Man*_*nos 1 html php mysql google-visualization
我是PHP和Google图表的新手.我想用它来制作谷歌图表与arduino数据存储在MYSL数据库中.到目前为止,我成功地将数据从arduino插入到mysql数据库,但我遇到了谷歌图表的困难.
这是我的PHP代码:
<?php
$mysqli =mysqli_connect('localhost', 'root', '', 'Arduino');
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: ".mysqli_connect_error();
}
$sql = mysqli_query($mysqli, 'SELECT * FROM analoog0');
if (!$sql) {
die("Error running $sql: " . mysql_error());
}
$results = array();
while($row = mysqli_fetch_assoc($sql))
{
$results[] = array(
'Date' => $row['Date'],
'Time' => $row['Time'],
'Temperature' => $row['Temperature']
);
}
$json = json_encode($results, JSON_PRETTY_PRINT);
echo $json;
?>
这是JSON输出:
[ { "Date": "2013-10-24", "Time": "18:15:49", "Temperature": "24" },
{ "Date": "2013-10-24", "Time": "18:16:19", "Temperature": "24" },
{ "Date": "2013-10-24", "Time": "18:16:49", "Temperature": "24" },
{ "Date": "2013-10-24", "Time": "18:17:19", "Temperature": "23" } ]
最后是HTM代码:
<html>
<head>
<!--Load the AJAX API-->
<script type="text/javascript" src="https://www.google.com/jsapi"></script>
<script type="text/javascript" src="//ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<script type="text/javascript">
// Load the Visualization API and the piechart package.
google.load('visualization', '1', {'packages':['corechart']});
// Set a callback to run when the Google Visualization API is loaded.
google.setOnLoadCallback(drawChart);
function drawChart() {
var jsonData = $.ajax({
url: "localhost/Charts/chart_ver2.php",
dataType:"json",
async: false
}).responseText;
// Create our data table out of JSON data loaded from server.
var data = new google.visualization.DataTable(jsonData);
// Instantiate and draw our chart, passing in some options.
var chart = new google.visualization.LineChart(document.getElementById('chart_div'));
chart.draw(data, {width: 400, height: 240});
}
</script>
</head>
<body>
<div id="chart_div"></div>
</body>
</html>
当我运行HTM代码时没有任何反应 - 屏幕保持空白.
任何帮助,指导或重定向将受到高度赞赏!提前致谢!!
您需要为Visualization API正确格式化数据.这应该使您的数据格式正确:
$results = array(
'cols' => array (
array('label' => 'Date', 'type' => 'datetime'),
array('label' => 'Temperature', 'type' => 'number')
},
'rows' => array()
);
while($row = mysqli_fetch_assoc($sql)) {
// date assumes "yyyy-MM-dd" format
$dateArr = explode('-', $row['Date']);
$year = (int) $dateArr[0];
$month = (int) $dateArr[1] - 1; // subtract 1 to make month compatible with javascript months
$day = (int) $dateArr[2];
// time assumes "hh:mm:ss" format
$timeArr = explode(':', $row['Time']);
$hour = (int) $timeArr[0];
$minute = (int) $timeArr[1];
$second = (int) $timeArr[2];
$results['rows'][] = array('c' => array(
array('v' => "Date($year, $month, $day, $hour, $minute, $second)"),
array('v' => $row['Temperature'])
));
}
$json = json_encode($results, JSON_NUMERIC_CHECK);
echo $json;
在你的javascript中,我建议重新安排你处理AJAX查询的方式.对于你这样做的方式,每次发言都没有错,但我认为这是一个更优雅的解决方案:
function drawChart() {
$.ajax({
url: 'localhost/Charts/chart_ver2.php',
dataType: 'json',
success: function (jsonData) {
// Create our data table out of JSON data loaded from server.
var data = new google.visualization.DataTable(jsonData);
// Instantiate and draw our chart, passing in some options.
var chart = new google.visualization.LineChart(document.getElementById('chart_div'));
chart.draw(data, {width: 400, height: 240});
}
});
}