2017-02-14 89 views
0

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]

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@Armin我编辑了你可以看看我的代码现在 – JMR

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你也可以在运行脚本后添加生成的代码吗? – Armin

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@Armin我发布了我的截图也 – JMR

回答

2

从您的代码从他们的网站删除var_dump($query)

要添加颜色,添加第三个参数例:

var data = google.visualization.arrayToDataTable([ 
     ["Element", "Density", { role: "style" } ], 
     ["Copper", 8.94, "#b87333"], 
     ["Silver", 10.49, "silver"], 
     ["Gold", 19.30, "gold"], 
     ["Platinum", 21.45, "color: #e5e4e2"] 
     ]); 
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如何给它们中的每一个添加颜色 – JMR

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检查更新的说明 – Armin

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它只是显示得分作为图例....如何添加传奇...感谢您的答案 – JMR