多线谷歌折线图,使用来自 mySQL db 的数据
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【中文标题】多线谷歌折线图,使用来自 mySQL db 的数据【英文标题】:Multi line Google Line Chart, using data from mySQL db 【发布时间】:2017-09-05 13:46:04 【问题描述】:我正在经历一些学习过程,以便为我的公司创建一个小型数据库支持的报告系统。
目的是使用谷歌图表绘制多折线图,基于mysql数据库。
我已经设法从 mysql 数据库中获取要回显的数据,但它没有生成图表。我得到的只是回声,以及图表应该所在的空白区域。显示回显是出于调试目的。
代码如下:
<?php include 'confile.php';
$qry = "SELECT time,p1,p2,p3,p4 from $db WHERE date = '2016-03-02' ORDER BY time ASC";
$result = $conn->query($qry);
if($result === FALSE)
echo mysql_errno($result) .": ". mysql_error($result) ."/n";
die(mysql_error());
$rows = array();
$table = array();
$table['cols'] = array(
array('label' => 'Time', 'type' => 'datetime'),
array('label' => 'Probe 1', 'type' => 'number'),
array('label' => 'Probe 2', 'type' => 'number'),
array('label' => 'Probe 3', 'type' => 'number'),
array('label' => 'Probe 4', 'type' => 'number')
);
while($r = mysqli_fetch_assoc($result))
$temp = array();
$temp[] = array($r['time']);
$temp[] = array($r['p1']);
$temp[] = array($r['p2']);
$temp[] = array($r['p3']);
$temp[] = array($r['p4']);
$rows[] = array('c' => $temp);
$table['rows'] = $rows;
$jsonTable = json_encode($table);
echo $jsonTable;
?>
<html>
<head>
<script type="text/javascript" src="https://www.gstatic.com/charts/loader.js"></script>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.2/jquery.min.js"></script>
<script type="text/javascript">
google.charts.load('current', 'packages':['corechart']);
google.charts.setOnLoadCallback(drawChart);
function drawChart()
var data = new google.visualization.DataTable(<?$jsonTable?>);
var options =
title: 'Recorded Temperatures',
legend: position: 'bottom' ,
width: 800,
height: 600
;
var chart = new google.visualization.Table(document.getElementById('curve_chart'));
chart.draw(data, options);
</script>
</head>
<body>
<div id="curve_chart" style="width: 900px; height: 500px"></div>
</body>
</html>
这是“回声”输出
"cols":["label":"Time","type":"datetime","label":"Probe 1","type":"number","label":"Probe 2","type":"number","label":"Probe 3","type":"number","label":"Probe 4","type":"number"],"rows":["c":[["03:02:07"],["270.26"],["298.40"],["111.54"],["228.06"]],"c":[["03:28:42"],["273.23"],["190.43"],["245.69"],["283.21"]],"c":[["07:26:04"],["144.33"],["217.26"],["206.53"],["167.68"]],"c":[["12:13:20"],["153.15"],["277.23"],["167.20"],["240.88"]]]
这是测试数据,在 db 上使用测试查询。一旦我了解了呈现图表的格式,它将被设置为允许用户选择查看哪个日期等。
这是我能找到的最接近的现有问题,但似乎没有回答这个问题。
Not able to generate a Google Chart using MySQL table data as the data source
按照@MickMackusa 的回答,我设法通过确保 mysql/php 数组以 Google Charts 可接受的方式输出来将其组合在一起以使其正常工作。
感谢@MickMacUSA 的帮助。
最终的工作代码如下。
<?php include 'confile.php';
$qry = "SELECT time,p1,p2,p3,p4 from $db WHERE date = '2016-04-16' ORDER BY time ASC";
$result = $conn->query($qry);
if($result === FALSE)
echo mysqli_errno($result) .": ". mysqli_error($result) ."/n";
die(mysqli_error());
$i = 0; //iteration counter - start at 0
$totalRows = mysqli_num_rows($result); // we need this to know when to change the output
$targetRows = $totalRows - 1; //row indies start from 0, not 1.
foreach ($result as $row)
$comTime = str_replace(":",",",$row['time']); // for each row, remove the : and put , in its place
if ($targetRows == $i) // if the index is the same value as the target (ie, it's the last row)...
$temp = "[[".$comTime."],".($row['p1']).",".($row['p2']).",".($row['p3']).",".($row['p4'])."]". PHP_EOL;
else
$temp = "[[".$comTime."],".($row['p1']).",".($row['p2']).",".($row['p3']).",".($row['p4'])."],". PHP_EOL;
$i = $i + 1;
$rows[] = $temp;
$table = $rows;
$data = implode($table); //format the table as a single string, with line returns
//echo $i;
//echo $data;
?>
<html>
<head>
<script type="text/javascript" src="https://www.gstatic.com/charts/loader.js"></script>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.2/jquery.min.js"></script>
</head>
<body>
<div id="chart" style="width: 900px; height: 500px"></div>
<script type="text/javascript">
google.charts.load('current', 'packages':['corechart']);
google.charts.setOnLoadCallback(drawChart);
function drawChart()
var data = new google.visualization.DataTable();
data.addColumn('timeofday','Time');
data.addColumn('number','Probe 1');
data.addColumn('number','Probe 2');
data.addColumn('number','Probe 3');
data.addColumn('number','Probe 4');
data.addRows([
<?php echo $data; ?> //dump the result into here, as it's correctly formatted
]);
var options =
title: 'Recorded Temperatures',
legend: position: 'bottom' ,
width: 900,
height: 500,
hAxis: format: 'hh:mm:ss'
;
var chart = new google.visualization.LineChart(document.getElementById('chart'));
chart.draw(data, options);
</script>
</body>
</html>
【问题讨论】:
尽量不要过多地修正你的问题,否则你会抢走我的风头,我的回答中的要点就没有意义了。我想请您回滚编辑,因为您无需更正原始消息即可讲述故事。 按要求回滚。 【参考方案1】:您的数字值的格式必须不同,并且您需要timeofday
而不是datetime
。
根据:https://developers.google.com/chart/interactive/docs/reference#dataparam
将您的数据格式化为如下所示:
cols:
[
"label":"Time","type":"timeofday",
"label":"Probe 1","type":"number",
"label":"Probe 2","type":"number",
"label":"Probe 3","type":"number",
"label":"Probe 4","type":"number"
],
rows:
[
c:[v:[03,02,07],f:'03:02:07',v:270.26,v:298.40,v:111.54,v:228.06],
c:[v:[03,28,42],f:'03:28:42',v:273.23,v:190.43,v:245.69,v:283.21],
c:[v:[07,26,04],f:'07:26:04',v:144.33,v:217.26,v:206.53,v:167.68],
c:[v:[12,13,20],f:'12:13:20',v:153.15,v:277.23,v:167.20,v:240.88]
]
而且你必须在 javascript 中回显它:
改变:
<?$jsonTable?>
到:
<?php echo $jsonTable; ?>
并将您的 javascript 代码块放在您的 </body>
标记之前。
这是我在服务器上测试的使用上述数据格式的完整工作代码:
<html>
<head>
<script type="text/javascript" src="https://www.gstatic.com/charts/loader.js"></script>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.2/jquery.min.js"></script>
</head>
<body>
<div id="curve_chart" style="width: 900px; height: 500px"></div>
<script type="text/javascript">
google.charts.load('current', 'packages':['corechart']);
google.charts.setOnLoadCallback(drawChart);
function drawChart()
var data = new google.visualization.DataTable(
cols:[
"label":"Time","type":"timeofday",
"label":"Probe 1","type":"number",
"label":"Probe 2","type":"number",
"label":"Probe 3","type":"number",
"label":"Probe 4","type":"number"
],
rows:[
c:[v:[03,02,07],f:'03:02:07',v:270.26,v:298.40,v:111.54,v:228.06],
c:[v:[03,28,42],f:'03:28:42',v:273.23,v:190.43,v:245.69,v:283.21],
c:[v:[07,26,04],f:'07:26:04',v:144.33,v:217.26,v:206.53,v:167.68],
c:[v:[12,13,20],f:'12:13:20',v:153.15,v:277.23,v:167.20,v:240.88]
]
);
var options =
title: 'Recorded Temperatures',
legend: position: 'bottom' ,
width: 900,
height: 500,
hAxis: format: 'hh:mm:ss'
;
var chart = new google.visualization.LineChart(document.getElementById('curve_chart'));
chart.draw(data, options);
</script>
</body>
</html>
这是一种替代格式,使用 mysqli 结果将更简单/更清晰/更容易构建/理解:
<html>
<head>
<script type="text/javascript" src="https://www.gstatic.com/charts/loader.js"></script>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.2/jquery.min.js"></script>
</head>
<body>
<div id="curve_chart" style="width: 900px; height: 500px"></div>
<script type="text/javascript">
google.charts.load('current', 'packages':['corechart']);
google.charts.setOnLoadCallback(drawChart);
function drawChart()
var data = new google.visualization.DataTable();
data.addColumn('timeofday','Time');
data.addColumn('number','Probe 1');
data.addColumn('number','Probe 2');
data.addColumn('number','Probe 3');
data.addColumn('number','Probe 4');
data.addRows([
[[03,02,07],270.26,298.40,111.54,228.06],
[[03,28,42],273.23,190.43,245.69,283.21],
[[07,26,04],144.33,217.26,206.53,167.68],
[[12,13,20],153.15,277.23,167.20,240.88]
]);
var options =
title: 'Recorded Temperatures',
legend: position: 'bottom' ,
width: 900,
height: 500,
hAxis: format: 'hh:mm:ss'
;
var chart = new google.visualization.LineChart(document.getElementById('curve_chart'));
chart.draw(data, options);
</script>
</body>
</html>
查看WhiteHat提供的SO Demo:
google.charts.load('current',
callback: drawChart,
packages: ['corechart', 'table']
);
function drawChart()
var data = new google.visualization.DataTable(cols: [
"label":"Time","type":"timeofday",
"label":"Probe 1","type":"number",
"label":"Probe 2","type":"number",
"label":"Probe 3","type":"number",
"label":"Probe 4","type":"number"
],
rows: [
c:[v:[03,02,07],f:'03:02:07',v:270.26,v:298.40,v:111.54,v:228.06],
c:[v:[03,28,42],f:'03:28:42',v:273.23,v:190.43,v:245.69,v:283.21],
c:[v:[07,26,04],f:'07:26:04',v:144.33,v:217.26,v:206.53,v:167.68],
c:[v:[12,13,20],f:'12:13:20',v:153.15,v:277.23,v:167.20,v:240.88]
]
);
var table = new google.visualization.Table(document.getElementById('chart_0'));
table.draw(data);
var options =
title: 'Recorded Temperatures',
legend: position: 'bottom' ,
width: 800,
height: 600,
hAxis:
format: 'hh:mm:ss'
;
var chart = new google.visualization.LineChart(document.getElementById('chart_1'));
chart.draw(data, options);
<script src="https://www.gstatic.com/charts/loader.js"></script>
<div id="chart_0"></div>
<div id="chart_1"></div>
【讨论】:
@StevenDavison 在 javascript 中实现我建议的回显后,您的 json 是否显示在页面的源代码中?请确认。 json数据总是通过echo $jsonTable展示出来的;在第 38 行。不再重复。 它并没有从代码的 javascript 部分回显,但我不希望它是,因为它是作为 Google 图表的数据传递的。 啊哈,我明白了。是的,它正在被回应。 对我来说看起来不错 -- 在问题中也注意到 -- 正在使用Table
图表,但尚未加载 'table'
包 --> var chart = new google.visualization.Table
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