Question

现在拔头发

我有一个查询，计算数组中所有相关的$ price值基本上，初始查询会将表检查到已完成但未开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（这些值在另一个表（workshop-items）中找到，并根据$ item数组值进行检查

总计正在计算中，我认为它与$ total的位置有关，因为它的累加所有返回的总数不是单个行总数

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>

Answer 1

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，请设置$price = [];或$price = array();或null ，因为如果您的最终价格是针对do while的每个查询而不是针对该洞的最终价格，您将所有以前的价格加起来作为最终价格，按照我说的做。

请记住执行$total +=而不是$total =，因为您将覆盖在主循环之外使用的变量，因此您将获得错误的总数。

array_sum计算结果查询中的所有记录，而不是单独计算每一行

1 个答案: