以下语句在SQLserver中可以通过：
select top 10 job_title from job
但在interbase变成无效，出错！
我的功能是要作排行榜！排行前100名，其余不显示！
但发现用interbase数据库即不能用！
是不是语法不一样，望指教！
在线等待！
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要前100条记录可以用下面的。
select job_title from job rows 100;或select job_title from job rows 1 to 100;
你下面说的功能一样可以通过rows来完成。
rows的具体用法:
ROWS
1<value>
2[TO <upper_value>]
3[BY <step_value>]
4[PERCENT][WITH TIES]
5• value is the total number of rows to return if used by itself
6
7• value is the starting row number to return if used with TO
8
9• value is the percent if used with PERCENT
10
11• upper_value is the last row or highest percent to return
12
13• If step_value = n, returns every nth row, or n percent rows
14
15• PERCENT causes all previous ROWS values to be interpreted as percents
16
17• WITH TIES returns additional duplicate rows when the last value in the ordered sequence is the same as values in subsequent rows of the result set; must be used in conjunction with ORDER BY
18
19The following examples reward the best performing sales people and terminate the least performing members of the sales team. The examples show how a Web developer, for example, could split the result set in half for display purposes.
20
21SELECT SALESMAN, SALES_DOLLARS, SALES_REGION
22FROM SALESPEOPLE
23ORDER BY SALES_DOLLARS DESC
24ROWS 1 TO 50;
25SELECT SALESMAN, SALES_DOLLARS, SALES_REGION
26FROM SALESPEOPLE
27ORDER BY SALES_DOLLARS DESC
28ROWS 50 TO 100 WITH TIES;
29
30Reward the best 100 performing salesmen with a 15 percent bonus:
31
32UPDATE SALESPEOPLE
33SET SALES_BONUS = 0.15 * SALES_DOLLARS
34ORDER BY SALES_DOLLARS DESC
35ROWS 100 WITH TIES;
36
37Eliminate the worst five percent of the sales force:
38
39DELETE FROM SALESPEOPLE
40ORDER BY SALES_DOLLARS
41ROWS 5 PERCENT WITH TIES;
42
43
44\---------------------------------------------------------------
45
46ib6.5 后提供了 Rows 语法，它是你想要的语法；
47FB1.01 也提供了 First 的语法，他于 ib6 兼容，也是开源免费使用项目。
48\---------------------------------------------------------------
49
50用 fb 吧.
51
52select first 10 job_title
53from job
54order by emp_id;
55
56
57select MyID, sum(ammount) as all_ammount
58from MyTable
59group by MyID
60order by sum(ammount)
61having sum(ammount) > 100;
62
63
64以上问题属interbase版本问题！6.5以后提供了更多的功能，要简化代码，请尽快升级吧！</step_value></upper_value></value>