-1
我想繪製一個圖表,其中的值將從MySQL數據庫中獲取。它不工作。但是如果我給出手動值,那麼圖表就會顯示出來。下面是我的代碼:需要來自mysql的數據繪製圖形
<?php
include "libchart/classes/libchart.php";
header("Content-type: image/png");
$chart = new LineChart();
$con=mysqli_connect("localhost","root","","bkash");
$result = mysqli_query($con,"SELECT time,trx_value FROM dialer_rate where mno='tnr'");
$serie1 = new XYDataSet();
while($row = mysqli_fetch_array($result))
{
$serie1->addPoint(new Point($row['time']->time, $row['trx_value']->trx_value));
}
$dataSet = new XYSeriesDataSet();
$dataSet->addSerie("TNR", $serie1);
$dataSet->addSerie("ROBI", $serie2);
$dataSet->addSerie("LNK", $serie3);
$dataSet->addSerie("AIR", $serie4);
$chart->setDataSet($dataSet);
$chart->setTitle("bKash Success/Failure for All MNO");
$chart->getPlot()->setGraphCaptionRatio(0.62);
$chart->render();
>
我沒有及時得到任何價值低於行& trx_value ...
while($row = mysqli_fetch_array($result))
{
$serie1->addPoint(new Point($row['time']->time, $row['trx_value']->trx_value));
}
,但如果我想用打印的價值?回聲低於輸出來
while($row = mysqli_fetch_array($result))
{ echo $row['time'];
echo $row['trx_value'];
};
然後輸出來了..
此圖也顯示了當我使用下面的手動值時。
$serie1->addPoint(new Point("06-01", 273));
$serie1->addPoint(new Point("06-02", 421));
$serie1->addPoint(new Point("06-03", 642));
$serie1->addPoint(new Point("06-04", 799));
$serie1->addPoint(new Point("06-05", 1009));
$serie1->addPoint(new Point("06-06", 1106));
根據您的建議更正。但仍然沒有輸出。 – Basudev 2015-02-11 14:06:04
@Basudev,你確定'$ row ['time']'和$ row ['trx_value']'包含你認爲他們做了什麼嗎? 'echo'出來,或者在'$ row'上使用'var_dump()'。 – Chris 2015-02-11 14:07:43
是的,我的數據在'時間'和'trx_value'列中。 – Basudev 2015-02-11 14:11:35