mysqli_fetch_assoc() 期望在尝试从不同的表中进行选择时mysqli_result参数 1

Question

我正在尝试从 2 个不同的表中选择 2 个值并收到此错误。 当我只从 1 个表中选择时，它起作用了。 这是我的代码（在查看此问题 n 行的答案时已经更改了它）：

$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

你需要写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.'";