Question

你好请帮助我解决以下问题

Mysql DATA表是

Sno Name    Subject Marks 
1   Test1   Sub1    20
2   Test1   Sub2    20
3   Test1   Sub3    20
4   Test1   Sub4    20
5   Test1   Sub5    20
6   Test2   Sub1    30
7   Test2   Sub2    30
8   Test2   Sub3    30
9   Test2   Sub4    30
10  Test2   Sub5    30
11  Test3   Sub1    40
12  Test3   Sub2    40
13  Test3   Sub3    40
14  Test3   Sub4    40
15  Test3   Sub5    40

我希望像这样显示

Sno Name    Marks
1   Test1   100
    Sub1    20
    Sub2    20
    Sub3    20
    Sub4    20
    Sub5    20
2   Test2   150
    Sub1    30
    Sub2    30
    Sub3    30
    Sub4    30
    Sub5    30
3   Test3   200
    Sub1    40
    Sub2    40
    Sub3    40
    Sub4    40
    Sub5    40

可以在MYSQL PHP中使用。

Answer 1

两种方法

1）找到所有唯一名称和标记总和

Select Name  AS Heading, SUM(Marks) AS Total FROM table GROUP BY Name

然后遍历结果并查询与名称

匹配的所有记录

Select * from table WHERE Name = $result['Heading']

2）获取所有记录，然后遍历结果以对它们进行分组

Select * from table WHERE

$headings = [];
$children = [];
foreach($allrecords as $result)
{
    if (!in_array($result['Name'], $headings)
    {
         $headings[$result['Name']]['Name'] = $result['Name'];
    }
    if (isset($headings[$result['Name']]['Marks']))
    {
        $headings[$result['Name']]['Marks'] += $result['Marks']
    }
    else
    {
        $headings[$result['Name']]['Marks'] = $result['Marks'] 
    }
    $children[$result['Name']][] = $result;
}

在一个表TD中显示两行数据

1 个答案: