我有一个问题,即使我有 2 个图表的代码,我也只能在我的网络中加载 1 个 zingchart。
代码将只生成最新的图表,在本例中为饼图并忽略条形图。
以下是我的代码
<?php
//getDBConnect function
require 'dbconnect.php';
//Get ID from form
$id = $_GET['staffid'];
//connect to database
$con = getDBConnect();
if(!mysqli_connect_errno($con)){
$sqlQueryStr =
"SELECT a.ai_Name, r.duration " .
"FROM report AS r, academicinstitution AS a " .
"WHERE r.ai_Id = a.ai_Id ";
$result = mysqli_query($con,$sqlQueryStr);
mysqli_close($con);
} else {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
//Get data into array
$emparray = array();
while ($row = mysqli_fetch_assoc($result)) {
$emparray[] = $row;
}
//Group array by ai_Name
$grouparray = array();
foreach($emparray as $item)
{
if(!isset($grouparray[$item["ai_Name"]]))
$grouparray[$item["ai_Name"]] = 0;
$grouparray[$item["ai_Name"]] += $item["duration"];
}
?>
<script>
var dataBar=[
<?php
foreach($grouparray as $keys => $value){
echo $value. ",";
}
?>];
window.onload=function(){
zingchart.render({
id:'chartBar',
height:400,
width:600,
data:{
"graphset":[
{
"type":"bar",
"title":{"text":"BarChart"},
"series":[
{
"values":dataBar
}
]
}
]
}
});
};
</script>
<script>
var dataPie=[
<?php
foreach($grouparray as $keys => $value){
echo '{';
echo '"text":"'.$keys.'","values":['.$value.']';
echo '},';
}
?>];
window.onload=function(){
zingchart.render({
id:'chartPie',
height:400,
width:600,
data:{
"graphset":[
{
"type":"pie",
"title":{"text":"PieChart"},
"series":dataPie
}
]
}
});
};
</script>
<div id="chartBar"></div>
<div id="chartPie"></div>
我应该怎么办?