Javascript Canvasjs柱形图通过json更新无效
我一直在试图找出这段代码的问题所在,但似乎还没有找到解决方案。所以我想让这个柱状图每秒刷新一次chartOne.php中的数据,图表会更新,但数据不会显示 下面是chartOne.php代码:Javascript Canvasjs柱形图通过json更新无效,javascript,php,jquery,json,canvasjs,Javascript,Php,Jquery,Json,Canvasjs,我一直在试图找出这段代码的问题所在,但似乎还没有找到解决方案。所以我想让这个柱状图每秒刷新一次chartOne.php中的数据,图表会更新,但数据不会显示 下面是chartOne.php代码: <?php include_once("connect.php"); $grabOne = mysqli_query($con, "SELECT * FROM tb_cndts WHERE id='1'") or die(mysql_error()); while($rows = mysqli_f
<?php
include_once("connect.php");
$grabOne = mysqli_query($con, "SELECT * FROM tb_cndts WHERE id='1'") or die(mysql_error());
while($rows = mysqli_fetch_array($grabOne)){
$oneCount = $rows['count'];
}
$grabTwo = mysqli_query($con, "SELECT * FROM tb_cndts WHERE id='2'") or die(mysql_error());
while($rows = mysqli_fetch_array($grabTwo)){
$twoCount = $rows['count'];
}
$grabThree = mysqli_query($con, "SELECT * FROM tb_cndts WHERE id='3'") or die(mysql_error());
while($rows = mysqli_fetch_array($grabThree)){
$threeCount = $rows['count'];
}
echo json_encode( array(
array("id"=>"0","count"=>$oneCount),
array("id"=>"1","count"=>$twoCount),
array("id"=>"2","count"=>$threeCount)
));
MySQL端的数据如下所示:
id | count
1 | 100
2 | 200
3 | 150
从页面加载一秒钟后,图表变成
我尝试了document.write();这一点和数据显示得很好。有人能帮我吗?提前感谢。问题在于如何将数据添加到图表中。重要的一点是:
chart.options.data[0].dataPoints[n] = { label: val.id, y: val.count };
下面是一个完整的工作示例,我不是在打印json数据,而是在远程获取它,但是如果您在$.get中交换回来,它应该可以正常工作:
<!DOCTYPE HTML>
<html>
<head>
<script src="https://code.jquery.com/jquery-1.12.2.min.js"></script>
<script src="http://canvasjs.com/assets/script/canvasjs.min.js"></script>
<script type="text/javascript">
<?php
$data = json_encode( array(
array("id"=>"Person One","count"=>20),
array("id"=>"Person Two","count"=>10),
array("id"=>"Person Three","count"=>15)
));
echo "var data = {$data}";
?>
window.onload = function () {
var dps = [
{label: "Person One", y: 0} ,
{label: "Person Two", y: 0},
{label: "Person Three", y: 0},
];
var chart = new CanvasJS.Chart("chartContainer",{
theme: "theme2",
title:{
text: "Students' Votes"
},
axisY: {
title: ""
},
legend:{
verticalAlign: "top",
horizontalAlign: "centre",
fontSize: 18
},
data : [{
type: "column",
showInLegend: true,
legendMarkerType: "none",
legendText: " ",
indexLabel: "{y}",
dataPoints: dps
}]
});
chart.render();
var updateInterval = 1000;
var updateChart = function () {
$.each(data, function(n, val) {
console.log(n);
console.log(val);
chart.options.data[0].dataPoints[n] = { label: val.id, y: val.count };
});
chart.render();
};
setInterval(function(){updateChart()}, updateInterval);
}
</script>
</head>
<body>
<div id="chartContainer" style="height: 300px; width: 100%;"></div>
</body>
</html>
window.onload=函数(){
var dps=[
{标签:“第一个人”,y:0},
{标签:“第二个人”,y:0},
{标签:“第三个人”,y:0},
];
var chart=new CanvasJS.chart(“chartContainer”{
主题:“主题2”,
标题:{
正文:“学生投票”
},
axisY:{
标题:“
},
图例:{
垂直排列:“顶部”,
水平对齐:“中心”,
尺寸:18
},
数据:[{
键入:“列”,
showInLegend:是的,
legendMarkerType:“无”,
传奇文本:“,
索引标签:“{y}”,
数据点:dps
}]
});
chart.render();
var updateInterval=1000;
var updateChart=函数(){
$.each(数据、函数(n、val){
控制台日志(n);
控制台日志(val);
chart.options.data[0].dataPoints[n]={label:val.id,y:val.count};
});
chart.render();
};
setInterval(函数(){updateChart()},updateInterval);
}
好的,多亏@oneskinnydj的帖子(console.log idea)。我发现MySQL查询返回的是字符串,而不是CanvasJS需要的整数。为此,您需要更改mysqli_fetch_数组();进入:
php部分的最终结果是:
<?php
include_once("connect.php");
$grabOne = mysqli_query($con, "SELECT * FROM tb_cndts WHERE id='1'") or die(mysql_error());
while($rows = mysqli_fetch_assoc($grabOne)){
$oneCount = intval($rows['count']);
}
$grabTwo = mysqli_query($con, "SELECT * FROM tb_cndts WHERE id='2'") or die(mysql_error());
while($rows = mysqli_fetch_assoc($grabTwo)){
$twoCount = intval($rows['count']);
}
$grabThree = mysqli_query($con, "SELECT * FROM tb_cndts WHERE id='3'") or die(mysql_error());
while($rows = mysqli_fetch_assoc($grabThree)){
$threeCount = intval($rows['count']);
}
$data = json_encode( array(
array("id"=>"Person One","count"=>$oneCount),
array("id"=>"Person Two","count"=>$twoCount),
array("id"=>"Person Three","count"=>$threeCount)
));
echo $data;
这个问题仍然存在。我相信问题出在php查询中。因为我需要它来查询,所以我将count更改为一个变量。现在,当变量为时,数据显示的结果是整数,但是如果我将变量更改为查询结果,数据将不会显示(与以前相同)。控制台日志显示查询中的变量是字符串而不是整数。很高兴听到这个消息,您能对我的答案进行投票吗,因为它对您有一些用处?
mysqli_fetch_assoc();
<?php
include_once("connect.php");
$grabOne = mysqli_query($con, "SELECT * FROM tb_cndts WHERE id='1'") or die(mysql_error());
while($rows = mysqli_fetch_assoc($grabOne)){
$oneCount = intval($rows['count']);
}
$grabTwo = mysqli_query($con, "SELECT * FROM tb_cndts WHERE id='2'") or die(mysql_error());
while($rows = mysqli_fetch_assoc($grabTwo)){
$twoCount = intval($rows['count']);
}
$grabThree = mysqli_query($con, "SELECT * FROM tb_cndts WHERE id='3'") or die(mysql_error());
while($rows = mysqli_fetch_assoc($grabThree)){
$threeCount = intval($rows['count']);
}
$data = json_encode( array(
array("id"=>"Person One","count"=>$oneCount),
array("id"=>"Person Two","count"=>$twoCount),
array("id"=>"Person Three","count"=>$threeCount)
));
echo $data;
chart.options.data[0].dataPoints[n] = { label: val.id, y: val.count };