【问题标题】:Google Charts datetime issue, not displaying JSON data nowGoogle Charts 日期时间问题,现在不显示 JSON 数据
【发布时间】:2015-06-25 21:30:49
【问题描述】:

基本上,当我尝试生成此 Google 图表时,我的 select 语句似乎不起作用。下面的 echo 语句不输出任何内容,就像没有进行选择一样。我已经盯着这个足够长的时间了,我知道错误就在那里,我只是在这一点上看不到它。 我最初遇到了一个问题,我只能将时间戳数据作为字符串处理,但是自从实现了下面的代码之后,就好像它没有选择任何东西一样。我没有收到任何错误,但是当我回显 JSON 数据和查询时,JSON 数据行是空的,并且 SQL 查询看起来格式正确。

这就是我通过 echo 语句得到的结果:

{"cols":[{"id":"Date","label":"Date","type":"datetime"},{"id":"Usage","label":"Usage","type":"number"}],"rows":[]} SELECT DATE(`dateRead`) as day, SUM(Ch3)*0.008 as totals FROM SNdata WHERE (dateRead >= 2012-01-01 AND dateRead <= 2012-01-15) and sn=5018 GROUP BY day

代码如下,以及数据库创建代码和一些示例数据。

谢谢。

谁能看出这是哪里出了问题?我可能只是太累了,看不出这里出了什么问题。我真的可以朝正确的方向轻推。

另外,请记住,此应用程序每次都会离线运行,我对 SQL 注入的担忧为零。

<?php
$serialNum = '5018';
$Chan = '3';
$Mult = '0.008';
$startDate = '2012-01-01';
$endDate ='2012-01-15';

$con=mysql_connect("localhost","user","password") or die("Failed to connect with database!!!!");
mysql_select_db("mydb", $con); 
$selectSQL="SELECT
                DATE(`dateRead`) as day,
                SUM(Ch$Chan)*$Mult as totals
            FROM SNdata
            WHERE (dateRead >= $startDate AND dateRead <= $endDate) AND sn=$serialNum
            GROUP BY day";
$sth = mysql_query($selectSQL); 
    $data = array (
        'cols' => array( 
            array('id' => 'Date', 'label' => 'Date', 'type' => 'datetime'), 
            array('id' => 'Usage', 'label' => 'Usage', 'type' => 'number')
    ),
        'rows' => array()
);
while ($res = mysql_fetch_assoc($sth)) {
// assumes dates are patterned 'yyyy-MM-dd hh:mm:ss'
preg_match('/(\d{4})-(\d{2})-(\d{2})\s(\d{2}):(\d{2}):(\d{2})/', $res['day'], $match);
$year = (int) $match[1];
$month = (int) $match[2] - 1; // convert to zero-index to match javascript's dates
$day = (int) $match[3];
$hours = (int) $match[4];
$minutes = (int) $match[5];
$seconds = (int) $match[6];
array_push($data['rows'], array('c' => array(
    array('v' => "Date($year, $month, $day, $hours, $minutes, $seconds)"), 
    array('v' => $res['totals'])
)));

}
echo json_encode($data, JSON_NUMERIC_CHECK);

echo  $selectSQL;
?>
<html>
  <head>
    <!--Load the Ajax API-->
    <script type="text/javascript" src="https://www.google.com/jsapi"></script>
    <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.2/jquery.min.js"></script>
    <script type="text/javascript">
       // Load the Visualization API and the chart package.
    google.load('visualization', '1', {'packages':['corechart']});
    google.setOnLoadCallback(drawChart);
    function drawChart() {

      var data = new google.visualization.DataTable(<?php echo json_encode($data); ?>);
      var options = {
        title: 'I am a potato',
        hAxis: {
          title: 'Date',
        },
        vAxis: {
          title: 'Useage',
        }
      };
      var chart = new google.visualization.ColumnChart(document.getElementById('chart_div'));
      chart.draw(data, options);
    }
    </script>
  </head>
  <body>    
    <div id="chart_div"></div>
  </body>
</html>

这是数据库的设置方式

CREATE TABLE `SNdata` (
    `SN` INT(10) UNSIGNED NULL DEFAULT '0',
    `dateRead` DATETIME NULL DEFAULT NULL,
    `Ch1` MEDIUMINT(8) UNSIGNED NULL DEFAULT '0',
    `Ch2` MEDIUMINT(8) UNSIGNED NULL DEFAULT '0',
    `Ch3` MEDIUMINT(8) UNSIGNED NULL DEFAULT '0',
    `Ch4` MEDIUMINT(8) UNSIGNED NULL DEFAULT '0',
    `Ch5` MEDIUMINT(8) UNSIGNED NULL DEFAULT '0',
    `Ch6` MEDIUMINT(8) UNSIGNED NULL DEFAULT '0',
    `Ch7` MEDIUMINT(8) UNSIGNED NULL DEFAULT '0',
    `Ch8` MEDIUMINT(8) UNSIGNED NULL DEFAULT '0',
    INDEX `DateINX` (`dateRead`),
    INDEX `SNINX` (`SN`)
)
COLLATE='latin1_swedish_ci'
ENGINE=MyISAM;

这是一个如何设置数据库的示例

5018,01/01/2012 00:15,0,77,73,84,0,0,3,62
5018,01/01/2012 00:30,0,100,45,77,0,0,3,67
5018,01/01/2012 00:45,0,96,62,73,0,0,2,61
5018,01/01/2012 01:00,0,81,79,85,0,0,3,56
5018,01/01/2012 01:15,0,79,47,73,0,0,2,45
5018,01/01/2012 01:30,0,72,54,69,0,0,3,65
5018,01/01/2012 01:45,0,121,100,77,0,0,3,61
5018,01/01/2012 02:00,0,77,52,87,0,0,2,61
5018,01/01/2012 02:15,0,100,99,87,0,0,3,61
5018,01/01/2012 02:30,0,87,110,107,0,0,2,60
5018,01/01/2012 02:45,0,91,86,79,0,0,3,69
5018,01/01/2012 03:00,0,91,75,67,0,0,3,61
5018,01/01/2012 03:15,0,93,99,77,0,0,2,66
5018,01/01/2012 03:30,0,87,97,72,0,0,3,64
5018,01/01/2012 03:45,0,89,73,80,0,0,3,55
5018,01/01/2012 04:00,0,92,80,69,0,0,2,46
5018,01/01/2012 04:15,0,83,94,84,0,0,3,54
5018,01/01/2012 04:30,0,99,81,66,0,0,2,67
5018,01/01/2012 04:45,0,77,89,65,0,0,3,60
5018,01/01/2012 05:00,0,85,85,79,0,0,3,

【问题讨论】:

标签: php mysql json datetime charts


【解决方案1】:

您的 SQL 从根本上被破坏了。您将日期值直接插入查询而不引用:

WHERE (dateRead >= $startDate AND dateRead <= $endDate) ...

变成了

WHERE (dateRead >= 2012-01-11 AND dateRead <= 2012-01-15) ...

由于它们没有被引用,它们只是算术表达式:你正在做 SUBTRACTION,而不是日期,所以你正在有效地运行

WHERE (dateRead >= 2011 AND dateRead <= 1996)

您的查询什么也不返回,因为没有匹配项。

你不应该再使用 mysql_*() 函数,也不应该直接将值转储到查询字符串中,但如果你坚持以这种方式继续,那么至少有

WHERE (dateRead >= '$startDate' AND dateRead <= '$endDate') ...

注意'-quotes...

【讨论】:

  • 感谢马克的建议。我添加了引号并更改为 PDO 连接,现在又收到另一个错误:警告:mysql_fetch_assoc() 期望参数 1 是资源所以我猜我的 SQL 语法仍然存在问题。我已将其替换为 code$sth = $conn->query("SELECT DATE(dateRead) as day, SUM(Ch$Chan)*$Mult as totals FROM SNdata WHERE (dateRead >= '$startDate ' AND dateRead code
  • 你不能混合数据库库。 PDO 和 mysql 是两个完全独立的系统,互不兼容
猜你喜欢
  • 2019-01-19
  • 1970-01-01
  • 1970-01-01
  • 1970-01-01
  • 1970-01-01
  • 1970-01-01
  • 1970-01-01
  • 1970-01-01
  • 2015-03-26
相关资源
最近更新 更多