Question

大家好，

   $sql = " SELECT prd.id as id,prd.weight as wgh,
           prd.srpid as srno,prd.dollarval as dolval,catid.id
           FROM productname as prd
           INNER JOIN catid as caid ON prd.catid = caid.id
           Where 1=1
           ORDER BY sid ASC "
  $sqlquery = mysql_query($sql);

其中id为主要，自动递增

以上价值低于：

id | wgh | srno| dolval 
1    1.05   1     155
2    1.00   3     300
3    0.67   2     170
4    0.01   3     140
5    0.20   2     50
6    0.35   1     300
7    2.04   4     190
8    3.0    5     200

做完PHP逻辑后，我得到以下结果作为输出（显示）：

echo"<table  width=\"100%\" border=\"1\">
         <tr>
            <td align=\"center\" valign=\"top\"><strong>Sr.No</strong></td>
            <td align=\"center\" valign=\"top\"><strong>Cur Wg</strong></td>
            <td align=\"center\" valign=\"top\"><strong>Dollar</strong></td>
         </tr>";

 $srno = 1;
 $pid = null;
 $sumnew = 0;
 $sumdollor = 0;

while($row = mysql_fetch_array($sqlquery)){

       $rowval =  $row['srno'];  // **** result is on this value.....

        if ($rowval == $pid || $pid === null){

           echo "<tr>";

           if($pid != $rowpkts){
              echo'<td align="center" valign="top">'.$srno++.'</td>';
           }else{
              echo'<td align="center" valign="top"></td>';
           }

           echo'<td align="center"  valign="top">'.number_format($row["wgh"],2,'.','').'</td>';
           echo'<td align="center" valign="top">'.round($row["dolval"],0).'</td>';

          echo "</tr>";  

          $pid = $rowval;        
          $sumnew += number_format($row["wgh"],2,'.','');
          $sumdollor+= round($row["dolval"],0);

      }else{

       echo "<tr>
         <td></td> 
         <td align=\"center\" valign=\"top\"><strong>".number_format($sumnew,2,'.','')."</strong></td>
         <td align=\"center\" valign=\"top\"><strong>".$sumdollor."</strong></td>
       echo "</tr>";

      <tr><td colspan='3' height='18px'></td></tr>";


          echo "<tr>";

           if($pid != $rowpkts){
              echo'<td align="center" valign="top">'.$srno++.'</td>';
           }else{
              echo'<td align="center" valign="top"></td>';
           }

           echo'<td align="center"  valign="top">'.number_format($row["wgh"],2,'.','').'</td>';
           echo'<td align="center" valign="top">'.round($row["dolval"],0).'</td>';

          echo "</tr>";  

          $sumnew = number_format($row["wgh"],2,'.','');
          $sumdollor = round($row["dolval"],0);
          $pid = $rowval;

      }
}

       echo "<tr>
         <td></td> 
         <td align=\"center\" valign=\"top\"><strong>".number_format($sumnew,2,'.','')."</strong></td>
         <td align=\"center\" valign=\"top\"><strong>".$sumdollor."</strong></td>
       echo "</tr>";

 echo"</table>";

从上面的PHP代码我得到以下结果：

Sr.No   Wgh     Dollar
 1      1.05      155
        0.35      300
-----------------------
                  455  //( 155+300=455)
------------------------
2       0.67      170
        0.20       50
------------------------
                  220
------------------------
3       1.00      300
        0.01      140
------------------------
                  440
------------------------
4       2.04      190
------------------------
                  190
------------------------
5       3.00      200
------------------------
                  200

但是我需要在第一次美元的总和中显示上述结果，那么总和的更高的美元（即445,440,220 ......等）应该先来，然后等等。 .. 下面需要

Sr.No Wgh Dollar 1 1.05 155 0.35 300 ----------------------- 455 //( 155+300=455) ----------------------- 2 1.00 300 0.01 140 ----------------------- 440 ----------------------- 3 0.67 170 0.20 50 ------------------------ 220 ------------------------ 4 3.00 200 ------------------------ 200 ------------------------ 5 2.04 190 ------------------------ 190

我如何获得以上结果请帮助我...如果有任何其他方式让我知道将尝试更改上述格式但需要结果...

编辑：如果某人有 groupby 的解决方案..这对我有帮助..但我没有使用分组，因为我需要显示所有8个值... 感谢

Answer 1

您可以使用此SQL，它会生成一个接近您想要生成的输出（和顺序）：

SELECT     prd.srpid as srno, 
           prd.weight as wgh,
           prd.dollarval as dolval,
           s.dollarsum
FROM       productname as prd
INNER JOIN catid as caid ON prd.catid = caid.id
INNER JOIN (select   srpid,
                     sum(dollarval) dollarsum
           from      productname
           group by  srpid) s ON s.srpid = prd.srpid
ORDER BY   s.dollarsum DESC,
           prd.srpid DESC,
           prd.weight DESC

这是SQL fiddle。

生成的输出是：

+-------+--------+-----------+-----------+
| srpid | weight | dollarval | dollarsum |
+-------+--------+-----------+-----------+
| 1     |  1.05  |  155      | 455       |
| 1     |  0.35  |  300      | 455       |
| 3     |  1     |  300      | 440       |
| 3     |  0.01  |  140      | 440       |
| 2     |  0.67  |  170      | 220       |
| 2     |  0.2   |  50       | 220       |
| 5     |  3     |  200      | 200       |
| 4     |  2.04  |  190      | 190       |
+-------+--------+-----------+-----------+

因此，在PHP中，您需要获取最后一列中返回的值，并在每次进入新部分时将其作为单独的“小计”行输出。

我不会那么远，但您甚至可以让SQL将这些小计作为行生成：

SELECT     prd.srpid as srno, 
           prd.weight as wgh,
           prd.dollarval as dolval,
           s.dollarsum
FROM       productname as prd
INNER JOIN catid as caid ON prd.catid = caid.id
INNER JOIN (select   srpid,
                     sum(dollarval) dollarsum
           from      productname
           group by  srpid) s ON s.srpid = prd.srpid
UNION
select     srpid,
           null,
           'sum:',
           sum(dollarval) dollarsum
from       productname
group by   srpid 
ORDER BY   4 DESC,
           1 DESC,
           2 DESC

输出：

+-------+--------+-----------+-----------+
| srpid | weight | dollarval | dollarsum |
+-------+--------+-----------+-----------+
| 1     |  1.05  |  155      | 455       |
| 1     |  0.35  |  300      | 455       |
| 1     |        |  sum:     | 455       |
| 3     |  1     |  300      | 440       |
| 3     |  0.01  |  140      | 440       |
| 3     |        |  sum:     | 440       |
| 2     |  0.67  |  170      | 220       |
| 2     |  0.2   |  50       | 220       |
| 2     |        |  sum:     | 220       |
| 5     |  3     |  200      | 200       |
| 5     |        |  sum:     | 200       |
| 4     |  2.04  |  190      | 190       |
| 4     |        |  sum:     | 190       |
+-------+--------+-----------+-----------+

SQL fiddle

没有groupby的ORDER BY SUM VALUE - php mysql

1 个答案: