【发布时间】:2012-12-09 04:17:58
【问题描述】:
我有两张桌子
表1
HTNO SUBJECTCODE INTERNALS EXTERNALS TOTAL
1 s1 20 58 78
1 s2 15 20 35
1 s3 10 60 70
2 s1 10 20 30
2 s2 12 30 42
2 s3 15 55 70
.
.
.
so on up to N
表 2
SUBJECTCODE SUBJECT NAME
s1 MATHS
s2 SCIENCE
s3 SOCIAL
我会给学生一张表格让学生输入门票号码
如果学生在表格中输入 1,那么结果应该是
Subjectcode SubjectName Internals Externals Total
s1 Maths 20 58 78
s2 Science 15 20 35
s3 Social 10 60 70
上面应该是输出
但是在这里我无法从结果中的 Table2 中检索 SubjectName
这是我正在使用的代码
<?PHP
$userInputEntities = htmlentities($userInput);
echo $userInputEntities;
$username = "admin";
$password = "123456";
$database = "test";
$server = "localhost";
$db = new PDO ("mysql:host=$server;dbname=$database", "$username", "$password");
if ($db) {
$id = $_GET['id'];
$SQL = $db->prepare("SELECT * FROM Table1 WHERE htno = :id");
$SQL -> execute(array(':id'=>$id));
$n = $SQL->rowCount();
echo "
<center><table class='dynamic styled with-prev-next' data-table-tools='{'display':true}' align=center>
<thead>
<tr>
<TH class='table-header dark' scope='col'>SUBJECT CODE</TH>
<TH class='table-header dark' scope='col'>SUBJECT NAME</TH>
<TH class='table-header dark' scope='col'>INTERNALS</TH>
<TH class='table-header dark' scope='col'>EXTERNALS</TH>
<TH class='table-header dark' scope='col'>TOTAL</TH>
</tr></thead><center>";
while ($db_field = $SQL->fetch(PDO::FETCH_ASSOC)) {
echo "<tr><tbody>";
echo "<td align=center>" . $db_field['SubjectCode'] . "</td>";
echo "<td align=center>" . $db_field['Internals'] . "</td>";
echo "<td align=center>" . $db_field['Externals'] . "</td>";
echo "<td align=center>" . $db_field['Total'] . "</td>";
echo "</tbody></tr>";
}
使用此代码,我无法获取学生特定主题代码的主题名称 实际上我没有写任何代码来从 Table2 中检索配音名称,我不知道如何写它
请帮帮我
【问题讨论】: