Question

我有以下PHP代码，用于通过调用存储过程将详细信息存储到表中：我希望以程序方式获取代码

$donpm = 0;
$cashCall = mysqli_prepare($dbc, 'call spCashDonation(?,@rno)');
mysqli_stmt_bind_param($cashCall, 'ssiid', $donFirstName, $donLastName, $donContactNo, $donpm, $donAmount);
mysqli_stmt_execute($cashCall) or die('Error recording Donation details.');

$getRno = mysqli_query($dbc, 'SELECT @rno');
$result = mysqli_fetch_assoc($getRno);
$ReceiptNo = $result['@rno'];

echo $ReceiptNo;

调用存储过程：

Call spCashDonation ('SAI','KUMAR',9967021655,0,10000,@rno);
select @rno as ReceiptNo;

我没有获得PHP代码的任何输出（收据号）。我第一次编写PHP代码来从MySQL调用过程。

请咨询

Answer 1

好的，经过一些研究后，问题似乎是mysqli_prepare语句：参数不正确。

以下是精炼代码：

$donpm = 0;
$cashCall = mysqli_prepare($dbc, 'call spCashDonation(?,?,?,?,?,@rno)');
mysqli_stmt_bind_param($cashCall, 'ssiid', $donFirstName, $donLastName, $donContactNo, $donpm, $donAmount);
mysqli_stmt_execute($cashCall) or die('Error recording Donation details.');

$getRno = mysqli_query($dbc, 'SELECT @rno');
$result = mysqli_fetch_assoc($getRno);
$ReceiptNo = $result['@rno'];

echo $ReceiptNo;
echo $donFirstName . ' ' . $donLastName . ' ' . $donContactNo . ' ' . $donpm . ' ' . $donAmount;

在Out存储过程PHP MySQLi中

1 个答案: