Google Chart Представление

Итак, мой план состоит в том, чтобы динамически связать данные моей формы и нанести их на график Google …
Моя форма будет иметь 3 отдельных столбца комбо
Мой план состоит в том, чтобы получить положение сотрудника из базы данных, взять среднее значение и представить его на графике …

Namely Intern-avg Feedback-1 combo(Including Attributes)chart
Employee-avg Feedback-1 combo(Including Attributes)chart
Team Leader-avg Feedback-1 combo(Including Attributes)chart

Это мой файл

 <html>
<head>
<?php
if(!empty('$_SESSION'))
{
$conn = mysql_connect("localhost", "root", "");
mysql_select_db('lothlorien');
session_start();

if (!$conn) {
echo "Error: Unable to connect to MySQL." . PHP_EOL;
echo "Debugging errno: " . mysqli_connect_errno() . PHP_EOL;
echo "Debugging error: " . mysqli_connect_error() . PHP_EOL;
exit;
}
$query1="SELECT `Productiveness`,`Work_Exp`,`Auth`,`Not_Biased`,`Emp_Name`,`Position` FROM `lothlorien`.`feedback`";
$result1 = mysql_query($query1,$conn);

}
else
{
echo"Problem in Fetching...";
}

?>
<script type="text/javascript" src="https://www.gstatic.com/charts/loader.js"></script>
<script type="text/javascript">
google.charts.load('current', {'packages':['corechart']});
google.charts.setOnLoadCallback(drawVisualization);function drawVisualization() {
// Some raw data (not necessarily accurate)
<?php
if(mysql_num_rows($result1)>0)
{
$index = 0;
$total1= 0;
$total2 = 0;
$total3= 0;
$total4= 0;
?>

var data = google.visualization.arrayToDataTable([
['Index', 'Productiveness', 'Work Experience', 'Authority/Discipline', 'Racist/Biased', 'Average'],

<?php
while($fetch1=mysql_fetch_assoc($result1))
{

$C_1=$fetch1['Productiveness'];
$C_2=$fetch1['Work_Exp'];
$C_3=$fetch1['Auth'];
$C_4=$fetch1['Not_Biased'];
$C_5=$fetch1['Emp_Name'];
$C_6=$fetch1['Position'];
if($C_6=='Intern')
{

$store1=$C_1;
$total1=$store1+$total1;
$avg1=$total1/$index+1;
echo $avg1;
$store2=$C_2;
$total2=$store2+$total2;
$avg2=$total2/$index+1;
echo $avg2;
$store3=$C_3;
$total3=$store3+$total3;
$avg3=$total3/$index+1;
echo $avg3;
$store4=$C_4;
$total4=$store4+$total4;
$avg4=$total4/$index+1;
echo $avg4;
$index++;
}
else if($C_6=='Employee')
{
$store11=$C_1;
$total11=$store11+$total11;
$avg11=$total11/$index+1;
echo $avg11;
$store12=$C_2;
$total12=$store2+$total2;
$avg12=$total2/$index+1;
echo $avg12;
$store13=$C_3;
$total13=$store13+$total13;
$avg13=$total13/$index+1;
echo $avg13;
$store14=$C_4;
$total14=$store14+$total14;
$avg14=$total14/$index+1;
echo $avg14;
$index++;

}
else if($C_6=='Team Leader')
{
$store21=$C_1;
$total21=$store21+$total21;
$avg21=$total21/$index+1;
echo $avg21;
$store22=$C_2;
$total22=$store22+$total22;
$avg22=$total22/$index+1;
echo $avg22;
$store23=$C_3;
$total23=$store23+$total23;
$avg23=$total23/$index+1;
echo $avg23;
$store24=$C_4;
$total24=$store24+$total24;
$avg24=$total24/$index+1;
echo $avg24;
$index++;

}
else
{
echo 'Error...';
}
?>
[<?php if($C_6=='Intern')
{
echo $avg1;
}
elseif ($C_6=='Employee')
{
echo $avg11;
}
elseif ($C_6=='Team Leader')
{
echo $avg21;
}
else
{
echo "error";
}
?>,
<?php if($C_6=='Intern')
{
echo $avg2;
}
elseif ($C_6=='Employee')
{
echo $avg12;
}
elseif ($C_6=='Team Leader')
{
echo $avg22;
}
else
{
echo "error";
}
?>,
<?php if($C_6=='Intern')
{
echo $avg3;
}
elseif ($C_6=='Employee')
{
echo $avg13;
}
elseif ($C_6=='Team Leader')
{
echo $avg23;
}
else
{
echo "error";
}
?>,
<?php if($C_6=='Intern')
{
echo $avg4;
}
elseif ($C_6=='Employee')
{
echo $avg14;
}
elseif ($C_6=='Team Leader')
{
echo $avg24;
}
else
{
echo "error";
}
?>,<?php echo (($C_1+$C_2+$C_3+$C_4) / 4); ?>]
<?php
$index++;

?>
]);

var options = {
title : 'Monthly rating of Manager',
vAxis: {title: 'Manager'},
hAxis: {title: 'Name'},
seriesType: 'bars',
series: {4: {type: 'line'}},
colors:['Orange','Red','Blue','Grey'],

};

var chart = new google.visualization.ComboChart(document.getElementById('chart_div'));
chart.draw(data, options);
}
</script>
</head>
<body>

<div id="chart_div" style="width: 900px; height: 500px;"></div>
</body>
</html>

Я получаю синтаксическую ошибку Parse
любая помощь будет оценена

1

Решение

Задача ещё не решена.

Другие решения

Других решений пока нет …

По вопросам рекламы ammmcru@yandex.ru
Adblock
detector