我有两张桌子
表 1 是
idno marks
1 12
1 13
1 22
2 32
2 35
2 11 and so on
表 2 是
idno marks
1 16
1 22
1 21
2 35
2 16
2 22 and so on
我正在为用户提供输入 idno 并提交的表单
如果用户在表单中输入“1”并提交,那么输出应该是
Total Marks
106
即表1中Idno 1的所有分数之和+表2中Idno 1的所有分数之和
(12+13+22)+(16+22+21) = 106
我正在使用以下代码
<form id="form" action="sum.php" method="post">
<td><p align="center"> IDNO : <input type="text" name="id" id="id" maxlength="10"></p></td>
<input type="submit" id="submit" class='btnExample' value="Click here to get your Result"
</form>
<?PHP
$user_name = "admin";
$password = "123456";
$database = "demo";
$server = "localhost";
$db_handle = mysql_connect($server, $user_name, $password);
$db_found = mysql_select_db($database, $db_handle);
if ($db_found) {
$id = mysql_real_escape_string($_POST['id']);
$add = "SELECT htno, SUM(tech)
FROM(SELECT htno, SUM(tm) AS tech FROM jbit WHERE htno='$id'
UNION ALL
SELECT htno, SUM(tm1) AS tech FROM hmm WHERE htno='$id') AS tech4 ";
$result3 = mysql_query($add);
echo "
<center><table id='mytable' cellspacing='0' border=3 align=center>
<tr>
<TH scope='col'>Total Marks</TH>
</tr><center>";
while ($row1 = mysql_fetch_assoc($result3)){
echo "<tr>";
echo "<td align=center>" . $row1['tech4']. "</td>";
echo "</tr>";
}
mysql_close($db_handle);
else {
print "Database NOT Found ";
mysql_close($db_handle);
}
但输出为空白
请帮帮我