如何使用phpmysql進行考勤查看？

-2

以下是我的出勤表，以使像this考勤觀點：

`CREATE TABLE IF NOT EXISTS `attendance` (
     `aid` int(11) NOT NULL AUTO_INCREMENT, 
     `Name_of_Student` varchar(100) COLLATE latin1_general_ci NOT NULL, 
     `Class` varchar(20) COLLATE latin1_general_ci NOT NULL, 
     `Roll_no` int(11) NOT NULL, 
     `Section` varchar(20) COLLATE latin1_general_ci NOT NULL, 
     `Status` binary(1) NOT NULL, 
     `time` varchar(20) COLLATE latin1_general_ci NOT NULL, 
     `Date` date NOT NULL, 
     PRIMARY KEY (`aid`) 
    ) ENGINE=MyISAM DEFAULT CHARSET=latin1 COLLATE=latin1_general_ci AUTO_INCREMENT=33 ; 

-- 
-- Dumping data for table `attendance` 
-- 

    INSERT INTO `attendance` (`aid`, `Name_of_Student`, `Class`, `Roll_no`, `Section`, `Status`, `time`, `Date`) VALUES 
    (1, 'Lalchhandami', 'X', 1, 'A', 'Y', 'Morning', '2013-07-01'), 
    (2, 'Zonundanga', 'X', 5, 'A', 'Y', 'Morning', '2013-07-01'), 
    (3, 'Lalchhandami', 'X', 1, 'A', 'Y', 'Morning', '2013-07-02'), 
    (4, 'Zonundanga', 'X', 5, 'A', 'Y', 'Morning', '2013-07-02'), 
    (5, 'Lalchhandami', 'X', 1, 'A', 'Y', 'Morning', '2013-07-03'), 
    (6, 'Zonundanga', 'X', 5, 'A', 'Y', 'Morning', '2013-07-03'), 
    (7, 'Lalchhandami', 'X', 1, 'A', 'Y', 'Morning', '2013-07-04'), 
    (8, 'Zonundanga', 'X', 5, 'A', 'Y', 'Morning', '2013-07-04'), 
    (9, 'Lalchhandami', 'X', 1, 'A', 'Y', 'Morning', '2013-07-05'), 
    (10, 'Zonundanga', 'X', 5, 'A', 'Y', 'Morning', '2013-07-05'), 
    (11, 'Lalchhandami', 'X', 1, 'A', 'Y', 'Morning', '2013-07-06'), 
    (12, 'Zonundanga', 'X', 5, 'A', 'Y', 'Morning', '2013-07-06'), 
    (13, 'Lalchhandami', 'X', 1, 'A', 'Y', 'Morning', '2013-07-07'), 
    (14, 'Zonundanga', 'X', 5, 'A', 'Y', 'Morning', '2013-07-07'), 
    (15, 'Lalchhandami', 'X', 1, 'A', 'Y', 'Morning', '2013-07-08'), 
    (16, 'Zonundanga', 'X', 5, 'A', 'Y', 'Morning', '2013-07-08'), 
    (17, 'Lalchhandami', 'X', 1, 'A', 'Y', 'Afternoon', '2013-07-01'), 
    (18, 'Zonundanga', 'X', 5, 'A', 'Y', 'Afternoon', '2013-07-01'), 
    (19, 'Lalchhandami', 'X', 1, 'A', 'Y', 'Afternoon', '2013-07-02'), 
    (20, 'Zonundanga', 'X', 5, 'A', 'Y', 'Afternoon', '2013-07-02'), 
    (21, 'Lalchhandami', 'X', 1, 'A', 'Y', 'Afternoon', '2013-07-03'), 
    (22, 'Zonundanga', 'X', 5, 'A', 'Y', 'Afternoon', '2013-07-03'), 
    (23, 'Lalchhandami', 'X', 1, 'A', 'Y', 'Afternoon', '2013-07-04'), 
    (24, 'Zonundanga', 'X', 5, 'A', 'Y', 'Afternoon', '2013-07-04'), 
    (25, 'Lalchhandami', 'X', 1, 'A', 'Y', 'Afternoon', '2013-07-05'), 
    (26, 'Zonundanga', 'X', 5, 'A', 'Y', 'Afternoon', '2013-07-05'), 
    (27, 'Lalchhandami', 'X', 1, 'A', 'Y', 'Afternoon', '2013-07-06'), 
    (28, 'Zonundanga', 'X', 5, 'A', 'Y', 'Afternoon', '2013-07-06'), 
    (29, 'Lalchhandami', 'X', 1, 'A', 'Y', 'Afternoon', '2013-07-07'), 
    (30, 'Zonundanga', 'X', 5, 'A', 'Y', 'Afternoon', '2013-07-07'), 
    (31, 'Lalchhandami', 'X', 1, 'A', 'Y', 'Afternoon', '2013-07-08'), 
    (32, 'Zonundanga', 'X', 5, 'A', 'N', 'Afternoon', '2013-07-08');`

請誰能幫助我？

來源

2013-07-04 Mawia HL

你至少可以去格式化你的代碼正確 – Bojangles

-1

可以是這樣的：

<table> 
<tr><th>Sr no</th><th > .....//all of your table heads </th> </tr> 
<?php 

    $query = "select * from `attendance`"; 
    $result = mysql_query($query); 
    $result = mysql_fetch_row($result); 
    foreach($result as $result) 
    { 
     echo "<tr><td>".$result['aid']."</td><td>all of id with index</td></tr>" 
    } 
    ?> </table>

來源

2013-07-04 11:29:35

我想要類似數據透視表的一些東西..我想使用日期行作爲列，每列下面將爲Y，表示現在，N表示缺席，我想按照

Name_of_Student

–

將它分組。這裏的參考http://stackoverflow.com/questions/14492552/pivot-dynamic-html-table-using-mysql-and-php –

演示在http://www.sqlfiddle.com/#!2/ec418/ 3 –

使用的mysql_query選擇所有數據，並將數據轉儲到您的HTML

來源

2013-07-04 11:27:23

我的努力新手......我想要的只是使用php來查看http://postimg.org/image/t46xc1laz/。 –

檢查我編輯了我的代碼 –

如何使用phpmysql進行考勤查看？

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