我有一个带有 3 个字段的 mysql 数据库:region(vachar)、new_customers(int) 和 old_customers(int)。
我正在尝试使用 google 图表和 php 创建条形图，但它不起作用。我认为我在 echo 上做错了。 这是我的代码:
<script type="text/javascript" src="https://www.google.com/jsapi"></script>
<script type="text/javascript">
google.load('visualization', '1', {packages: ['corechart', 'bar']});
google.setOnLoadCallback(drawMaterial);
function drawMaterial() {
var data = google.visualization.arrayToDataTable([
['Region', 'New Customers', 'Old Customers'],
<?php
$query = "SELECT sum(new_customers) AS new, sum(old_customers) AS old, region FROM daily GROUP BY region";
$exec = mysqli_query($con,$query);
while($row = mysqli_fetch_array($exec)){
echo "['".$row['region']."',";
echo "['".$row['new']."',";
echo "['".$row['old']."',";
}
?>
]);
var options = {
title: 'Country wise new and returned visitors',
bars: 'horizontal'
};
var material = new google.charts.Bar(document.getElementById('barchart'));
material.draw(data, options);
}
</script>
最佳答案
您遇到的问题是图表没有采用数据变量中的三个参数。
目前你有这样的:['Region', 'New Customers', 'Old Customers'], .
将其更改为: ['Region', 'New Customers'] ，。
在 while loop ，PHP 无法识别任何名为 $row['new'] 的列或$row['old'] 。相反，使用索引并将 echo 更改为:
<强> echo "['".$row[2]." [".$row[0]."]', ".$row[1]."], ”；
我还建议您在函数之外进行 sql 查询，并且仅对 sql 结果进行迭代( while loop 和 echo )。
此外，您没有正确加载图表。您忘记添加“图表”一词:
google.charts.load('visualization', '1', {packages: ['corechart', 'bar']});
google.charts.setOnLoadCallback(drawMaterial);
这是最终的解决方案:
<?php
include("YOUR PHP CONNECT FILE");
$query = "SELECT sum(new_customers) AS new, sum(old_customers) AS old, region FROM daily GROUP BY region";
$exec = mysqli_query($YOURCONNECTIONVARIABLE ,$query);
?>
<!DOCTYPE html>
<head>
<script type="text/javascript" src="https://www.gstatic.com/charts/loader.js"></script>
<script type="text/javascript">
google.charts.load('visualization', '1', {packages: ['corechart', 'bar']});
google.charts.setOnLoadCallback(drawMaterial);
function drawMaterial() {
var data = google.visualization.arrayToDataTable([
['Region', 'New Customers'],
<?php
while($row = mysqli_fetch_array($exec)){
echo "['".$row[2]." [".$row[0]."]', ".$row[1]."],";
}
?>
]);
var options = {
title: 'Country wise new and returned visitors',
bars: 'horizontal'
};
var material = new google.charts.Bar(document.getElementById('barchart'));
material.draw(data, options);
}
</script>
</head>
<body>
<div id="barchart" style="width: 100%; height: 40em;"></div>
</body>
</html>
如果您有任何疑问，请告诉我。
关于php - 来自 Mysql 数据库的 Google 条形图，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/42921138/