ดึงข้อมูลจากตารางมาแสดง แล้วต้องการนำค่าที่ได้จากตารางมาคำนวนเงื่่อนไข
Code (PHP) <?require("Function/PageScript.php");?>
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<table width="100%" border="1" cellspacing="1" cellpadding="3">
<tr>
<th scope="col">COURSEID</th>
<th scope="col">COURSENAME</th>
<th scope="col">GROUPSUBJECTSNAME</th>
<th scope="col">COURSENAME</th>
<th scope="col">UNIT</th>
<th scope="col">TEACHERNAME</th>
<th scope="col">GRADE</th>
</tr>
<?php
$conn = mysql_connect("localhost", "root", "1234"); //เชื่อมต่อกับฐานข้อมูล
$objDB = mysql_select_db("ge");//เลือกใช้ฐานข้อมูลชื่อ company
$objDB = mysql_query("SET NAMES utf8");
$sql ="SELECT
registration.COURSEID,
cur_regcourse.COURSENAME,
sub_group.GROUPSUBJECTSNAME,
cur_regcourse.UNIT,
registration.STUDENTID,
studen.STUDENTNAME,
teacher.TEACHERNAME,
registration.GRADE
FROM
((sub_group
Inner Join cur_regcourse ON sub_group.GROUPSUBJECTSID = cur_regcourse.GROUPSUBJECTSID)
Inner Join studen
Inner Join registration ON '' = '' AND studen.STUDENTID = registration.STUDENTID AND cur_regcourse.COURSEID = registration.COURSEID)
Inner Join teacher ON teacher.TEACHERID = cur_regcourse.TEACHER
WHERE
registration.STUDENTID='54104410102'
";
$objQuery = mysql_query($sql) or die ("Error Query [".$sql."]");
$Num_Rows = mysql_num_rows($objQuery);
$Per_Page = 10; // Per Page
$Page = $_GET["Page"];
if(!$_GET["Page"])
{
$Page=1;
}
$Prev_Page = $Page-1;
$Next_Page = $Page+1;
$Page_Start = (($Per_Page*$Page)-$Per_Page);
if($Num_Rows<=$Per_Page)
{
$Num_Pages =1;
}
else if(($Num_Rows % $Per_Page)==0)
{
$Num_Pages =($Num_Rows/$Per_Page) ;
}
else
{
$Num_Pages =($Num_Rows/$Per_Page)+1;
$Num_Pages = (int)$Num_Pages;
}
$sql .=" order by registration.STUDENTID ASC LIMIT $Page_Start , $Per_Page";//
$objQuery = mysql_query($sql);
?>
<?
//แสดงรหัส ชื่อ ของนักศึกษาที่เรียกค้นหา
if ($e==0){
$sql = "select * from studen where studentid ='54104410102'";
$rstudent = mysql_query($sql);
$ostudent = mysql_fetch_array($rstudent);
echo "<tr bgcolor=#dddddd> $ostudent[STUDENTID] $ostudent[STUDENTNAME]";}
while ($data = mysql_fetch_array($objQuery))
{
// ส่วนการสลับสีของตาราง
$i=0;
$i++;
if($i%2==0)
{
$bg = "#CCCCCC";
}
else
{
$bg = "#FFFFFF";
}
?>
<tr bgcolor="<?=$bg;?>">
<td><?echo $data['COURSEID']?></td>
<td><?echo $data['COURSENAME']?></td>
<td><?echo $data['GROUPSUBJECTSNAME']?></td>
<td><?echo $data['COURSENAME']?></td>
<td><?echo $data['UNIT']?></td>
<td><?echo $data['TEACHERNAME']?></td>
<td><?echo $data['GRADE']?></td>
</tr>
<?
}
?>
</table>
Total <?= $Num_Rows;?> Record
<?
$pages = new Paginator;
$pages->items_total = $Num_Rows;
$pages->mid_range = 10;
$pages->current_page = $Page;
$pages->default_ipp = $Per_Page;
$pages->url_next = $_SERVER["PHP_SELF"]."?QueryString=value&Page=";
$pages->paginate();
echo $pages->display_pages()
?>
<?
mysql_close($conn);
?>
</body>
</html>
Tag PHP, MySQL
ประวัติการแก้ไข
Date :
2012-01-09 21:36:45By :
motokoView :
1561Reply :
3
แล้วคุณใช้การคิดเกรดแบบไหนหละ
Date :
2012-01-09 21:48:00By :
ozma
ตรง loop while ที่แสดงข้อมูลแต่ละ record
Date :
2012-01-10 08:04:19By :
amuropao08
ทำได้ล่ะครับ
Date :
2012-01-10 15:12:34By :
motoko
Server 00
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