我尝试使用CanvasJS和PHP创建范围区域图表以从数据库加载数据。
我创建了php并从数据库返回值。 这是php:
<?php
header('Content-Type: application/json');
$con = mysqli_connect("127.0.0.1","root","pwd1","db");
// Check connection
if (mysqli_connect_errno($con))
{
echo "Failed to connect to DataBase: " . mysqli_connect_error();
}else
{
$data_points = array();
$result = mysqli_query($con, "select (CalYear+1) as CalYear, concat('[',REPLACE(Year1PercWC,',','.'),',',REPLACE(Year1PercBC,',','.'),']') as ResultSet, concat('Sessies: ',calyear) as Help FROM table where cat='1' and (CalYear+1)<year(now())");
while($row = mysqli_fetch_array($result))
{
$point = array("x" => $row['CalYear'] , "y" => $row['ResultSet'],"name" => $row['Help']);
array_push($data_points, $point);
}
echo json_encode($data_points);
}
mysqli_close($con);
?>
结果:
[{&#34; X&#34;:&#34; 2007&#34;&#34; Y&#34;:&#34; [35.94,35.94]&#34;&#34;姓名&#34;:&#34; Sessies:2006&#34;},{&#34; x&#34;:&#34; 2008&#34;,&#34; y&#34;:&#34; [ 27.67,27.67]&#34;,&#34;名称&#34;:&#34; Sessies:2007&#34;},...,...]
问题是x和y值中的引号(=字符串值)。 CanvasJS仅采用数字来创建图形。所以输出应该是:
[{&#34; x&#34;:2007,&#34; y&#34;:[35.94,35.94],&#34; name&#34;:&#34; Sessies 2006&#34;} ,{&#34; x&#34;:2008,&#34; y&#34;:[27.67,27.67],&#34; name&#34;:&#34; Sessies 2007&#34;},.. 。,...]
<!DOCTYPE html>
<html xmlns="http://www.w3.org/1999/xhtml" >
<head>
<title></title>
<script type="text/javascript" src="jquery-1.11.2.min.js"></script>
<script type="text/javascript" src="jquery.canvasjs.min.js"></script>
<script type="text/javascript">
$(document).ready(function () {
$.getJSON("TestGraf.php", function (result) {
var chart = new CanvasJS.Chart("chartContainer", {
axisX: {
intervalType: "number",
title: "Year",
interval: 1,
valueFormatString: "#"
},
data: [
{
type: "rangeArea",
dataPoints: result
[{"x":2007,"y":[35.94,35.94],"name":"Sessies 2006"},{"x":2008,"y":[27.67,27.67],"name":"Sessies 2007"}] -- This is working fine
}
]
});
chart.render();
});
});
</script>
</head>
<body>
<div id="chartContainer" style="width: 800px; height: 380px;"></div>
</body>
</html>
我确定必须有一种方法来调整我的php,以便x和y作为数字而不是字符串传递,但我在php(第一次)时真的很新,并且可以&# 39;找到解决方案,尤其是第二部分(y)。
有谁能告诉我对php和/或html文件做出哪些改编?
THX,
答案 0 :(得分:0)
这应该对你有用:
json_encode($arr, JSON_NUMERIC_CHECK);
答案 1 :(得分:0)
对我的回答来说可能有点太迟了但是像Alex回答的那样,你应该使用echo json_encode($ data_points,JSON_NUMERIC_CHECK);
数字检查是一个选项。见http://php.net/manual/en/json.constants.php
然而,更多地了解范围区域图表的工作原理似乎是个问题。范围区域图表具有X和2 Y值。绘制范围需要2 Y值。如果您只是在浏览器中转到testgraf.php文件,那么您的JSON结果应为: [{x:somevalue,y:[low_value,high_value]}]
您可能必须更改sql语句以获取另一个y值。你可以用你想做的事情。无论如何,这是你应该为你的PHP代码做的事情:
更改
$point = array("x" => $row['CalYear'] , "y" => $row['ResultSet'],"name" => $row['Help']);
为:
$point = array("x" => $row['CalYear'] , "y" => $row['ResultSet', 'some_value'],"name" => $row['Help']);
如果您的2 Y值没有变化,您可能看不到图表线。线图会更合适吗?
答案 2 :(得分:0)
经过一些试验和错误后,我找到了以下解决方案:
$result1 = mysqli_query($con, "select (CalYear+1) as CalYear, Year1PercWC, Year1PercBC, calyear as Help FROM table_2 where cat='1' and (CalYear+1)<year(now())");
while($row = mysqli_fetch_array($result1))
{
$point = array("x" => floatval($row['CalYear']),"y" => array(floatval($row['Year1PercWC']),floatval($row['Year1PercBC'])),"name" => floatval($row['Help']));
array_push($data_points, $point);
}
echo json_encode($data_points);
问题是我需要在数组中为数组中的Y值创建一个数组。在这个数组中,我可以存储图形所需的2个值。
完成此操作后,我需要将数值转换为float_val,以便值周围的引号消失。
感谢所有人:)