从派生表中提取列并将它们汇总到一个 MySQL SELECT 语句中

Question

我需要从两个表中提取列数据，对数据运行计算并将结果保存为别名，然后将这些结果汇总到其他别名中以显示在 php 表中。我试图通过在我的 SELECT 语句中创建一个派生表来实现这一点，但它不起作用。我没有收到任何错误，但我的表格只显示列标题。

代码：

$sql = "SELECT x.company, x.stagestatus, x.shippeddate, SUM(x.totprice) as totalprice, SUM(x.sgtotquantity) as sgtotqty, SUM(x.sgtotalsqft) as sgtotsqft, SUM(x.avgsqftrev) as avgsqftrevenue, SUM(x.avgunitrev) as avgunitrevenue FROM (SELECT t1.company, t1.stagestatus, t1.shippeddate, t1.id, FORMAT(TRIM(LEADING '$' FROM t1.totalprice), 2) AS totprice, t2.invoiceid, SUM(t2.quantity) AS sgtotqauntity, FORMAT(SUM(t2.width * t2.height * t2.quantity ) /144, 2) AS sgtotalsqft, FORMAT((TRIM(LEADING '$' FROM t1.totalprice)/(SUM(t2.width * t2.height * t2.quantity ) /144)), 2) as avgsqftrev, FORMAT((TRIM(LEADING '$' FROM t1.totalprice) / SUM(t2.quantity)), 2) AS avgunitrev
  FROM invoices AS t1 INNER JOIN lineitems AS t2 ON t1.id = t2.invoiceid
  WHERE (t2.invoiceid = t1.id)
  GROUP BY t1.id) x
WHERE x.stagestatus='Complete'
GROUP BY x.company ASC";

此代码中断，但当我单独使用较小的部分时，它可以正常工作。

前：

$sql="SELECT invoices.id, invoices.orderdate, invoices.stagestatus, FORMAT(TRIM(LEADING '$' FROM invoices.totalprice), 2) AS totalprice, clients.company, lineitems.invoiceid, SUM(lineitems.quantity) AS sgtotqty, FORMAT(SUM(lineitems.width * lineitems.height * lineitems.quantity ) /144, 2) AS sgtotsqft, FORMAT((TRIM(LEADING '$' FROM invoices.totalprice)/(SUM(lineitems.width * lineitems.height * lineitems.quantity ) /144)), 2) as avgsqftrevenue, FORMAT((TRIM(LEADING '$' FROM invoices.totalprice) / SUM(lineitems.quantity)), 2) AS avgunitrevenue
FROM clients
INNER JOIN invoices ON clients.id = invoices.clientid
INNER JOIN lineitems ON invoices.id = lineitems.invoiceid
WHERE (lineitems.invoiceid = invoices.id) AND invoices.orderdate BETWEEN '".$revenuefrom."' AND '".$revenueto."' AND invoices.stagestatus IN (". implode(',', array_map(function($item) {return '"' . $item . '"'; }, $revenue_check)) .")
GROUP BY invoices.id DESC";

此代码运行良好，并按 invoices.id 对所有数据进行分组。但是，项目需求进行了调整，现在所有内容都必须按 invoices.company 分组。当我尝试按 invoices.company 而不是 invoices.id 进行分组时，我的表格完成了，但每个公司行的值都非常不准确，（不是 sum()ing 正确的）。

建表的PHP代码：

$result = $conn->query($sql);


    echo "<table id='revenueReportA' align='center' class='report_DT'>
    <thead>
    <tr>

    <th>Customer</th>
    <th>Total Revenue</th>
    <th>Total SQ FT</th>
    <th>AVG Revenue Per SQ FT</th>
    <th>Total Number of Units</th>
    <th>AVG Revenue Per Unit</th>
    </tr>
    </head>";


 if ($result = $conn->query($sql)) {

   // fetch associative array 
  while ($row = $result->fetch_assoc()) {

  echo "<tbody>";
  echo "<tr>";
  echo "<td>" . $row['company'] . "</td>";
  echo "<td>" ."$". $row['totalprice'] . "</td>";
  echo "<td>" . $row['sgtotsqft'] ."&nbsp;&nbsp;". "ft<sup>2</sup>". "</td>";
  echo "<td>" ."$". $row['avgsqftrevenue'] . "</td>";
  echo "<td>" . $row['sgtotqty'] . "</td>";
  echo "<td>" ."$". $row['avgunitrevenue'] . "</td>";
  echo "</tr>";
  echo "</tbody>";
  } 

   echo "</table>";

echo "<BR>";

感谢所有帮助。

谢谢，

Answer 1

我遇到了拼写错误和格式问题。通过格式化最终数据而不是在嵌入的 SELECT 语句中格式化，我的表数据是准确的。

成功代码：

$sql = "SELECT x.company, x.stagestatus, x.shippeddate, FORMAT(SUM(x.totprice), 2) as totalprice, FORMAT(SUM(x.sgtotquantity), 2) as sgtotqty, FORMAT(SUM(x.sgtotalsqft), 2) as sgtotsqft, FORMAT(SUM(x.avgsqftrev), 2) as avgsqftrevenue, FORMAT(SUM(x.avgunitrev), 2) as avgunitrevenue FROM (SELECT t1.company, t1.stagestatus, t1.shippeddate, t1.id, TRIM(LEADING '$' FROM t1.totalprice) AS totprice, t2.invoiceid, SUM(t2.quantity) AS sgtotquantity, SUM(t2.width * t2.height * t2.quantity ) /144 AS sgtotalsqft, (TRIM(LEADING '$' FROM t1.totalprice)/(SUM(t2.width * t2.height * t2.quantity ) /144)) as avgsqftrev, (TRIM(LEADING '$' FROM t1.totalprice) / SUM(t2.quantity)) AS avgunitrev
FROM invoices AS t1 INNER JOIN lineitems AS t2 ON t1.id = t2.invoiceid
WHERE (t2.invoiceid = t1.id)
GROUP BY t1.id) x
WHERE x.stagestatus='Complete'
GROUP BY x.company ASC";

谢谢！！！