<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']."'],"; } }
1 回答
慕的地6264312
TA贡献1817条经验 获得超6个赞
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>
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