我有一个课程数据库,我可以按照我想要的方式返回课程。我无法确定如何超链接课程的名称,以便我只能在新页面上显示课程数据(full.php)。我想获取课程ID并在full.php中使用这个作为我的WHERE语句来显示其他字段。echo中的超链接将数据传递到新的PHP页面
下面是我曾尝试:
echo "<table>
<tr>
<th>Course ID</th>
<th>Course Name</th>
<th>Provider Type</th>
<th>Audience</th>
<th>Provider Track</th>
<th>Course Delivery</th>
</tr>";
while($row = mysql_fetch_array($result)) {
echo "<tr>";
echo "<td>" .$row['courseID'] . "</td>";
echo "<td> <p> <strong>Course Name:</strong><a href=Full.php?name='".$courseID."'>". $row['courseName'] ."</a> </p>
<p> <strong>Course Number:</strong>".$row['courseNumber']." </p
<p> ".$row['courseDescription'] . " </p>
<p> <strong>Course Length:</strong> " .$row['courseLength'] . " </p>
</td>";
echo "<td>" .$row['courseProviderType'] . "</td>";
echo "<td>" .$row['courseAudience'] . "</td>";
echo "<td>" .$row['courseTrack'] . "</td>";
echo "<td>" .$row['courseDelivery'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysql_close($con);
?>
". $row['courseName'] ." 取代它 ". $row['courseName'] ." –
$ courseID没有在你的脚本中定义,尝试用替换它:$行[ 'courseID'] –
谢谢!一旦我找到替代courseId的名字它的作品 - 我对PHP很新颖(我们将其称为学习曲线) –