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?
发布于 2018-06-23 17:13:02
首先,使用sys.parameters检查参数的类型和顺序。然后使用结果设置params:
<?php
...
$sql =
"SELECT [name], [is_output]
FROM sys.parameters
WHERE object_id = object_id('[MTP].[Process_07a_create_a_contract_version_wrapper]')
ORDER BY parameter_id";
$res = sqlsrv_query($conn, $sql);
if ($res === false) {
echo "Error (sqlsrv_query): ".print_r(sqlsrv_errors(), true);
exit;
}
while ($row = sqlsrv_fetch_array($res, SQLSRV_FETCH_ASSOC)) {
echo 'Name: '.$row['name'].', ';
echo 'Output: '.($row['is_output'] == 1 ? 'Yes' : 'No');
echo '</br>';
}
...
?>
当@contract_id参数不是OUTPUT参数时,可能会引发错误消息“形式参数"@contract_id”未声明为output参数,但实际传入请求的输出参数“。所以你可以尝试这样做(使用你的代码):
<?php
...
$sql = "{call [MTP].[Process_07a_create_a_contract_version_wrapper](?, ?)}";
$procedure_params = array(
array($params['contract_id'], SQLSRV_PARAM_IN),
array(&$params['major_version'], SQLSRV_PARAM_INOUT)
);
$stmt = sqlsrv_query($conn, $sql, $procedure_params);
if( $stmt === false ) {
echo "Error (sqlsrv_query): ".print_r(sqlsrv_errors(), true);
exit;
}
// Output parameters are available after consuming all resultsets.
while ($row = sqlsrv_fetch_array($stmt, SQLSRV_FETCH_ASSOC)) {
}
...
?>
https://stackoverflow.com/questions/50998295
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