我使用Google图表通过下面的代码绘制图表,但问题是我的数据是在sql表上,我如何将表格变量放入此图表?
<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([
['Year', 'Sales', 'Expenses'],
['2004', 1000, 400],
['2005', 1170, 460],
['2006', 660, 1120],
['2007', 1030, 540]
]);
var options = {
title: 'Company Performance'
};
var chart = new google.visualization.LineChart(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:0)
通过查询数据库并将其转换为JSON字符串,从表中检索数据。然后将json字符串传递给变量
var data=/*your json string*/
答案 1 :(得分:0)
$db = new mysqli('localhost', 'user', 'pass', 'demo');
if($db->connect_errno > 0){
die('Unable to connect to database [' . $db->connect_error . ']');
}
$sql = "SELECT attempt, login FROM result";
if(!$result = $db->query($sql)){
die('There was an error running the query [' . $db->error . ']');
}
echo "var data = google.visualization.arrayToDataTable([['Attempt', 'Login']";
while($row = $result->fetch_assoc()){
echo ",['".$row['attempt'] ."','".$row['attempt']."']";
}
echo "]);";
$result->free();
$db->close();
答案 2 :(得分:0)
<?php
$result = mysql_query("SELECT year,sales,expenses from tablename");
$value=array();
while($r = mysql_fetch_assoc($result)) {
$year=$r['year'];
$sales=$r['sales'];
$expenses=$r['expenses'];
$val="[".$year.",".$sales.",".$expenses."]";
array_push($value,$val );
}
$final_value = implode(",", $value);
?>
将此$ final_value放入Google图表
function drawChart() {
var data = google.visualization.arrayToDataTable([
<?php echo $final_value?>
]);
var options = {
title: 'Company Performance'
};
var chart = new google.visualization.LineChart(document.getElementById('chart_div'));
chart.draw(data, options);
}