如何使用PDO和SQL Server获取最后一个INSERTed行的ID？

Question

在其他情况下,我可能会被诱惑使用

$result = mssql_query("INSERT INTO table (fields) VALUES (data); 
                       SELECT CAST(scope_identity() AS int)");

但由于我将插入用户提交的数据,我想继续使用PDO,它返回一个空数组.

不幸的是,我在Linux服务器上运行PHP并使用dblib与不支持的Microsoft SQL Server连接PDO::lastInsertID().

请帮忙!

更新以包含代码示例

这是我正在使用的代码:col1是一个类型的字段,int identitycol2是datetime一个默认值为的字段getdate().

//  Connect to db with PDO
$pdo = new PDO( 'dblib:host=' . $host . ';dbname=' . $database . ';', $username, $password, array(PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION) );
//  Connect to db with MSSQL
$sql = mssql_connect( $host, $username, $password );
mssql_select_db( $database, $sql );
//  Create SQL statement
$query = "INSERT INTO [table] ( col3, col4, col5 )
          VALUES ( 'str1', 'str2', 'str3' );
          SELECT SCOPE_IDENTITY() AS theID;";
//  Run with MSSQL
echo "Using MSSQL...\n";
$result = mssql_query( $query );
$the_id = mssql_result( $result, 0, 'theID' );
echo "Query OK. Returned ID is " . $the_id . "\n";
// Run with PDO
echo "\nUsing PDO...\n";
$stmt = $pdo->query( $query );
$result = $stmt->fetchAll( PDO::FETCH_ASSOC );
print_r( $result );

这就是显示的内容:

Using MSSQL...
Query OK. Returned ID is 149

Using PDO...
Array
(
)

我很想知道我做了一些愚蠢的事情,而不是遇到一个可怕的死胡同:)

Answer 1

你有几个选择:

SELECT @@IDENTITY - return the last ID created by actions of the current connection, regardless of table/scope

SELECT SCOPE_IDENTITY() - last ID produced by the current connection, in scope, regardless of table

SELECT IDENT_CURRENT('name_of_table'); - last ID produced on that table, regardless of table/scope/connection

在这三个中,SCOPE_IDENTITY()是最佳候选者.