我正在繪製一張顯示溫度,日期和時間的圖表。到目前爲止,所有工作都很好,但爲了從我的MySQL數據庫中獲取新值,以便完全刷新我的頁面。 我希望能夠在不按刷新按鈕的情況下更新圖形。 這是代碼:如何在按下按鈕時更新D3.js中的圖形?
<?php require($_SERVER['DOCUMENT_ROOT'] . '/assets/php/getTemp.php'); ?>
<!DOCTYPE html>
<meta charset="utf-8">
<head>
<title>RPi</title>
</head>
<style>
div.tooltip {
position: absolute;
text-align: center;
width: 120px;
height: 28px;
padding: 3px;
font: 12px sans-serif;
background: lightgrey;
border: 0px;
border-radius: 8px;
pointer-events: none;
}
body { font: 14px Arial;}
path {
stroke: #FF8C00;
stroke-width: 3;
fill: none;
}
.axis path,
.axis line {
fill: none;
stroke: black;
stroke-width: 2;
shape-rendering: crispEdges;
}
.grid .tick {
stroke: grey;
stroke-opacity: 0.3;
shape-rendering: geometricPrecision;
}
.grid path {
stroke-width: 0;
}
</style>
<body>
<div id="option">
<input name="updateButton"
type="button"
value="Update"
onclick="updateData()"
/>
</div>
<!-- load the d3.js library -->
<script src="http://d3js.org/d3.v3.min.js"></script>
<script>
var margin = {top: 30, right: 20, bottom: 30, left: 50},
width = 800 - margin.left - margin.right,
height = 270 - margin.top - margin.bottom;
var parseDate = d3.time.format("%Y-%m-%d %H:%M:%S").parse;
var formatTime = d3.time.format("%d-%m-%Y %H:%M:%S");
var x = d3.time.scale().range([0, width]);
var y = d3.scale.linear().range([height, 0]);
var xAxis = d3.svg.axis().scale(x)
.orient("bottom");
var yAxis = d3.svg.axis().scale(y)
.orient("left").ticks(5);
var valueline = d3.svg.line()
.interpolate("basis")
.x(function(d) { return x(d.datetime); })
.y(function(d) { return y(d.temperature); });
var div = d3.select("body").append("div")
.attr("class", "tooltip")
.style("opacity", 0);
var svg = d3.select("body")
.append("svg")
.attr("width", width + margin.left + margin.right)
.attr("height", height + margin.top + margin.bottom)
.append("g")
.attr("transform",
"translate(" + margin.left + "," + margin.top + ")");
function make_x_axis() {
return d3.svg.axis()
.scale(x)
.orient("bottom")
.ticks(5)
}
function make_y_axis() {
return d3.svg.axis()
.scale(y)
.orient("left")
.ticks(5)
}
<?php echo "data=".$json_data.";" ?>
data.forEach(function(d) {
d.datetime = parseDate(d.datetime);
d.temperature = +d.temperature;
});
x.domain(d3.extent(data, function(d) { return d.datetime; }));
y.domain([0, d3.max(data, function(d) { return d.temperature; })]);
svg.append("path")
.attr("class", "line")
.attr("d", valueline(data));
svg.selectAll("dot")
.data(data)
.enter().append("circle")
.attr("r", 4)
.attr("cx", function(d) { return x(d.datetime); })
.attr("cy", function(d) { return y(d.temperature); })
.on("mouseover", function(d) {
div.transition()
.duration(200)
.style("opacity", 1);
div.html(formatTime(d.datetime) + "<br/>" + d.temperature + " ℃")
.style("left", (d3.event.pageX + 16) + "px")
.style("top", (d3.event.pageY + 16) + "px")
.style("position", "absolute");
})
.on("mouseout", function(d) {
div.transition()
.duration(50)
.style("opacity", 0);
});
svg.append("g")
.attr("class", "grid")
.attr("transform", "translate(0," + height + ")")
.call(make_x_axis()
.tickSize(-height, 0, 0)
.tickFormat("")
)
svg.append("g")
.attr("class", "grid")
.call(make_y_axis()
.tickSize(-width, 0, 0)
.tickFormat("")
)
svg.append("g")
.attr("class", "x axis")
.attr("transform", "translate(0," + height + ")")
.call(xAxis);
svg.append("g")
.attr("class", "y axis")
.call(yAxis);
</script>
</body>
我試着這樣做:
function updateData() {
//d3.json("/assets/php/getTemp.php", function(error, data) {
<?php echo "data=".$json_data.";" ?>
data.forEach(function(d) {
d3.select("body").selectAll("svg")
d.datetime = parseDate(d.datetime);
d.temperature = +d.temperature;
});
x.domain(d3.extent(data, function(d) { return d.datetime; }));
y.domain([0, d3.max(data, function(d) { return d.temperature; })]);
var svg = d3.select("body").transition();
svg.select(".line")
.duration(750)
.attr("d", valueline(data));
svg.select(".x.axis")
.duration(750)
.call(xAxis);
svg.select(".y.axis")
.duration(750)
.call(yAxis);
};
但沒有任何反應,甚至不是一個錯誤。
如果它的事項,這是用來獲取溫度和時間形式的MySQL的PHP代碼:
<?php
$hostname = 'localhost';
$username = 'root';
$password = 'admin';
try {
$dbh = new PDO("mysql:host=$hostname;dbname=temp_database",
$username, $password);
$sth = $dbh->prepare("
SELECT `datetime`, `temperature` FROM `tempLog`
");
$sth->execute();
$result = $sth->fetchAll(PDO::FETCH_ASSOC);
$dbh = null;
}
catch(PDOException $e)
{
echo $e->getMessage();
}
$json_data = json_encode($result);
?>
我需要做什麼?
看起來是對我的。你可以用一個你正在使用的數據類型的例子把它粘在一個jsfiddle中嗎? – BHouwens
當然:https://jsfiddle.net/zzmoL7Le/4/我希望鏈接是好的。 –
問題似乎是您在'updateData'函數中對數據執行的格式。如果您在代碼運行時註釋該部分 – BHouwens