編輯記錄PHP表單
[英]Edit Records PHP Form
問題描述
我目前有這些PHP頁面，可讓我向數據庫添加記錄。 （在這種情況下，它的成員）在可以添加，刪除和查看的意義上說，它可以完美工作。 但是我不確定如何使edit（或UPDATE）功能正常工作。
這是我的數據庫連接代碼：
<?php
// Server Info
$server = 'localhost';
$username = 'root';
$password = '';
$database = 'gamgam';
// Connect to database
$connection = new mysqli($server, $username, $password, $database);
?>
這是我的添加代碼：
<!DOCTYPE html>
<html>
<head><title>Insert Users</title></head>
<body>
<h2>Insert User Confirmation</h2>
<form action="<?php $_SERVER['PHP_SELF']?>" method="post"/> <br>
<?php
require_once('connection.php');
echo "<label for='memberID' >Member ID:</label>";
echo "<input type='text' name='memberID' id='memberID' />";
echo "<br /><br />";
echo "<label for='username' >Username:</label>";
echo "<input type='text' name='username' id='username' />";
echo "<br /><br />";
echo "<label for='password' >Password:</label>";
echo "<input type='password' name='password' id='password' />";
echo "<br /><br />";
echo "<label for='fName' >Firstname:</label>";
echo "<input type='text' name='fName' id='fName' />";
echo "<br /><br />";
echo "<label for='lName' >Lastname:</label>";
echo "<input type='text' name='lName' id='lName' />";
echo "<br /><br />";
echo "<label for='address' >Address:</label>";
echo "<input type='text' name='address' id='address' />";
echo "<br /><br />";
echo "<label for='email' >Email:</label>";
echo "<input type='text' name='email' id='email' />";
echo "<br /><br />";
echo "<input type='submit' name='submit' value='Submit' />";
echo "<input type='reset' value='Clear' />";
echo "<br /><br />";
?>
</form>
</section>
<p><a href='login.php'>Login</a></p>
<?php
if(!isset($_POST['submit'])) {
echo 'Please Register';
} else {
$memberID = $_POST['memberID'];
$username = $_POST['username'];
$password = $_POST['password'];
$fName = $_POST['fName'];
$lName = $_POST['lName'];
$address = $_POST['address'];
$email = $_POST['email'];
$query = "INSERT INTO `members`
(MemberID, Username, Password, FirstName, LastName,
StreetAddress, Email)
VALUES ('$memberID', '$username', '$password', '$fName',
'$lName', '$address', '$email')";
mysqli_query($connection, $query)
or die(mysqli_error($connection));
$rc = mysqli_affected_rows($connection);
if ($rc==1)
{
echo '<h4>The database has been updated with the following details: </h4> ';
echo 'MemberID: '.$memberID.'<br />';
echo 'Username: '.$username.'<br />';
echo 'Password: '.$password.'<br />';
echo 'First Name: '.$fName.'<br />';
echo 'Last Name: '.$lName.'<br />';
echo 'Address: '.$address.'<br />';
echo 'Email: '.$email.'<br />';
} else {
echo '<p>The data was not entered into the database this time.</p>';
}
}
?>
</body>
</html>
這是我的查看代碼：
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">
<html>
<head>
<title>View Records</title>
</head>
<body>
<table border="1" style="width:100%" >
<?php
/*
VIEW.PHP
Displays all data from 'players' table
*/
// connect to the database
include('connection.php');
// get results from database
$result = mysqli_query($connection, "SELECT * FROM members")
or die(mysqli_error());
// loop through results of database query, displaying them in the table
while($row = mysqli_fetch_array( $result )) {
// echo out the contents of each row into a table
echo "<tr>";
echo '<td>' . $row['MemberID'] . '</td>';
echo '<td>' . $row['Username'] . '</td>';
echo '<td>' . $row['Password'] . '</td>';
echo '<td>' . $row['FirstName'] . '</td>';
echo '<td>' . $row['StreetAddress'] . '</td>';
echo '<td>' . $row['Email'] . '</td>';
echo '<td><a href="edit.php?MemberID=' . $row['MemberID'] . '">Edit</a></td>';
echo '<td><a href="delete.php?MemberID=' . $row['MemberID'] . '">Delete</a></td>';
echo "</tr>";
}
// close table>
echo "</table>";
?>
<p><a href="insert_user.php">Add a new record</a></p>
</body>
</html>
這是刪除代碼：
<?php
// Connect to the database
include('connection.php');
// Confirm that the 'code' variable has been set
if (isset($_GET['MemberID']))
{
// Get the 'MemberID' variable from the URL
$MemberID = $_GET['MemberID'];
// Delete record from database
if ($stmt = $connection->prepare("DELETE FROM members WHERE MemberID = ? LIMIT 1")) {
$stmt->bind_param("i",$MemberID);
$stmt->execute();
$stmt->close();
} else {
echo "ERROR: could not prepare SQL statement.";
}
$connection->close();
// Redirect user after delete is successful
header("Location: view.php");
} else {
// If the 'code' variable isn't set, redirect the user
header("Location: view.php");
}
?>
我在網上瀏覽了許多基本的php表單模板，試圖結合他們為獲得結果而做的事情，但沒有取得任何成功。 我的網站需要編寫什么代碼才能具有編輯已經在數據庫中創建的記錄的功能，而無需通過phpmyadmin。 任何幫助都非常感謝。
1 個解決方案
解決方案1
0 2014-10-07 15:29:50
Edit就像Add一樣，但是您需要先讀取記錄並填充字段值。
從add的代碼開始，然后執行以下操作：
<?php $MemberID = (int) $_GET['MemberID']; ?>
<form action="<?php $_SERVER['PHP_SELF']?>" method="post"/> <br>
<input type="hidden" name="MemberID" value="<?php echo $MemberID; ?>"
<?php
require_once('connection.php');
$result = mysqli_query($connection, "SELECT * FROM members where MemberID = $MemberID") or die(mysqli_error());
// loop through results of database query, displaying them in the table
$row = mysqli_fetch_assoc($result);
extract($row);
echo "<label for='memberID' >Member ID:</label>";
echo "$memberID"; // member ID should not be editable
echo "<br /><br />";
echo "<label for='username' >Username:</label>";
echo "<input type='text' name='username' id='username' value="$username" />";
echo "<br /><br />";
PHP代碼將有一個類似的查詢
`UPDATE `members` SET `username` = '$username' ... WHERE `MemberID` = '$MemberID'"
編輯書籍記錄PHP
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