我正在尝试使用CanvasJS和PHP创建范围图,以从数据库中加载数据。
我已经创建了php,它从数据库返回了值。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);
?>
结果如下:
[{“ x”:“ 2007”,“ y”:“ [35.94,35.94]”,“ name”:“ Sessies:2006”},{“ x”:“ 2008”,“ y”:“ [27.67, 27.67]“,” name“:” Sessies:2007“},...,...]
问题是x和y值(=字符串值)中的引号。CanvasJS仅使用数字来创建图形。所以输出应该像这样:
[{“ x”:2007,“ y”:[35.94,35.94],“ name”:“ Sessies 2006”},{“ x”:2008,“ y”:[27.67,27.67],“ name”:“ Sessies 2007“},...,...]
<!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适应我的php,以便x和y以数字而不是字符串的形式传递,但是我真的是php的新手(有史以来第一次),并且找不到解决方案,特别是对于第二部分(y)。
谁能告诉我要对php和/或html文件进行哪些修改?
谢谢,
经过一番尝试和错误后,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);
问题是我需要为数据点数组中的Y值创建一个数组。在此数组中,我可以存储图形所需的2个值。
完成此操作后,我需要将所有数值转换为float_val,以便值周围的引号消失。
谢谢大家的帮助:)
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