Google Charts 饼图显示其他 100%
<html>
<head>
<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(drawChart);
function drawChart() {
var data = google.visualization.arrayToDataTable([
['Task', 'Hours per Day'],
<?php
$query = "SELECT name, score FROM `Users` order by score desc limit 5";
$result = mysqli_query($conn, $query);
if ($result->num_rows > 0) {
while($row = mysqli_fetch_array($result))
{
echo "['".$row['name']."', '".$row['score']."'],";
}
}
?>
]);
var options = {
title: 'My Daily Activities'
};
var chart = new google.visualization.PieChart(document.getElementById('piechart'));
chart.draw(data, options);
}
</script>
</head>
<body>
<div id="piechart" style="width: 900px; height: 500px;"></div>
</body>
</html>
我的代码有什么问题吗？我读到我的值存储为字符串，以便图表显示 100% 其他。
我该如何让它发挥作用？我尝试使用(int)$varPHP 函数，但这还不能使它工作..
while($row = mysqli_fetch_array($result))
{
$score = $row['score'];
$int = (int)$score;
echo "['".$row['name']."', '".$row['score']."'],";
}
}
慕侠23898041回答
-
慕的地6264312
不要自己生成json。json_encode 始终使用：<?php$query = "SELECT name, score FROM `Users` order by score desc limit 5";$result = mysqli_query($conn, $query);$chartData = [ ['Task', 'Hours per Day'],];if ($result->num_rows > 0) { while($row = mysqli_fetch_array($result)) { $chartData[] = [$row['name'], $row['score']]; } }?><script type="text/javascript"> google.charts.load('current', {'packages':['corechart']}); google.charts.setOnLoadCallback(drawChart); function drawChart() { var data = google.visualization.arrayToDataTable(<?php echo json_encode($chartData)?>); var options = { title: 'My Daily Activities' }; var chart = new google.visualization.PieChart(document.getElementById('piechart')); chart.draw(data, options); }</script>
随时随地看视频慕课网APP
PHP