Question

sqlsrv_fetch_array(): 19 is not a valid ss_sqlsrv_stmt resource in

请问，我用php链接sqlserver查询，出现以上错误。请问有大神知道是怎么回事吗？
sqlsrv_query()方法的返回值打印出来是Resource id #17

Answer 1

//$stmt = sqlsrv_query($sql);
// 上面这句有问题，类似如下写法：
$serverName = "serverName\sqlexpress" ;
$connectionInfo = array( "Database" => "dbName" , "UID" => "username" , "PWD" => "password" );
$conn = sqlsrv_connect ( $serverName , $connectionInfo );
if( $conn === false ) {
<a href="https://www.baidu.com/s?wd=die&tn=44039180_cpr&fenlei=mv6quAkxTZn0IZRqIHckPjm4nH00T1Y3ny7WrjubrjT4nycvPvDv0ZwV5Hcvrjm3rH6sPfKWUMw85HfYnjn4nH6sgvPsT6KdThsqpZwYTjCEQLGCpyw9Uz4Bmy-bIi4WUvYETgN-TLwGUv3ErHD4nHbznj0YnjDvnjDdnjD4" target="_blank" class="baidu-highlight">die</a>( print_r ( sqlsrv_errors (), true ));
}
$stmt = sqlsrv_query($conn, $sql);

追问

我是写的 $stmt = sqlsrv_query($conn, $sql); 只不过是放在另一个方法里。这里是调用的这个方法