现在，这就是我所拥有的：
$query = "INSERT INTO COMMENTS VALUES ('$user', '$comment', '$star')";
mssql_query($query, $connection);
$commentIDQuery = "SELECT SCOPE_IDENTITY() AS ins_id";
$CI = mssql_query ($commentIDQuery, $connection);
$commentID = mssql_fetch_row($CI);
$idQuery = "SELECT recipeid FROM t_recipe WHERE recipename = '$recipeName'";
$RID = mssql_query($idQuery, $connection);
$recipeID = mssql_fetch_row($RID);
$rcQuery = "INSERT INTO COMMENT_RECIPE VALUES ('$commentID[0]', '$recipeID[0]')";
mssql_query($rcQuery, $connection);
那么我该如何获取ins_id呢？
它将它添加到第一个表中，该表是注释，但不是关系表。
使用sql server 2008
这个。。。。。。
$query = "DECLARE @NewID INT
INSERT INTO COMMENTS VALUES ('$user', '$comment', '$star');
SELECT @NewID = SCOPE_IDENTITY();
INSERT INTO COMMENTS_RECIPE VALUES (@NewID, '$recipeid')";
$stmt = sqlsrv_query($conn,$query);