[英]Google Charts using JSON
我以前没有使用谷歌图表,但我正在尝试使用位于房子周围的传感器进行一些温度图表,但是我不断得到一个例外。 我很确定它是因为JSON的格式不正确,但是在它需要什么格式以及如何让我的脚本以这种格式生成JSON方面苦苦挣扎。
PHP脚本从数据库生成JSON
<?php
require_once ("config.php");
$array = array();
$res = mysqli_query($con, "SELECT * FROM sensors WHERE vera_variable='CurrentTemperature'");
while ($row = mysqli_fetch_array($res)) {
$sensor_id = $row['sensor_id'];
$sensor_name = $row['sensor_name'];
$res2 = mysqli_query($con, "SELECT * FROM logs WHERE sensor_id='$sensor_id'");
while ($row2 = mysqli_fetch_array($res2)) {
$time = strtotime($row2['log_time']);
$formattedTime = date("m-d-y g:i", $time);
$sensor_value = $row2['sensor_value'];
$array[$formattedTime][$sensor_name] = $sensor_value;
}
}
$json = json_encode($array, JSON_PRETTY_PRINT);
echo "<pre>" . $json . "</pre>";
?>
上述脚本的示例输出。 您可以看到有一个日期,多个传感器及其相应的值
{
"12-12-15 8:35": {
"Living Room Temperature": "18.3",
"Outside Temperature": "-5",
"Mud Room Temperature": "16.0",
"Basement Temperature": "14.0"
},
"12-12-15 8:40": {
"Living Room Temperature": "18.3",
"Outside Temperature": "-5",
"Mud Room Temperature": "16.0",
"Basement Temperature": "13.0"
},
"12-12-15 8:45": {
"Living Room Temperature": "18.3",
"Outside Temperature": "-5",
"Mud Room Temperature": "16.0",
"Basement Temperature": "13.0"
},
"12-12-15 8:50": {
"Living Room Temperature": "18.3",
"Outside Temperature": "-5",
"Mud Room Temperature": "16.0",
"Basement Temperature": "13.0"
},
"12-12-15 8:55": {
"Living Room Temperature": "18.3",
"Outside Temperature": "-5",
"Mud Room Temperature": "16.0",
"Basement Temperature": "13.0"
},
"12-12-15 9:00": {
"Living Room Temperature": "17.8",
"Outside Temperature": "-5",
"Mud Room Temperature": "16.0",
"Basement Temperature": "13.0"
}
}
以下是我所拥有的(只是一个使用json的简单示例图表)
<html>
<head>
<!-- Load jQuery -->
<script language="javascript" type="text/javascript"
src="http://ajax.googleapis.com/ajax/libs/jquery/1.7.0/jquery.min.js"></script>
<!-- Load Google JSAPI -->
<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 jsonData = $.ajax({
url : "json_temp.php",
dataType : "json",
async : false
}).responseText;
var obj = window.JSON.stringify(jsonData);
var data = google.visualization.arrayToDataTable(obj);
var options = {
title : 'Graph'
};
var chart = new google.visualization.LineChart(document.getElementById('chart_div'));
chart.draw(data, options);
}
</script>
</head>
<body>
<div id="chart_div" style="width: 900px; height: 500px;"></div>
</body>
</html>
编辑:以下是Chrome在尝试加载图表时显示的错误:
未捕获错误:不是arraylha @ format + en,默认+ en,ui + en,corechart + en.I.js:191bha @ format + en,默认+ en,ui + en,corechart + en.I.js:193drawChart @ temperature.php:22
由于输入JSON数据格式( obj
变量)与Google Chart数据JSON格式不兼容 ,因此会发生此错误。
您可以将输入数据转换为支持的格式,如下所示:
var chartData = [];
chartData.push(['Time','Living Room Temperature','Outside Temperature','Mud Room Temperature','Basement Temperature']);
for (var key in obj) {
var item = obj[key];
chartData.push([new Date(key),parseFloat(item['Living Room Temperature']),parseFloat(item['Outside Temperature']),parseFloat(item['Mud Room Temperature']),parseFloat(item['Basement Temperature'])]);
}
var data = google.visualization.arrayToDataTable(chartData);
工作实例
对数据的加载方式进行了一些更改,特别是因为将同步调用
async
设置为true
被认为是一种不好的做法。 此外,请求通过promises处理。
google.load("visualization", "1", { packages: ["corechart"] }); google.setOnLoadCallback(drawChart); function drawChart() { $.ajax({ url: "https://gist.githubusercontent.com/vgrem/e08a3da68a5db5e934a1/raw/2f971a9d1524d0457a6aae4df861dc5f0af0a2ef/data.json", //json_temp.php dataType: "json" }) .done(function (data) { var chartData = []; chartData.push(['Time','Living Room Temperature','Outside Temperature','Mud Room Temperature','Basement Temperature']); for (var key in data) { var item = data[key]; chartData.push([new Date(key),parseFloat(item['Living Room Temperature']),parseFloat(item['Outside Temperature']),parseFloat(item['Mud Room Temperature']),parseFloat(item['Basement Temperature'])]); } var dataTable = google.visualization.arrayToDataTable(chartData); var options = { title: 'Graph' }; var chart = new google.visualization.LineChart(document.getElementById('chart_div')); chart.draw(dataTable, options); }); }
<script type="text/javascript" src="http://code.jquery.com/jquery-1.11.3.min.js"></script> <script type="text/javascript" src="https://www.google.com/jsapi"></script> <div id="chart_div" style="width: 900px; height: 500px;"></div>
看起来你正在传递一个Object
而不是预期的Array
。
试试这个:
var obj = window.JSON.stringify(jsonData);
var arr = [];
Object.keys(obj).forEach(function(key){
var o = obj[key];
o.time = key;
arr.push(o);
});
var data = google.visualization.arrayToDataTable(arr);
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