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回复 #13 wildlily980 的帖子
stu_num的值在select返回后就会取得了,这是因为在select语句中“as stu_num”得到的,其他地方都是这样做的,我实在想不通为什么这个地方会有问题。
- for($i=0;$i<$dd;$i++) {
- $row = mysql_fetch_array($result);
-
- /* Get student number of each course */
- //$query = "Select count(*) as stu_num from selected where Course_id='$row[Course_id]'";
- //$query = "Select count(*) as stu_num from selected where Course_id=".$row['Course_id'];
- $query = "Select count(*) as stu_num from selected where Course_id='{$row['Course_id']}'";
- $res = mysql_query($query) or die("Invalid query: " . mysql_error());
- /********************************************************/
- echo $query, "\n", $row['Course_id'], "\n", $res['stu_num'], "\n", "xxx
- /********************************************************/
- echo "<tr bgcolor=\"#CCFF99\"><td>".$row["Course_id"]."</td><td>".$row["Course_name"]." </td><td>".$row["Teacher_name"]. "</td><td>".$row["Course_time"]."</td><td>".$row["Course_place"]."</td><td>".$row["Credit"]."</td><td>".$stu_num."</td><td><a href=\"delete.php?courseid=".$row["Course_id"]."\">³·Ïû</a></td></tr>";
- }
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修改过的部分如上所示,还是没有值。页面显示如下:
- Select count(*) as stu_num from selected where Course_id='1' 1 xxx
- Select count(*) as stu_num from selected where Course_id='3' 3 xxx
- Select count(*) as stu_num from selected where Course_id='2' 2 xxx
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[ 本帖最后由 NewCore 于 2009-2-22 09:22 编辑 ] |
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