javascript - 对象时间(2014,2,17 11 :02:08, 2014-03-17 11,02,08)没有方法'getTimezoneOffset

Question

正在处理从 phpmyadmin 获取数据的谷歌图表。我在时间/日期格式方面遇到问题:Object Time(2014, 2, 17 11:02:08, 2014-03-17 11, 02, 08) 没有方法 'getTimezoneOffset.

我的数据库:

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' => 'datetime'),
    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['Time'];
    $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];

    $temp = array();
    $temp[] = array('v' => "Time($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/JavaScript 代码:

   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);


      }

Answer 1

这是在做什么...

$dateString = $r['Time'];
$dateArray = explode('-', $dateString);

...正在占用您的时间:

2014-03-17 11:02:08

...进入:

'2014'
'03'
'17 11:02:08'

..其中最后一项不是您想要的 $day。

同样，

$timeArray = explode(':', $timeString);

...正在分块为:

'2014-03-17 11'
'02'
'08'

...第一个项目在几个小时内都不是您想要的。

但是，更严重的是，我不确定是否:

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

...只会构建一个字符串，将被 Google's DataTable constructor 接受，但看起来并非如此。

虽然您有正确的想法来构建 JSON 对象以获得对 JavaScript 有用的格式，但遗憾的是 JSON 不支持日期，因此我相信一旦您在 PHP 端获取了日期和小时，您将需要迭代通过 JavaScript 端的表格将这些字符串转换为 Date() 对象，而不是您构建的 Time(...) 字符串。