เกี่ยวกับการสร้างกราฟค่ะ...แทนค่ะปีไป แต่มันไม่เจอค่า พอลองแทนค่าปีไป จะแสดงผล ไม่ทราบว่าเกิดจากอะไร รบกวนผู้เชี่ยวชาญด้วยค่ะ
Code (PHP) <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=windows-874" />
<title></title>
</head>
<?
//include"../config.php";
$y=date("Y")+543;
$year1=$y;
$year2=$y-1;
$year3=$y-2;
?>
<script type="text/javascript" src="graph/jquery1.8.min.js"></script>
<script type="text/javascript">
$(function () {
$('#container').highcharts({
chart: {
type: 'bar'
},
title: {
text: 'จำนวนผู้ไปพัฒนาตนเอง (ย้อนหลัง 3 ปี)'
},
subtitle: {
text: 'คณะ'
},
xAxis: {
categories: ['ปีงบประมาณ <?=$year3?>', 'ปีงบประมาณ <?=$year2?>', 'ปีงบประมาณ <?=$year1?>'],
title: {
text: null
}
},
yAxis: {
min: 0,
title: {
text: 'จำนวนบุคลากร(ครั้ง)',
align: 'high'
},
labels: {
overflow: 'justify'
}
},
tooltip: {
valueSuffix: ' คน'
},
plotOptions: {
bar: {
dataLabels: {
enabled: true
}
}
},
legend: {
layout: 'vertical',
align: 'right',
verticalAlign: 'top',
x: -10,
y: 300,
floating: true,
borderWidth: 1,
backgroundColor: '#EEEEE1',
shadow: true
},
credits: {
enabled: false
},
series: [{
<?
//-------------------------
$data=array();
for($i=$year1;$i>=$year3;$i--){
//$sql="select DISTINCT per_a_id from project p,person pe WHERE pe.per_id=p.pro_per_id and p.pro_yy_card='$i' and pe.per_prog_id=002";
$sql="select SUM(p.pro_budget3) from project p,person pe WHERE pe.per_id=p.pro_per_id and p.pro_yy_card='$i' and pe.per_prog_id=002";
$res=mysql_query($sql);
$ln=mysql_fetch_array($res);
$data[]=$ln[0];
//$rows=mysql_num_rows($res);
//$data[]=$rows;
}
?>
name: 'สำนักงานคณบดี',
data: [<?=$data[2]?>, <?=$data[1]?>, <?=$data[0]?>]
}, {
<?
//-------------------------
$data=array();
for($i=$year1;$i>=$year3;$i--){
$sql="select SUM(p.pro_budget3) from project p,person pe WHERE pe.per_id=p.pro_per_id and p.pro_yy_card='$i' and pe.per_prog_id=003";
$res=mysql_query($sql);
$ln=mysql_fetch_array($res);
$data[]=$ln[0];
}
?>
name: 'ภาษาไทย',
data: [<?=$data[2]?>, <?=$data[1]?>, <?=$data[0]?>]
}, {
<?
//-------------------------
$data=array();
for($i=$year1;$i>=$year3;$i--){
$sql="select SUM(p.pro_budget3) from project p,person pe WHERE pe.per_id=p.pro_per_id and p.pro_yy_card='$i' and pe.per_prog_id=004";
$res=mysql_query($sql);
$ln=mysql_fetch_array($res);
$data[]=$ln[0];
}
?>
name: 'อังกฤษ',
data: [<?=$data[2]?>, <?=$data[1]?>, <?=$data[0]?>]
}, {
<?
//-------------------------
$data=array();
for($i=$year1;$i>=$year3;$i--){
$sql="select SUM(p.pro_budget3) from project p,person pe WHERE pe.per_id=p.pro_per_id and p.pro_yy_card='$i' and pe.per_prog_id=005";
$res=mysql_query($sql);
$ln=mysql_fetch_array($res);
$data[]=$ln[0];
}
?>
name: 'รปศ',
data: [<?=$data[2]?>, <?=$data[1]?>, <?=$data[0]?>]
}, {
<?
//-------------------------
$data=array();
for($i=$year1;$i>=$year3;$i--){
$sql="select SUM(p.pro_budget3) from project p,person pe WHERE pe.per_id=p.pro_per_id and p.pro_yy_card='$i' and pe.per_prog_id=006";
$res=mysql_query($sql);
$ln=mysql_fetch_array($res);
$data[]=$ln[0];
}
?>
name: 'สังคมศาสตร์.',
data: [<?=$data[2]?>, <?=$data[1]?>, <?=$data[0]?>]
}]
});
});
</script>
<link rel="stylesheet" href="style.css">
<!--begin css-->
<link rel="stylesheet" href="menu/menu_style.css" type="text/css" />
<!--end css-->
<style>
.tb2 th{
text-align:center;
border:1px solid steelblue;
}
.tb2 td{
padding:5px;
}
</style>
<body>
<table width="100%" border="0" cellpadding="0" cellspacing="0">
<tr>
<td height="35"><!--begin menu--><!--end menu--></td>
</tr>
<tr>
<td height="500" valign="top" style="padding:10px;">
<script src="graph/highcharts.js"></script>
<script src="graph/exporting.js"></script>
<div id="container" style="min-width: 310px; height: 600px; margin: 0 auto;"></div>
</td>
</tr>
</table>
</body>
</html>
Tag PHP, MySQL, HTML/CSS, JavaScript, Ajax, jQuery
Date :
2017-04-03 15:04:10By :
moo@tooView :
751Reply :
3
แทนค่ะปีไป แต่มันไม่เจอค่า พอลองแทนค่าปีไป จะแสดงผล
Date :
2017-04-03 15:41:18By :
apisitp
ลอง Echo ตัว Query แล้วเอาไปรันใน phpMyAdmin ครับ
Date :
2017-04-03 17:54:14By :
mr.win
Server 01
Load balance : Server 01
