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Home Backend Development PHP Tutorial 怎么样查找同一姓名的数据?

怎么样查找同一姓名的数据?

Jun 23, 2016 pm 01:38 PM

a表:
id       name      banji_id               //banji_id即为b表的id
1         张龙          1
2         张龙          2
3         李四          1
3         李四          5
5         王五          3
6         赵虎          2
7         赵虎          1
8         赵虎          4

b表:
id          km             sj1          sj2
1        珠心算        2015      春季
2           口才         2015      春季
3          作文          2015      春季
4          数学          2015      春季
5          英语          2014      秋季

张龙是在2015春季学了珠心算和口才;
赵虎是在2015春季学了珠心算和口才和数学;
李四是在2014秋季学的英语,2015春学的珠心算;
我想做一个按纽,点击后搜索出张龙、赵虎(即在同一期学多科的学员);
而李四虽然是不同科目,但李四由于是不同学期,所以不在显示之列


回复讨论(解决方案)

select name from (select a.*, b.km, b.sj1, b.sj2, count(*) as cou from a join b on a.banji_id = b.id group by name, sj1, sj2) as c where cou >= 2;
Copy after login


不知道有没有更简单的方法。。


GROUP BY xy.name,bj.shijian_1,bj.shijian_2 HAVING COUNT(*)>1

GROUP BY xy.name,bj.kemu_1 HAVING COUNT(*)<=1

怎么把这两句合在一起用??

select a.id, name from a, b where a.banji_id=b.id group by name,sj1,sj2 having count(*) > 1
Copy after login
Copy after login

select a.id, name from a, b where a.banji_id=b.id group by name,sj1,sj2 having count(*) > 1
Copy after login
Copy after login



用这个可以了,但还有两个问题:
1、在分页时显示不正确,
function get_xueyuan_count($duoke.......省略)
{
global $fdyu,$db;
        ......省略
        if($duoke!=0)
{
$sql_where .= " and xy.cur_banji_id=bj.banji_id group by xy.name,bj.shijian_1,bj.shijian_2 having count(*) > 1";
}
        $sql = "SELECT COUNT(distinct xy.xy_id) FROM ".$fdyu->table('oa_xueyuan')." as xy left join 
".$xfsql.
$fdyu->table('oa_banji') . " as bj on bj.banji_id=xy.cur_banji_id left join ".
$fdyu->table('oa_banji') . " as bj_1 on bj_1.banji_id=xy.pre_banji_id 
".$xiashu.
$sql_where;
        $count = $db->getOne($sql);
        return $count;
}

function get_xueyuan_list($duoke.......省略)
{
        global $fdyu,$db;
        ......省略
        if($duoke!=0)
{
$sql_where .= " and xy.cur_banji_id=bj.banji_id group by xy.name,bj.shijian_1,bj.shijian_2 having count(*) > 1";
}
        $sql = "SELECT COUNT(distinct xy.xy_id) FROM ".$fdyu->table('oa_xueyuan')." as xy left join 
".$xfsql.
$fdyu->table('oa_banji') . " as bj on bj.banji_id=xy.cur_banji_id left join ".
$fdyu->table('oa_banji') . " as bj_1 on bj_1.banji_id=xy.pre_banji_id 
".$xiashu.
$sql_where;
        $res = $db->selectLimit($sql, $size, ($page-1) * $size);
$arr = array();
        if ($res)
        {
while ($row = $db->fetchRow($res))
               {
                      ......省略
               }
        }
}

2、我想把学多科的名单都显示,这句 and xy.cur_banji_id=bj.banji_id group by xy.name,bj.shijian_1,bj.shijian_2 having count(*) > 1应该怎么改
张龙
张龙
赵虎
赵虎
赵虎
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