<html>
<head>
<script type="text/javascript" src="https://www.google.com/jsapi"></script>
<script type="text/javascript">
google.load("visualization", "1", {packages:["corechart"]});
google.setOnLoadCallback(drawChart);
function drawChart() {
var data = google.visualization.arrayToDataTable([
['Task', 'Hours per Day'],
['Work', 11],
['Eat', 2],
['Commute', 2],
['Watch TV', 2],
['Sleep', 7]
]);
var options = {
title: 'My Daily Activities'
};
var chart = new google.visualization.PieChart(document.getElementById('chart_div'));
chart.draw(data, options);
}
</script>
</head>
<body>
<div id="chart_div" style="width: 900px; height: 500px;"></div>
</body>
</html>
答案 0 :(得分:3)
首先,您需要设计一个数据库结构。例如:
** Activities **
| ID | Task | Hours |
| 1 | Work | 11 |
| 2 | Eat | 2 |
| 3 | Cummute | 2 |
| 4 | Watch TV | 2 |
| 5 | Sleep | 7 |
现在您需要建立数据库连接。 (最简单的方法,有更好的方法来做到这一点,这只是一个例子)
<?php
$link = mysql_connect('localhost', 'mysql_user', 'mysql_password');
if (!$link) {
die('Not connected : ' . mysql_error());
}
// make foo the current db
$db_selected = mysql_select_db('your_db_name', $link);
if (!$db_selected) {
die ('Can\'t use database : ' . mysql_error());
}
?>
您需要动态构建此代码:
var data = google.visualization.arrayToDataTable([
['Task', 'Hours per Day'],
['Work', 11],
['Eat', 2],
['Commute', 2],
['Watch TV', 2],
['Sleep', 7]
]);
在此示例中,标题是静态的
var data = google.visualization.arrayToDataTable([
['Task', 'Hours per Day'],
<?php
$query = mysql_query("SELECT * FROM Activities");
$count = mysql_row_count($query);
$i = 0;
$last = ',';
while ($row = mysql_fetch_array($result)) {
$i++;
if($i == $count) { $last = ''; )
echo "['". $row['Task'] .", ". $row['Hours'] ." '] " . $last
}
?>
]);
答案 1 :(得分:-1)
尝试以下方法:
如下所述分配PHP变量:
['Task', 'Hours per Day'],
['Work', <?php echo $work; ?>],
['Eat', <?php echo $eat; ?>],
['Commute', <?php echo $commute; ?>],
['Watch TV', <?php echo $watchTv; ?>],
['Sleep', <?php echo $sleep; ?>]
变量:
$work
$eat
$commute
$watchTv
$sleep
是包含DB
中相应值的PHP变量我希望这会有所帮助。