简单的php登录代码-freejs.net代码演示

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简单的php登录代码

本例无演示，因为就下面这些简单的代码

PHP Code

<?php
session_start();
$conn = mysqli_connect("localhost","root","123456789","freejs");
$message="";
if(!emptyempty($_POST["login"])) {
$result = mysqli_query($conn,"SELECT * FROM users WHERE user_name='" . $_POST["user_name"] . "' and password = '". $_POST["password"]."'");
$row = mysqli_fetch_array($result);
if(is_array($row)) {
$_SESSION["user_id"] = $row['user_id'];
} else {
$message = "Invalid Username or Password!";
}
}
if(!emptyempty($_POST["logout"])) {
$_SESSION["user_id"] = "";
session_destroy();
}
?>

XML/HTML Code

<?php if(empty($_SESSION["user_id"])) { ?>
<form action="" method="post" id="frmLogin">
<div class="error-message"><?php if(isset($message)) { echo $message; } ?></div>
<div class="field-group">
<div><label for="login">Username</label></div>
<div><input name="user_name" type="text" class="input-field"></div>
</div>
<div class="field-group">
<div><label for="password">Password</label></div>
<div><input name="password" type="password" class="input-field"> </div>
</div>
<div class="field-group">
<div><input type="submit" name="login" value="Login" class="form-submit-button"></span></div>
</div>
</form>
<?php
} else {
$result = mysqlI_query($conn,"SELECT * FROM users WHERE user_id='" . $_SESSION["user_id"] . "'");
$row = mysqli_fetch_array($result);
?>
<form action="" method="post" id="frmLogout">
<div class="member-dashboard">Welcome <?php echo ucwords($row['user_name']); ?>, You have successfully logged in!<br>
Click to <input type="submit" name="logout" value="Logout" class="logout-button">.</div>
</form>
</div>
</div>
<?php } ?>