我希望讲师选择他们的名字和一年,以便显示该年的统计表。我有不同的讲师。我为每位讲师创建了不同的表格。当我选择第一个讲师时,系统显示正确的表格,但是当我选择第二个讲师系统时,它会显示第一个讲师的表格。另外,当我选择lec1
和2006
时,它会显示lec1表,但会显示2005年的分数。
form.php的
<form name="myform" action="lecturer.php" method="POST" >
<b>Lecturers:<b/>
<select name="lecturer">
<option value="Choose">Please select..</option>
<option value="lec1">Lec 1</option>
<option value="lec2">Lec 2</option></select><br/><br/>
<b>Year:<b/>
<select name="year">
<option value="Choose">Please select..</option>
<option value="2005">2005</option>
<option value="2006">2006</option>
<option value="2007">2007</option>
<option value="2008">2008</option>
<option value="2009">2009</option>
<option value="2010">2010</option></select><br/><br/>
<br/>
<input type="submit" name="submit" value="Submit">
<input type="reset" name="reset" value="Clear">
</form>
lecturer.php
<?php
switch($_POST['lecturer']&& $_POST['year']){
case 'lec1' && '2005':
include 'href="/../../statistics/lec12005.php"';
break;
case 'lec1' && '2006':
include 'href="/../../statistics/lec12006.php"';
break;
case'lec2' && '2006':
include 'href="/../../statistics/lec22006"';
}
?>
lec1.php
<?php
include 'connect.php';
$sql="SELECT unit_name,a1,a2,a3,l1,l2,r1,r2,u1,u2,u3 FROM lec1 WHERE year=2005";
$result=mysql_query($sql);
?>
<html>
<body>
<table width="900" border="1" cellspacing="1">
<tr>
<td>Unit Name</td>
<td>A1 </td>
<td>A2 </td>
<td>A3 </td>
<td>L1 </td>
<td>L2 </td>
<td>R1 </td>
<td>R2 </td>
<td>U1 </td>
<td>U2 </td>
<td>U3 </td>
</tr>
<?php
while($unit=mysql_fetch_assoc($result)){
echo "<tr>";
echo "<td>".$unit['unit_name']."</td>";
echo "<td>".$unit['a1']."</td>";
echo "<td>".$unit['a2']."</td>";
echo "<td>".$unit['a3']."</td>";
echo "<td>".$unit['l1']."</td>";
echo "<td>".$unit['l2']."</td>";
echo "<td>".$unit['r1']."</td>";
echo "<td>".$unit['r2']."</td>";
echo "<td>".$unit['u1']."</td>";
echo "<td>".$unit['u2']."</td>";
echo "<td>".$unit['u3']."</td>";
echo "</tr>";
}
?>
</table>
</body>
</html>
答案 0 :(得分:1)
你在WHERE子句中有2005年的硬编码。它将始终返回2005年的结果。
另外,
将切换句更改为:
switch($_POST['lecturer'] . $_POST['year']){
case 'lec1' . '2005':
include 'href="/../../statistics/lec12005.php"';
break;
case 'lec1' . '2006':
include 'href="/../../statistics/lec12006.php"';
break;
case'lec2' . '2006':
include 'href="/../../statistics/lec22006"';
}
答案 1 :(得分:1)
您的switch
和include
逻辑完全错误。
您应该验证$_POST
变量(非常重要!)并替换:
switch($_POST['lecturer']&& $_POST['year']){
case 'lec1' && '2005':
include 'href="/../../statistics/lec12005.php"';
break;
case 'lec1' && '2006':
include 'href="/../../statistics/lec12006.php"';
break;
case'lec2' && '2006':
include 'href="/../../statistics/lec22006"';
}
使用:
include "path/to/statistics/" . $_POST['lecturer'] . $_POST['year'] . ".php";
答案 2 :(得分:1)
这不是编写代码的好方法。而不是你做了什么,只需加上讲话文件:
<?php
include 'connect.php';
$year = $_POST['year'];
$lecturer = $_POST['lecturer']; // Don't forget to handle the SQL Injections ...
$years = array(
2005,
2006,
2007
);
$lecturers = array(
'lec1',
'lec2'
);
if (in_array($lecturer, $lecturers) && in_array($year, $years)) {
$sql = "SELECT unit_name,a1,a2,a3,l1,l2,r1,r2,u1,u2,u3 FROM $lecturer WHERE year=$year";
$result = mysql_query($sql);
}
else {
// Handle the error
}
?>
<html>
<body>
...