $table = array();

    $table['cols'] = array(

    array('label' => 'Time', 'type' => 'date'),
    array('label' => 'Date', 'type' => 'date'),
    array('label' => 'PH',      'type' => 'number'),
    array('label' => 'temperature','type' => 'number'), 
    array('label' => 'Chlorine','type' => 'number'),
    );

    $rows = array();

    while($r = mysql_fetch_assoc($sth)) {

    // assumes dates are in the format "yyyy-MM-dd"
    $dateString = $r['Date'];
    $dateArray = explode('-', $dateString);
    $year = $dateArray[0];
    $month = $dateArray[1] - 1; // subtract 1 to convert to javascript's 0-indexed months
    $day = $dateArray[2];

    // assumes time is in the format "hh:mm:ss"
    $timeString = $r['Time'];
    $timeArray = explode(':', $timeString);
    $hours = $timeArray[0];
    $minutes = $timeArray[1];
    $seconds = $timeArray[2];
    echo $dateString."<br>";
    echo $timeString."<br>";
    $temp = array();
    $temp[] = array('v' => "Date($year, $month, $day, $hours, $minutes, $seconds)"); 
    $temp[] = array('v' => (string) $r['PH']);
    $temp[] = array('v' => (string) $r['temperature']);
    $temp[] = array('v' => (string) $r['Chlorine']);

    $rows[] = array('c' => $temp);

    }

    $table['rows'] = $rows;
    $jsonTable = json_encode($table);
     echo $jsonTable;   

?>


<html>
  <head>

    <script type="text/javascript" src="https://www.google.com/jsapi"></script>
    <script type="text/javascript">
      google.load("visualization", "1", {packages:["corechart"]});
      google.setOnLoadCallback(drawChart);

      function drawChart() {
        var data = new google.visualization.DataTable(<?=$jsonTable?>);



        var options = {
        /*width: 900, height: 900, */
          title: 'Visualization',
          curveType: 'function', 
           legend: { position: 'bottom' },
           pointSize: 12,
        vAxis: {title: "Values", titleTextStyle: {italic: false}},
        hAxis: {title: "Time", titleTextStyle: {italic: false}},
        explorer: { 
                actions: ['dragToZoom', 'rightClickToReset'], 
                axis: 'vertical'
            }


        };

        var chart = new google.visualization.LineChart(document.getElementById('chart_div'));
        chart.draw(data, options);


      }
    </script>

Answer 1

问题是因为Google Chart API想要一个JavaScript Date对象。你给图表的是一个字符串对象，尽管它的值似乎是Date（）。

您需要在没有引号的情况下分配该属性，因为它需要Date对象，所以它是这样的：

    "v": new Date("2014-02-17 15:03:14")

Answer 2

$temp[] = array('v' => "Date($year, $month, $day, $hours, $minutes, $seconds)");

如下所示：

$temp[] = array('v' => 'Date('.date('Y',strtotime($r['Date'])).','.(date('n',strtotime($r['Date'])) - 1).','.date('d',strtotime($r['Date'])).','.date('H',strtotime($r['Time'])).','.date('i',strtotime($r['Time'])).','.date('s',strtotime($r['Time'])).')');

Google图表时间/日期格式问题

2 个答案: