Question

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

我正在提供一个表格供用户输入idno并提交

如果用户在表单中输入“1”并提交，则输出应为

        Total Marks
            106

，即表1中Idno 1的所有商标的总和+表2中Idno 1的所有商标的总和

（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

Answer 1

WHERE htno='$id) AS tech4 "; 
               ^----here is the problem  should be htno='$id'

Answer 2

在WHERE子句之后，您似乎在联合的两个部分中都缺少 GROUP by htno 。或者您可以考虑从union的SELECT子句中删除 htno 字段。

 select sum(sm) from
   (select Sum(marks) sm from sums1 where idno=1
   union
   select Sum(marks) sm from sums2 where idno=1 ) ss

在mysql上测试它并且它可以正常工作

Answer 3

SELECT SUM(tech) AS tech4 FROM (
(SELECT SUM(tm) AS tech FROM jbit WHERE htno='$id')
UNION ALL
(SELECT SUM(tm) AS tech FROM hmm WHERE htno='$id')
) t1

MySQL - 一个表中一列与另一个表中其他列的总和

3 个答案: