Question

我有一个表，其中dat_eb_registrants的值为行（例如$row[1]）和水平数组中的值，从dat_eb_field_values中提取，我想将它们拆分为我可以把所有东西都放到桌子上，我想要它。

如何将数据放入我的表格中：

$count = 0;
while ($row = mysql_fetch_row($result))
{
echo "<tr>";
$myArray[] ="<tr><td>" . $row[9] . "</td><td>"; echo $myArray[$count];
$count++;
echo "</tr>"; 
}

如何从数据库中提取所有数据（是的，我知道它已经过时了）：

SELECT dr.id, dr.first_name, dr.last_name, dr.email, dr.comment, dr.amount, dr.published, dr.transaction_id, dr.register_date, GROUP_CONCAT(df.field_value SEPARATOR '</td><td>') 
   FROM dat_eb_registrants dr 
   LEFT JOIN dat_eb_field_values df 
    ON dr.id=df.registrant_id
   WHERE `event_id` >= 20 AND `event_id` <= 25
   GROUP BY dr.id
   ORDER BY '".$sort."', '".$ascdsc."'

现在，我想在第一个表（例如$row[1]）和（.eg。$row[2]）中垂直放置一些行，位于数组的中间。我怎么能这样做？

因为数组一次填充我的表，并使用$rows，你可以简单地告诉哪些行显示在哪里......

这就是我想要的（代码可能不正确）：

while ($row = mysql_fetch_row($result)) {
  echo "<tr>";
echo "<td>"'.$row_table_1[1]'"</td>";
echo "<td>"'.$row_table_1[2]'"</td>";
echo "<td>"'.$row_table_2[35]'"</td>";
echo "<td>"'.$row_table_2[45]'"</td>";
echo "<td>"'.$row_table_1[5]'"</td>";
echo "<td>"'.$row_table_2[6]'"</td>";
echo "</tr>";
}

我以前做过：

echo "<td>"; $result24 = mysql_query("SELECT field_id, field_value FROM dat_eb_field_values WHERE (field_id = 88) AND (registrant_id = $row[0])"); $r24 = mysql_fetch_row($result24); echo $r24[1]; echo "</td>";
echo "<td>"; $result25 = mysql_query("SELECT field_id, field_value FROM dat_eb_field_values WHERE (field_id = 57) AND (registrant_id = $row[0])"); $r25 = mysql_fetch_row($result25); echo $r25[1]; echo "</td>";
echo "<td>" . $row[5] . "</td>";
echo "<td>" . $row[6] . "</td>";

我想这样做：

echo "<td>" . $rowfromsecondtable[1] . "</td>";
echo "<td>" . $rowfromsecondtable[2] . "</td>";
echo "<td>" . $row[5] . "</td>";
echo "<td>" . $row[6] . "</td>";

（使用我上面给出的代码）

预览dat_eb_registrants：

    |   id   |   first_name   |   last_name   |   email               |
------------------------------------------------------------------------
    |    1   |      Mike      |     Doe       |   mikedoe@hotmail.com |
    |    2   |      John      |     Smith     |   j_smith@hotmail.com |

预览dat_eb_field_values：

field 1 = fav.sport
field 2 = fav. color
field 3 = fav. food

    |   registrant_id   |   field_id  |   field_value    |
----------------------------------------------------------
    |    1              |      1      |     tennis       |
    |    1              |      2      |     green        |
    |    1              |      3      |     spagetti     |
    |    2              |      1      |     hockey       |
    |    2              |      2      |     red          |
    |    2              |      3      |     fish         |

我需要：

first_name  |   id   |   fav.sport   |   last_name   |   fav.food   |
---------------------------------------------------------------------
Mike        |    1   |      Tennis   |     Doe       |   spagetti   |
John        |    2   |      Hockey   |     Smith     |   fish       |

Answer 1

也许我错过了一些东西，但为什么不在SQL中执行此任务。这基本上是 pivot 。 MySQL没有数据透视表，但您可以使用带有CASE语句的聚合函数：

select r.first_name,
  r.id,
  r.last_name,
  max(case when f.field_id =1 then f.field_value else null end) As FavSport,
  max(case when f.field_id =2 then f.field_value else null end) As FavColor,
  max(case when f.field_id =3 then f.field_value else null end) As FavFood
from dat_eb_registrants r
left join dat_eb_field_values f
  on r.id = f.registrant_id
group by r.first_name, r.id, r.last_name
order by r.id

请参阅SQL Fiddle with Demo

查询的结果是您想要的输出：

| FIRST_NAME | ID | LAST_NAME | FAVSPORT | FAVCOLOR |  FAVFOOD |
----------------------------------------------------------------
|       Mike |  1 |       Doe |   tennis |    green | spagetti |
|       John |  2 |     Smith |   hockey |      red |     fish |

或者您可以在dat_eb_field_values表上使用多个联接：

select r.first_name,
  r.id,
  r.last_name,
  fSport.field_value FavSport,
  fColor.field_value FavColor,
  fFood.field_value FavFood
from dat_eb_registrants r
left join dat_eb_field_values fSport
  on r.id = fSport.registrant_id
  and fSport.field_id = 1
left join dat_eb_field_values fColor
  on r.id = fColor.registrant_id
  and fColor.field_id = 2
left join dat_eb_field_values fFood
  on r.id = fFood.registrant_id
  and fFood.field_id = 3
order by r.id

见SQL Fiddle with Demo。它产生相同的结果。

将数组拆分为行

1 个答案: