我是MSSQL Server的新手,有一个由一些开发人员创建的存储进程,我所需要的就是从我的PHP代码中运行该进程。但是我得到了下面的错误:形式参数"@contract_id“没有声明为输出参数,而是传递到请求输出中的实际参数。
下面是我的代码
$params['contract_id'] = '00990007';
$params['major_version'] = '1';
$procedure_params = array(
array(&$params['contract_id'], SQLSRV_PARAM_OUT),
array(&$params['major_version'], SQLSRV_PARAM_OUT)
);
$sql = "EXEC [MTP].[Process_07a_create_a_contract_version_wrapper] @contract_id = ?, @major_version = ?";
$stmt = sqlsrv_prepare($conn, $sql, $procedure_params);
if( !$stmt ) {
die( print_r( sqlsrv_errors(), true));
}
if(sqlsrv_execute($stmt)){
while($res = sqlsrv_next_result($stmt)){
// make sure all result sets are stepped through, since the output params may not be set until this happens
}
// Output params are now set,
}else{
die( print_r( sqlsrv_errors(), true));
}
有人能在这方面给我指点一下吗?
https://stackoverflow.com/questions/50998295
复制相似问题