Javascript 谷歌饼图
添加3D饼图(谷歌图表)的JavaScript代码。饼图不会显示在视图中。如何解决 Javascript:Javascript 谷歌饼图,javascript,php,pie-chart,pygooglechart,Javascript,Php,Pie Chart,Pygooglechart,添加3D饼图(谷歌图表)的JavaScript代码。饼图不会显示在视图中。如何解决 Javascript: <script type="text/javascript" src="https://www.gstatic.com/charts/loader.js"></script> <script type="text/javascript"> $(function() { <?php
<script type="text/javascript" src="https://www.gstatic.com/charts/loader.js"></script>
<script type="text/javascript">
$(function() {
<?php
$bookingData = array();
$i=0;
foreach($bookingCounts as $booking){
$bookingData[$i]['label'] = $booking['Name'];
$bookingData[$i]['data'] = $booking['total'];
$i++;
}
?>
var data = <?php echo json_encode($bookingData, JSON_NUMERIC_CHECK);?>;
console.log(data);
var options = {
title: 'Booking',
is3D: true,
};
var chart = new google.visualization.PieChart(document.getElementById('flot-pie-chart'));
chart.draw(data, options);
});
<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() {
<?php
$bookingData = array();
$i=1;
$bookingData[0][0] = 'Task';
$bookingData[0][1] = 'Hours per Day';
foreach($bookingCounts as $booking){
$bookingData[$i][0] = $booking['Name'];
$bookingData[$i][1] = $booking['total'];
$i++;
}
?>
var data = <?php echo json_encode($bookingData, JSON_NUMERIC_CHECK);?>;
console.log(data);
var data2 = google.visualization.arrayToDataTable(data);
var options = {
title: '',
is3D: true,
};
var chart = new google.visualization.PieChart(document.getElementById('flot-pie-chart'));
chart.draw(data2, options);
};
$(函数(){
var数据=;
控制台日志(数据);
变量选项={
标题:"预订",,
is3D:是的,
};
var chart=new google.visualization.PieChart(document.getElementById('flot-pie-chart');
图表绘制(数据、选项);
});
我的看法是:
<div class="flot-chart">
<div class="flot-chart-content" id="flot-pie-chart">
</div>
</div>
您正在使用jQuery而不支持它(使用
$()
函数);快速修复方法是在主脚本之前添加此元素:
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
.我注意到您似乎没有添加列标题。只需在脚本标记中执行以下操作即可轻松完成此操作:
var data = new google.visualization.DataTable();
data.addColumn('Type', 'ColName');
data.addRows([ <?php PHP ?> ]);
Javascript:
<script type="text/javascript" src="https://www.gstatic.com/charts/loader.js"></script>
<script type="text/javascript">
$(function() {
<?php
$bookingData = array();
$i=0;
foreach($bookingCounts as $booking){
$bookingData[$i]['label'] = $booking['Name'];
$bookingData[$i]['data'] = $booking['total'];
$i++;
}
?>
var data = <?php echo json_encode($bookingData, JSON_NUMERIC_CHECK);?>;
console.log(data);
var options = {
title: 'Booking',
is3D: true,
};
var chart = new google.visualization.PieChart(document.getElementById('flot-pie-chart'));
chart.draw(data, options);
});
<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() {
<?php
$bookingData = array();
$i=1;
$bookingData[0][0] = 'Task';
$bookingData[0][1] = 'Hours per Day';
foreach($bookingCounts as $booking){
$bookingData[$i][0] = $booking['Name'];
$bookingData[$i][1] = $booking['total'];
$i++;
}
?>
var data = <?php echo json_encode($bookingData, JSON_NUMERIC_CHECK);?>;
console.log(data);
var data2 = google.visualization.arrayToDataTable(data);
var options = {
title: '',
is3D: true,
};
var chart = new google.visualization.PieChart(document.getElementById('flot-pie-chart'));
chart.draw(data2, options);
};
load(“当前”{packages:[“corechart”]});
google.charts.setOnLoadCallback(drawChart);
函数绘图图(){
var数据=;
控制台日志(数据);
var data2=google.visualization.arrayToDataTable(数据);
变量选项={
标题:“”,
is3D:是的,
};
var chart=new google.visualization.PieChart(document.getElementById('flot-pie-chart');
图表绘制(数据2,选项);
};
什么是
console.log(数据)代码>打印?你能给我们看一个php代码的HTML输出示例吗?我希望我的代码是这样的格式:另外,你有没有一个脚本标记链接到jQuery,而你没有在这里显示?因为$(function(){})
是jQuery。@Saclt7您的问题得到解决了吗?@Saclt7您能从浏览器中获得源HTML代码吗?在大多数浏览器中,您只需在URL之前写入查看源代码:
。它显示了你的PHP输出,帮助你调试。我不明白你在说什么mean@Saclt7如果您知道代码不起作用,则必须在浏览器中进行检查。如果你是那个浏览器的话。例如,如果浏览器是Chrome或Firefox,只需转到URL查看源代码:
{your URL here}(例如:查看源代码:https://www.google.com
将向您显示https://www.google.com
)。这就是PHP代码生成并发送给客户端的HTML:您可以直接将其与前面链接的Google饼图源代码示例进行比较。