Question

我有一张订单表。我希望用总天数填充另一张表。对于一年中的每一天，我想要总共一行。

我写了这个，我认为最大的问题是日期格式。

我问语法是否正确

我需要在这个例子中做一些事情，因为在完整版本中我需要更改一些变量：$somma_paid如果货币发生变化..

我使用“if”并从一年的第一天开始，然后使用strtotime

进行修改

我需要帮助来完成它。

最终结果将填充一个包含365行的新表格（data，totale，merce，servizi，numero）。每行总共有一天。

$date1 = 2013-01-01;
$somma_paid = 0;
$somma_products = 0;
$somma_shipping = 0;


$query55 = "SELECT id_order, total_paid, payment, total_products, id_currency, total_shipping, shipping_number, delivery_date FROM ps_orders WHERE delivery_date LIKE '%2013%'";
$result55 = mysql_query($query55) or die(mysql_error());

 while ($row55 = mysql_fetch_array($result55))
{
    $currency = $row55['id_currency'];
    $id = $row55['id_order'];
    $date = $row55['delivery_date'];

    if ($date == $date1)  {

        echo '
        <tr>
            <td style="width: 120px;">' . $id . '</td>
            <td style="width: 120px;">' . $row55['delivery_date'] . '</td>
            <td style="width: 170px;">' . $row55['payment'] . '</td>
            <td style="width: 100px;">' . $row55['total_paid'] . '</td>
            <td style="width: 100px;">' . $row55['total_products'] . '</td>
            <td style="width: 100px;">' . $row55['total_shipping'] . '</td>
            <td style="width: 158px;">' . $row55['shipping_number'] . '</td>
        </tr>   
        ';

        $somma_paid += $row55['total_paid'];
        $somma_products += $row55['total_products'];
        $somma_shipping += $row55['total_shipping'];
    }
    else 
    {
        $query1122 = "REPLACE INTO calendar (data, totale, merce, servizi, numero) VALUES('$date', '$somma_paid', '$somma_products', '$somma_shipping')";
        $result1122 = mysql_query($query1122) or die(mysql_error());

        $somma_paid = 0;
        $somma_products = 0;
         somma_shipping = 0;

        $date1 = strtotime($date."+ 1 day");
    }
}

Answer 1

您应该更改查询以获取已格式化的数据，每天使用一组来为一年中的每一天添加一行，将您的选择替换为：

SELECT date(delivery_date) as data, SUM(total_paid) as totale, SUM(total_products) as merce, SUM(total_shipping) as servizi, COUNT(DISTINCT id_order) as numero 
FROM ps_orders 
WHERE YEAR(delivery_date) = '2013' 
GROUP BY DAY(delivery_date) ORDER BY data

这样您就可以获得只需要导出的所有数据。（您可以使用INSERT INTO日历（数据，totale，merce，servizi，numero）SELECT ...）或在您的行上执行foreach并使用php插入来制作它，就像您已经在做的那样，处理“无效”日你没有任何数据的地方。

<强>更新

如果您需要在插入新表之前执行转换，您可以执行按日期选择，然后循环遍历行以在需要时转换信息。转换后将所有数据存储在使用给定数据作为键的数组中（同样，如果您想要一年中的所有日期，您可以使用for（）初始化数组并创建所有365个条目，即使某些条目仍为空

这是一个基于您如何获取和订购数据的代码的示例：

$query55 = "SELECT date(delivery_date) as data, total_paid as totale, total_products as merce, total_shipping as servizi, id_order as numero, id_currency as valuta 
FROM ps_orders 
WHERE YEAR(delivery_date) = '2013' 
ORDER BY data ASC";
$result55 = mysql_query($query55) or die(mysql_error());

$calendario_array = array();

while ($row = mysql_fetch_array($result55))
    {

    if(!isset($calendario_array[$row['data']]))
        $calendario_array[$row['data']] = array(
            'data' => $row['data'], //redundant but useful for fast insert
            'totale' => 0,
            'merce' => 0,
            'servizi' => 0,
            'numero' => 0

        );


    switch($row['valuta'])
    {
        case(1): //convert value based on your id_currency
            $row['merce'] = convertValue($row['merce']);
            $row['totale'] = convertValue($row['totale']);
            $row['servizi'] = convertValue($row['servizi']);
            break;
        case(2): //convert value based on your id_currency 
            $row['merce'] = convertValue($row['merce']);
            $row['totale'] = convertValue($row['totale']);
            $row['servizi'] = convertValue($row['servizi']);
            break;
    }
        $calendario_array[$row['data']]['numero'] = $calendario_array $row['data']]['numero'] + 1;
        $calendario_array[$row['data']]['merce'] = $calendario_array[$row['data']]['merce'] + $row['merce'];
        $calendario_array[$row['data']]['totale'] = $calendario_array[$row['data']]['totale'] + $row['totale'];
        $calendario_array[$row['data']]['servizi'] = $calendario_array[$row['data']]['merce'] + $row['servizi'];

    }

//$calendario_array now contains all your data, you can cycle on it to perform the insert
foreach($calendario_array as $row)
    //insert or update data in calendario table

更新v2

至于最后一部分，要在新表中插入数据，我建议执行从$calendar_array到查询字符串的转换，它将立即执行所有插入。这样我们就不会遇到性能问题。请记住，我不打算截断表格以清除以前的数据我会留给您选择。至于数据字段，我想你将它存储为日期时间，如果没有，你将不得不编辑数据以获得所需的格式。

$insert_string = array(); //lets create a clear array that will contain a string for each entry. this will represent our insert value row

foreach($calendario_array as $row){ //cycle through the calendario array, store the data as a VALUES ( ) string
    $insert_string[] = '("'.$row['data'].' 00:00:00",'.$row['merce'].','.$row['totale'].','.$row['servizi'].','.$row['numero'].')';
}

$insert_query = 'INSERT INTO ps_calendar (data, merce, totale, servizi, numero) VALUES '.implode(',', $insert_string).';'; //implode the array containing all the data.

$result = mysql_query($insert_query )or die(mysql_error());

在prestashop中，导出总天数总计。使用PHP和MYSQL

1 个答案: