Question

我有点紧张。我正在使用WHMCS并构建自定义报告，但我的一个查询遇到了麻烦。以下是概述：

我有2张桌子; tblinvoices，其中包含发票和tblinvoiceitems的小计，税金，总额等，其中包含发票上显示的各个行项目。我想运行一个查询，返回我能够做的所有单个行项目及其价格。当我做“GROUP BY＆＃39;并按发票编号对结果进行分组，然后它只返回每个发票的第一个行项目。我想按发票编号对它们进行分组，因此报表中只有1行。这是我的疑问：

$query = "SELECT date_format(tblinvoices.datepaid,'%m-%d-%Y') AS datepaid,
    tblinvoices.userid,
    tblinvoices.id,
    tblinvoices.subtotal,
    tblinvoices.credit,
    tblinvoices.tax,
    tblinvoices.tax2,
    tblinvoices.total,
    tblinvoices.taxrate,
    tblinvoices.taxrate2,
    tblinvoiceitems.description,
    tblinvoiceitems.amount,
    tblinvoices.status,
    FROM tblinvoices
    INNER JOIN tblinvoiceitems ON tblinvoices.id = tblinvoiceitems.invoiceid
    GROUP BY tblinvoices.id";
$result = mysql_query($query);

# Math Operations
$statement = array();
$count = 0;

if ($result !== false) {
    while ($data = mysql_fetch_object($result)) {
        $invoiceid = $data->id;
        $datepaid = $data->datepaid;
        $description = $data->description;
        $item_amount = $data->item_amount;
        $subtotal = $data->subtotal;
        $credit = $data->credit;
        $tax = $data->tax;
        $tax2 = $data->tax2;
        $total = $data->total;

        if ($export != true) {
            $client_link = '<a href=clientssummary.php?userid='.$data->userid.'>'.$data->userid;
            $invoice_link = '<a href=invoices.php?action=edit&id='.$data->id.'>'.$data->id;
        }
        else {
            $client_link = $data->userid;
            $invoice_link = $data->id;
        }
        if (strpos($description, 'Setup') !== false) {
            $setup = $item_amount;
        }
        else {
            $setup = 0;
        }
        if (strpos($description, 'Addon') !== false) {
            $addon = $item_amount;
        }
        else {
            $addon = 0;
        }
        if (strpos($description, 'Tax Guide: No => Yes') !== false) {
            $taxguide = $item_amount;
        }
        else {
            $taxguide = 0;
        }
        if (strpos($description, 'Reading Rack Bundle') !== false) {
            $reading = $item_amount;
        }
        else {
            $reading = 0;
        }
        if (strpos($description, 'Toolkit Bundle') !== false) {
            $toolkit = $item_amount;
        }
        else {
            $toolkit = 0;
        }
        $hosting = $subtotal - $setup - $addon - $taxguide - $reading - $toolkit;

        $statement[$invoiceid."_".$count] = array($datepaid,$client_link,$promo,$dtn,$company,$state,$invoice_link,$setup,$addon,$taxguide,$reading,$toolkit,$hosting,$subtotal,$credit,$tax+$tax2,$total);
        $count++;
    }
}

foreach ($headings AS $k=>$v) {
    $reportdata["tableheadings"][] = $v;
}

//ksort($statement);
foreach ($statement AS $invoiceid=>$entry) {
    $reportdata["tablevalues"][] = array(
        $entry[0], // datepaid
        $entry[1], // clientid
        $entry[2], // promocode
        $entry[3], // dtn
        $entry[4], // companyname
        $entry[5], // state
        $entry[6], // invoiceid
        formatCurrency($entry[7]), // setup
        formatCurrency($entry[8]), // addon
        formatCurrency($entry[9]), // taxguide
        formatCurrency($entry[10]), // reading
        formatCurrency($entry[11]), // toolkit
        formatCurrency($entry[12]), // hosting
        formatCurrency($entry[13]), // subtotal
        formatCurrency($entry[14]), // credit
        formatCurrency($entry[15]), // tax
        formatCurrency($entry[16]) // total
    );
}
mysql_free_result($result);

如果有帮助，我很乐意提供任何其他信息/代码。我虽然这可能是一个更普遍的类型问题...谢谢！

Answer 1

因此，请尝试使用invoice.id和invoice.userid进行分组，然后尝试invoice.subtotal ....

如果您以后决定聚合任何内容，可能会遇到一些问题，但是应该这样做。

Answer 2

结果是您描述的结果集如下：

userid | id | subtotal | credit | tax | status | desc | amount
______________________________________________________________

     1 |  1 |       20 |      0 |  .7 |      1 | item1|     10
     1 |  1 |       20 |      0 |  .7 |      1 | item2|     10
     1 |  2 |       30 |      0 | 2.1 |      1 | item3|     15
     1 |  2 |       30 |      0 | 2.1 |      1 | item3|     15
     1 |  3 |       10 |      0 |  .7 |      0 | item1|     10
     1 |  4 |        1 |      0 | .07 |      0 | item4|      1

这是你在找什么？

合并SQL查询的结果

2 个答案: