我使用谷歌图表,发现语法错误不能纠正语法错误的PHP
意外的标记{
<script type="text/javascript" src="https://www.google.com/jsapi"></script>
<script type="text/javascript">
google.load("visualization", "1", {packages:["corechart"]});
google.setOnLoadCallback(drawChart);
function drawChart() {
var data = google.visualization.arrayToDataTable([
['name', 'score'],
<?php
global $wpdb;
$query = $wpdb->get_results("select t2.name, count(t1.id) as score from wp3_wpsp_custom_status as t2 left join wp3_wpsp_ticket as t1 on t2.name = t1.status group by t2.name");
var_dump($query);
foreach($query as $row){
$object_array =(array)$row;
echo "['".$object_array['name']."',".$object_array['score']."],";
}
?>
]);
var options = {
title: 'Date wise visits'
};
var chart = new google.visualization.ColumnChart(document.getElementById("columnchart"));
chart.draw(data, options);
}
</script>
上面的代码是用于创建柱形图的基本语法。看看值之后,“VAR数据= google.visualization.arrayToDataTable([”这其中有静态值。但为了显示我们的统计,我们需要动态地从我们的数据库和PHP加载这些值。
<body>
<h3>Column Chart</h3>
<div id="columnchart" style="width: 900px; height: 500px;"></div>
</body>
错误[![在这里输入的形象描述] [1] [1]
@Armin我编辑了你可以看看我的代码现在 – JMR
你也可以在运行脚本后添加生成的代码吗? – Armin
@Armin我发布了我的截图也 – JMR