当尝试从不同的表中进行选择时，mysqli_fetch_assoc（）期望参数1为mysqli_result

Question

我试图从2个不同的表中选择2个值并得到此错误。当我只从1张桌子中选出它的时候。这是我的代码（在查看这个问题的答案后已经改变它了）：

$apartmentNum = $_GET['apartmentNumber'];



$getBillsSumQuery = "SELECT a.totalBillsAmount , b.count(id) FROM apartments AS a , bills AS b WHERE a.number = "'.$apartmentNum.'" AND b.apartmentNumber = '.$apartmentNum.'";

$totalSum = mysqli_query($con,$getBillsSumQuery);

$row = mysqli_fetch_assoc($totalSum); //this is where the program crushes

$newSum = $row['totalBillsAmount'];

$billsCount = $row['count(id)'];


echo '{"billSum":' . $newSum . ' , "billCount":' . $billsCount .'}';

Answer 1

您需要写count(b.id)而不是b.count(id)。使用以下查询。

$getBillsSumQuery = "SELECT a.totalBillsAmount , count(b.id) FROM apartments AS a , bills AS b 
WHERE a.number = "'.$apartmentNum.'" AND b.apartmentNumber = '.$apartmentNum.'";