问题与SUM加倍结果

Question

我一直在这工作了两天，无法得到它，我需要一些帮助。

目标 - 别的不说......

• SUM所有项目的时间表项
• SUM所有项目发票

集团通过这些项目（专案编号），并显示在一个表。

在与上述表格相同的页面中，有一种表格允许用户输入新的时间表条目。刷新时，运行和表格显示更新的时间表总数。

当前情况 - 提交时间表格（例如：项目X为1.25小时）时，发生三件事。

1. 表单数据发布到数据库。这工作完美。数据输入完全是应该的。
2. 页面刷新和项目X的时间表条目更新2.5小时（应该增加1.25）
3. 发票总额也会增加该项目的总发票金额。即如果为项目X开发了5000美元，那么添加一个新的时间表条目将推动该项目达到10000美元和15000美元......等等。

查询 - 如下：

<?php 
    $query = "SELECT tsm_projects.projectName AS projectName, tsm_projects.projectID AS projectID, tsm_projects.value AS value, tsm_projects.estHours AS estHours, tsm_clients.clientName AS clientName, tsm_projects.estHours - SUM(tsm_timesheets.time) AS remaining, SUM(tsm_invoices.invoiceValue) AS invoiceValue, SUM(tsm_timesheets.time) AS totalTime FROM tsm_projects 
    LEFT JOIN tsm_timesheets ON tsm_projects.projectID = tsm_timesheets.projectID 
    LEFT JOIN tsm_clients ON tsm_clients.clientID = tsm_projects.clientID 
    LEFT JOIN tsm_invoices ON tsm_invoices.projectID = tsm_projects.projectID 
    WHERE projectType = 'active' 
    GROUP BY tsm_timesheets.projectID 
    ORDER BY tsm_projects.projectName"; 
    $result = mysql_query($query) or die(mysql_error()); 
    while($row = mysql_fetch_array($result)){ 
    echo "<tr><td>". $row['projectName'] . " [" . $row['clientName'] . "]</td><td>$" . number_format($row[value], 2, '.', ',') . " [" . $row['estHours'] . "]</td><td>$" . $row['invoiceValue'] . "</td><td>" . number_format($row[totalTime], 2, '.', ',') ." ["; 
    if($row["remaining"] <= 0) { 
    echo "<span class=\"redText\">" . $row['remaining'] . "</span>"; } 
    else { 
    echo "<span class=\"greenText\">+" . $row['remaining'] . "</span>"; } 
    echo "]</td></tr>"; } 
?> 

SQL - 我猜时间表和/ orinvoicing表很可能相关：

TABLE `tsm_timesheets` (
    `id` int(11) NOT NULL AUTO_INCREMENT, 
    `projectID` varchar(10) NOT NULL, 
    `activity` varchar(20) NOT NULL, 
    `date` date NOT NULL, 
    `time` decimal(4,2) NOT NULL, 
    `timesheetID` varchar(10) NOT NULL, 
    `memberID` varchar(20) NOT NULL, 
    PRIMARY KEY (`id`) 
) 

TABLE `tsm_invoices` (
    `id` int(11) NOT NULL AUTO_INCREMENT, 
    `projectID` varchar(10) NOT NULL, 
    `month` varchar(15) NOT NULL, 
    `notes` varchar(255) NOT NULL, 
    `invoiceValue` decimal(10,2) NOT NULL DEFAULT '0.00', 
    `gstValue` decimal(10,2) NOT NULL DEFAULT '0.00', 
    `fee` decimal(6,2) NOT NULL DEFAULT '0.00', 
    `costs` decimal(6,2) NOT NULL DEFAULT '0.00', 
    `invoiceNumber` varchar(15) NOT NULL, 
    `dateSent` date NOT NULL, 
    `dateDeposit` date NOT NULL, 
    `dateAdded` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP, 
    `addedBy` varchar(20) NOT NULL, 
    `invoiceID` varchar(10) NOT NULL, 
    PRIMARY KEY (`id`) 
) 

希望有人可以提供帮助。提前致谢。

rrfive

Answer 1

聚合函数计算上不每个表列每结果行的基础。

您需要执行单独分组：

LEFT JOIN (
    SELECT projectID, SUM(invoiceValue) AS SumInvoiceValue 
    FROM tsm_invoices 
    GROUP BY projectID) i ON i.projectID = tsm_projects.projectID 

整个查询：

SELECT p.projectName AS projectName, p.projectID AS projectID, p.value AS value, 
    p.estHours AS estHours, c.clientName AS clientName, 
    p.estHours - t.SumTime AS remaining, 
    i.SumInvoiceValue AS invoiceValue, 
    t.SumTime AS totalTime 
FROM tsm_projects p 
    LEFT JOIN tsm_clients c ON c.clientID = p.clientID 
    LEFT JOIN (
     SELECT projectID, SUM(time) AS SumTime 
     FROM tsm_timesheets 
     GROUP BY projectID 
    ) t ON p.projectID = t.projectID 
    LEFT JOIN (
     SELECT projectID, SUM(invoiceValue) AS SumInvoiceValue 
     FROM tsm_invoices 
     GROUP BY projectID 
    ) i ON i.projectID = p.projectID 
WHERE projectType = 'active' 
GROUP BY p.projectID 
ORDER BY p.projectName