dongqiang5865 2013-03-05 20:10
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在Highcharts中添加系列时,Javascript循环不通过sqlite_fetch_array()设置PHP值

I have a SQLite database and am trying to graph data on a linechart using Highcharts, PHP, and Javascript. I am graphing one series per user (a user being a text value per tuple), but am running into trouble with retrieving a subsequent user via a PHP loop.

$dbhandle = sqlite_open('db/test.db', 0666, $error);

if (!$dbhandle) die ($error);
$query5 = "SELECT DISTINCT User FROM Events";

$ok0 = sqlite_query($dbhandle, $query5, $error_msg);

if (!$ok0) 
{
    die("dead" . $error_msg);
}
$rows = sqlite_num_rows($ok0);

echo 
"for(var i=2; i<$rows; i++) // start of JS loop. 
                           //$rows is 4; I am graphing 2 series here
{";

  $array = sqlite_fetch_array($ok0, SQLITE_ASSOC); // $ok0 is the unique list
// of users. After graphing one series, I want to grab the next user to graph


        echo "chart.addSeries({
                name: '{$array["User"]}',
                data: [";
          for($i=0; $i<$diff+1; $i++)
          {
            $target = date("D, j M", (strtotime($_GET["start"]) + $i * 24 * 3600));
            $query6 = "SELECT * FROM Events WHERE User = '{$array["User"]}' AND Start LIKE '%{$target}%'";
            $result6 = sqlite_query($dbhandle, $query6);
            if (!$result6) die("Cannot execute query.");
            $num = sqlite_num_rows($result6);            
            if($i==($diff))
            {
              echo $num;
            }              
            else
              echo $num . ", ";
          }?>],

                pointStart: 


                 <?php
$date = DateTime::createFromFormat('D M d Y', urldecode($_GET["start"])); 
echo $date->getTimestamp()*1000;?>,
            pointInterval: 24 * 3600 * 1000 // one day

              });
      <?php echo "}";?> // end of JS loop

The result does graph two additional series, but they are both the same data from the same user. It doesn't look like the loop with sqlite_fetch_array() correctly returns the next user. Can anyone see the problem here? Maybe something with the way I'm integrating Javascript with PHP?

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2条回答 默认 最新

  • dongping4901 2013-03-06 20:01
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    I've figured it out. I believe that due to PHP being executed server-side, the block of code within the echoed Javascript loop results in the same values when the browser executes the Javascript. The server simply echoes the string that represents the js loop, and will run the code block after. When everything is sent to the browser, the Javascript will loop normally and simply output the results of the code block twice. I've just forgone the JS loop via PHP and replaced it with a PHP loop.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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