Question

我在数据库中有3个列的表'mags_sold'：

date，name of magazine，number sold。

每天我都会运行一条cronjob，其中插入3条新线，当天就是杂志的名称和那天销售的数量。（我有3本杂志）。

我想在表格中显示数据。

我已经设置了4列。 Date和mag1，mag2，mag3。

我有一个查询SELECT * FROM 'mags_sold'，但我不知道如何遍历结果以显示它们，因此日期只显示一次，销售数字显示在每个标题下。

<?php
$values = $wpdb->get_results($wpdb->prepare("SELECT * FROM 'mags_sold'"));
?>
<table id="email_subscription">
<tr>
<th>Date</th>
<th>mag1</th>   
<th>mag2</th>
<th>mag3</th>
</tr>
<?php
foreach($values as $email_signup){
$id = $email_signup->id;
$list_name = array($email_signup->list_name);
$date = $email_signup->date;
echo       '<tr><td>' . $date . '</td>';
$sold_count = array($email_signup->sold_count);
       echo '<td>' . $sold_count[0] . '</td>';
       echo '<td>' . $sold_count[1] . '</td>';  
       echo '<td>' . $sold_count[2] . '</td>';  
 echo    '</tr>';
 }
 ?>   
</table>

但是此代码显示了每个日期和销售计数的单独行。

var_dump使用代码：

array (size=3)
  0 => 
    object(stdClass)[637]
      public 'date' => string '2015-05-05' (length=10)
      public 'B' => string '132467' (length=6)
      public 'D' => string '2' (length=1)
      public 'F' => string '15330' (length=5)
  1 => 
    object(stdClass)[549]
      public 'date' => string '2015-05-04' (length=10)
      public 'B' => string '132467' (length=6)
      public 'D' => string '2' (length=1)
      public 'F' => string '15330' (length=5)
  2 => 
    object(stdClass)[547]
      public 'date' => string '2015-05-03' (length=10)
      public 'B' => string '132467' (length=6)
      public 'D' => string '2' (length=1)
      public 'F' => string '15330' (length=5)
array (size=3)
  0 => 
    object(stdClass)[548]
      public 'date' => string '2015-05-05' (length=10)
      public 'B' => string '132462' (length=6)
      public 'D' => string '98577' (length=5)
      public 'F' => string '15343' (length=5)
  1 => 
    object(stdClass)[467]
      public 'date' => string '2015-05-04' (length=10)
      public 'B' => string '132462' (length=6)
      public 'D' => string '98577' (length=5)
      public 'F' => string '15343' (length=5)
  2 => 
    object(stdClass)[468]
      public 'date' => string '2015-05-03' (length=10)
      public 'B' => string '132462' (length=6)
      public 'D' => string '98577' (length=5)
      public 'F' => string '15343' (length=5)
array (size=3)
  0 => 
    object(stdClass)[547]
      public 'date' => string '2015-05-05' (length=10)
      public 'B' => string '132468' (length=6)
      public 'D' => string '6' (length=1)
      public 'F' => string '15349' (length=5)
  1 => 
    object(stdClass)[549]
      public 'date' => string '2015-05-04' (length=10)
      public 'B' => string '132468' (length=6)
      public 'D' => string '6' (length=1)
      public 'F' => string '15349' (length=5)
  2 => 
    object(stdClass)[637]
      public 'date' => string '2015-05-03' (length=10)
      public 'B' => string '132468' (length=6)
      public 'D' => string '6' (length=1)
      public 'F' => string '15349' (length=5)

Answer 1

您需要的是一个MYSQL查询，它为每个唯一日期返回以下行：

date         magazine_1      magazine_2      magazine_3
2015-04-05   *number_sold*  *number_sold*    *number_sold*

所以 number_sold 等于name_of_magazine等于magazine_1,2 3等的销售数量的COUNT。

当你的查询从mags_sold表中选择所有内容时，foreach正在完全按照它应该单独返回每一行的方式。

见下面的MYSQL小提琴： -

SQL FIDDLE

<强>解

好的，你可以试试这个： -

<?php
$values = $wpdb->get_results($wpdb->prepare("SELECT DISTINCT `date` FROM mags_sold"));
?>
<table id="email_subscription">
<tr>
<th>Date</th>
<th>mag1</th>   
<th>mag2</th>
<th>mag3</th>
</tr>
<?php

    foreach($values as $email_signup){

        $date = $email_signup->date;

        $sold_values = $wpdb->get_results($wpdb->prepare(
        "SELECT DISTINCT `date`,
        (select number_sold from mags_sold where name_of_magazine = 'weekly news' AND date = $date) as 'weekly_news',
        (select number_sold from mags_sold where name_of_magazine = 'monthly news' AND date = $date) as 'monthly_news',
        (select number_sold from mags_sold where name_of_magazine = 'todays news' AND date = $date) as 'todays_news'
        from mags_sold  
        WHERE date = $date "));

        echo '<tr>';
            echo '<td>' . $date . '</td>';
            echo '<td>' . $sold_values->weekly_news. '</td>';
            echo '<td>' . $sold_values->monthly_news . '</td>';  
            echo '<td>' . $sold_values->todays_news . '</td>';  
        echo '</tr>';

    }
 ?>   
</table>

将虚拟杂志的名称，即weekly_news, monthly_news替换为杂志的名称，这应该有效。

在表格中显示查询结果

1 个答案: