我有以下PHP代码,通过调用存储过程将详细信息存储到表中:我更希望以过程方式编写代码
$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;我没有得到任何输出的PHP代码(收据编号)。这是我第一次编写一个PHP代码来从MySQL调用一个过程。
请指点
发布于 2016-05-19 04:38:11
好的,在做了一些研究之后,问题似乎是关于mysqli_prepare语句:不正确的参数。
以下是完善的代码:
$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;https://stackoverflow.com/questions/37298716
复制相似问题