array_sum 计算结果查询中的所有记录,而不是单独计算每一行

array_sum counting all records in result query not each row individually

正在拔头发

我有一个计算数组中所有相关 $price 值的查询 基本上初始查询检查 table 到已完成但未开具发票的作业 第二个查询(在初始查询循环内)获取所有需要加起来的项目(这些值在另一个 table ( workshop-items ) 中找到,并根据 $item 数组值

检查

总计计算正常,我认为这与 $total 的放置位置有关,因为它将所有返回的总计相加而不是单个行的总计

下面的代码

<ul class="list-group">
<?php 
    $uninvoicedq = mysqli_query($con,"SELECT * FROM `workshop-jobs` WHERE completed = '1' AND invoiced = '0' AND wscid !='0' ORDER BY workstartdate ASC");
    $uninvoiced = mysqli_fetch_assoc($uninvoicedq);
    if($uninvoiced) {
        do { 
            // User Query
            $wscid = $uninvoiced['wscid'];
            $userq = mysqli_query($cona,"SELECT * FROM `users` WHERE userid = '$wscid'");
            $user = mysqli_fetch_assoc($userq);
            $wtbdq = mysqli_query($con,"SELECT * FROM `workshop-jobs` WHERE wsjid = '$uninvoiced[wsjid]'");
            $wtbdr = mysqli_fetch_assoc($wtbdq);
                do {
                    $wtbd = explode(":",$wtbdr['worktobedone']);
                    foreach($wtbd as $item) 
                    {
                        $priceq = mysqli_query($con,"SELECT * FROM `workshop-items` WHERE wsiid = '$item'");
                        $pricer = mysqli_fetch_assoc($priceq);

                        $price[] = $pricer['incvat'];
                        $items[] = $pricer['description'];
                        //echo $item.' - '. $pricer['incvat'].'<br>';
                    }

                    $total = array_sum($price);
                } while($wtbdr = mysqli_fetch_assoc($wtbdq));
?>
    <li class="list-group-item text-right" style="border:none;" title="<?php echo $itemview;?>"><span class="badge pull-left" style="background-color:#F00;">Not Invoiced</span><?php echo '&pound;'.$total.' - '; echo $user['forename'].' '.$user['surname'].' - ' .$uninvoiced['summary'];?> </li>
<?


                $itemList = implode(":",$items);
                $itemview = str_replace(":","\n",$itemList);
?>

<?          } while($uninvoiced = mysqli_fetch_assoc($uninvoicedq));
        } else { 
            echo "No Jobs Waiting To Invoiced";
        }
?>
        </ul>

如果您的意思是 do while 的每一行应该是不同的总数,那么,当 do 开始时,设置 $price = [];$price = array();null,因为你会将之前的所有价格加起来作为最终价格,如果你的最终价格是针对do while的每个查询而不是针对漏洞,请按我说的去做。

记得做 $total += 而不是 $total = 因为你会覆盖你在主循环外使用的变量,所以你会得到一个错误的总数。