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[英]Selecting records from a database table in PHP using CodeIgniter and pass to a view
[英]Filtering records from database table to be view and update
我目前正在尝试从数据库中提取记录,并将其显示在我的php表单上,以供用户更新。 在我的数据库表中,有一个名为stage的列。 在阶段之下,有基础,中间和先进。 这是一个例子:
有效的XHTML http://img259.imageshack.us/img259/2461/databasei.png 。 有效的XHTML http://imageshack.us/photo/my-images/849/profileint.png/ 。
因此,当我按php形式(第二张图片)中的按钮时,列表中将仅显示其阶段为基础的记录(以及他们的姓名,联系电话,电子邮件和学生证)。可以做到
您是否正在寻找这样的东西?
<?php
$stage = $_GET['stage'];
$sql = 'SELECT * FROM {$tablename} WHERE stage=:stage';
$db = new PDO('mysql:host=localhost;dbname=yourdbname', $user, $pass);
try {
$statement = $db->prepare($sql);
$statement->execute(array(':stage', db->quote($stage)));
foreach($statement->fetchAll() as $row) {
// Do things here.
}
} catch (PDOException $e) {
print "Error!: " . $e->getMessage() . "<br/>";
die();
}
<?php
include('connect-db2.php');
// get results from database
$result = mysql_query("SELECT * FROM players2 WHERE stage = 'foundation'")
or die(mysql_error());
echo "<table border='1' cellpadding='10'>";
echo "<tr> <th>S/N</th> <th>Student_ID</th> <th>Name</th> <th>Contacts No.</th> <th>Email</th><th>Stage</th></tr>";
while($row = mysql_fetch_array( $result )) {
// echo out the contents of each row into a table
echo "<tr>";
echo '<td>' . $row['id'] . '</td>';
echo '<td>' . $row['student_id'] . '</td>';
echo '<td>' . $row['name'] . '</td>';
echo '<td>' . $row['contact_no'] . '</td>';
echo '<td>' . $row['email'] . '</td>';
echo '<td>' . $row['stage'] . '</td>';
echo '<td><a href="edit2.php?id=' . $row['id'] . '">Edit</a></td>';
echo '<td><a href="delete2.php?id=' . $row['id'] . '">Delete</a></td>';
echo "</tr>";
}
?>
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