【发布时间】:2018-10-13 04:05:20
【问题描述】:
<?php
include('classes/DB.php');
$male = DB::query('SELECT COUNT(*) AS MTOTAL FROM users WHERE gender=\'male\';');
$female = DB::query('SELECT COUNT(*) AS FTOTAL FROM users WHERE gender=\'female\';');
?>
<html>
<head>
<script type="text/javascript" src="https://www.gstatic.com/charts/loader.js"></script>
<script type="text/javascript">
google.charts.load('current', {'packages': ['corechart']});
google.charts.setOnLoadCallback(drawChart);
function drawChart() {
var data = google.visualization.arrayToDataTable([
['Task', 'Number of Male and Female visitors'],
['Male', <?php while ($s = $male->fetch()) {echo $s ['count(*)'];}?>],
['Female', <?php while ($s = $female->fetch()) {echo $s ['count(*)'];}?>],
]);
var chart = new google.visualization.PieChart(document.getElementById('piechart'))
chart.draw(data, options);} </script>
</head>
<body>
<div id="piechart" style="width: 900px; height: 500px;"></div>
</body>
</html>
我想知道为什么我得到空数据。当我调用 echo json_encode($male) 和 json_encode($female) 时,我有真正的价值观。在这种情况下,我的数据库中有一个男性用户和三个女性用户,所以我看到输出为 [{"MTOTAL":"1","0":"1"}] [{"FTOTAL":"3"," 0":"3"}]。我想知道为什么我不能在脚本中获取这些值。
【问题讨论】:
-
您可以将查询括在
"中,这样您就不必转义' -
<?php while ($s = $male->fetch()) {echo $s ['count(*)'];}?>尝试将其添加到$female = DB::query下方,看看是否会导致任何错误。也可以按照guide 启用错误 -
我收到了这个错误,未定义的索引:COUNT(*)
标签: php mysql charts google-visualization