我有一個課程數據庫,我可以按照我想要的方式返回課程。我無法確定如何超鏈接課程的名稱,以便我只能在新頁面上顯示課程數據(full.php)。我想獲取課程ID並在full.php中使用這個作爲我的WHERE語句來顯示其他字段。echo中的超鏈接將數據傳遞到新的PHP頁面
下面是我曾嘗試:
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沒有在你的腳本中定義,嘗試用替換它:$行[ 'courseID'] –
謝謝!一旦我找到替代courseId的名字它的作品 - 我對PHP很新穎(我們將其稱爲學習曲線) –