php应用sql数据库，读取关联表数据

$sql = mysql_query("SELECT * FROM characters left join departments on departments.departmentid=characters.departmentid ORDER BY salary DESC LIMIT 0,20"); while ($data = mysql_fetch_array($sql)){ ?>






? }?>

姓名:

薪金：

年龄:

部门：

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有两个表，人物表characters和部门表departments，要求在网页中显示人物姓名、年龄、薪金和其所在的部门，并且以薪金从高到低显示出前20位的人物。
现在的问题是显示不出人物所在的部门名称
请高手指教，多谢！！！

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查询改为:

$sql = mysql_query("SELECT characters.*,departments.name as depname FROM characters left join departments on departments.departmentid=characters.departmentid ORDER BY salary DESC LIMIT 0,20");

部门：

如果还不行，那么吵瞎你直接在mysql客户端执行一下

SELECT characters.*,departments.name as depname FROM characters left join departments on departments.departmentid=characters.departmentid ORDER BY salary DESC LIMIT 0,20

如果depname没有值的话者配，那么就说明数据库里就没有值了。你自然也取首碰指不出来。

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是因为你的人物表和部门表都有name字段,导致departments的namef无法访问.
语岩源纤句改一裂伍下试试:
SELECT characters .*,departments.name as dept_name FROM characters left join departments on departments.departmentid=characters.departmentid ORDER BY salary DESC LIMIT 0,20

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在while循环中部门使用 $date['粗仿dept_name'];

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