Question

select查询在特定日期输出每个basestation_id的值，由avg排序。我想从Mysql查询结果中显示php中的表，如下所示。

basestation_id  12-Jun-18   13-Jun-18   14-Jun-18   15-Jun-18   16-Jun-18
    5            value1      value2     value3      value4      value5
    3            value1      value2     value3      value4      value5
    4            value1      value2     value3      value4      value5
    1            value1      value2     value3      value4      value5
    2            value1      value2     value3      value4      value5

但代码显示结果为

date            12-Jun-18   12-Jun-18   12-Jun-18   12-Jun-18   13-Jun-18 ...    
basetstaion_id  5             3          4            1           2      ...
values          value1      value1      value1      value1     value2   ...

PHP： -

<?
foreach($result as $key => $row) {
   foreach($row as $field => $value) {
       $recNew[$field][] = $value;
   }
}
//This creates a new array composed/transposed with the field names as keys 
//and the "rowed" values as sub-arrays.

echo "<table>\n";

foreach ($recNew as $key => $values) // For every field name (basestation_id, 
// , dates, values)
{
  echo "<tr>\n"; // start the row
  echo "\t<td>" . $key . "</td>\n" ; // create a table cell with the field 
name
    foreach ($values as $cell) // for every sub-array iterate through all 
values
    {
       echo "\t<td>" . $cell . "</td>\n"; // write cells next to each other
    }
echo "</tr>\n"; // end row

}

echo "</table>";
?>

示例数据： -

delivery_date   col_head    basestation_id  receptions  daily_avg_stats
16-Jun-18      Sat 16 Jun   152              0           under_avg
16-Jun-18      Sat 16 Jun   13               0           under_avg
16-Jun-18      Sat 16 Jun   140              0           under_avg
16-Jun-18      Sat 16 Jun   12               1093        under_avg
16-Jun-18      Sat 16 Jun   9                0           under_avg
16-Jun-18      Sat 16 Jun   1                50          under_avg
16-Jun-18      Sat 16 Jun   5                250         under_avg
.   .   .      .                                         
15-Jun-18      Fri 15 Jun   152              0           under_avg
15-Jun-18      Fri 15 Jun   13               0           under_avg
15-Jun-18      Fri 15 Jun   140              0           under_avg
15-Jun-18      Fri 15 Jun   12               1734        under_avg
15-Jun-18      Fri 15 Jun   9                0           under_avg
15-Jun-18      Fri 15 Jun   1                0           under_avg

SQL查询以＆＃34;在avg＆＃34;下显示结果首先列出的值： -

SELECT 
    *
FROM
    (SELECT 
        DATE(r.ad_date) AS delivery_date,
            DATE_FORMAT(r.ad_date, '%a %d %b') AS col_head,
            r.basestation_id,
            r.receptions,
            (CASE
                WHEN r.receptions > FLOOR(avg.monthly_receptions_avg * 1.3) THEN 'over_avg'
                WHEN r.receptions < FLOOR(avg.monthly_receptions_avg * 0.7) THEN 'under_avg'
            END) AS daily_avg_stats
    FROM
        receptions r
    JOIN receptions_avg avg ON r.basestation_id = avg.basestation_id
    WHERE
        r.ad_date > DATE_SUB(CURDATE(), INTERVAL 30 DAY)
            AND avg.ad_date = CURDATE() - INTERVAL 1 DAY
    ORDER BY basestation_id , delivery_date) abc
ORDER BY delivery_date DESC , FIELD(abc.daily_avg_stats, 'under_avg') DESC

Answer 1

使用此代码

<?
foreach($result as $key => $row) {
   foreach($row as $field => $value) {
       $recNew[$field][] = $value;
   }
}
//This creates a new array composed/transposed with the field names as keys 
//and the "rowed" values as sub-arrays.

echo "<table>";

foreach ($recNew as $key => $values) // For every field name (basestation_id, 
// , dates, values)
{
  echo "<tr>"; // start the row
  echo "<td>" . $key . "</td>" ; // create a table cell with the field 
name
    foreach ($values as $cell) // for every sub-array iterate through all 
values
    {
       echo "<td>" . $cell . "</td>"; // write cells next to each other
    }
echo "</tr>"; // end row

}

echo "</table>";
?>

PHP根据id和group by datewise显示来自MySQL查询的行

1 个答案: