我有这个发布到自己的php页面然后检查天气是否登录某人。我遇到的问题是，如果它登录...然后它仍然显示用户名和密码文本框..但如果我刷新它们就会消失，现在由于会话而出现欢迎的事情。
我想要的是一旦点击提交并将其记录下来，立即不显示文本框（用户名，密码）并显示欢迎消息。现在我必须刷新。
请注意我是PHP新手，非常感谢任何明智的建议。
<?php
echo "<form method=\"post\" action=\"index.php?form_type=$page_vals\">";
echo "<body>";
//Start session
session_start();
//Check whether the session variable SESS_MEMBER_ID is present or not
if(!isset($_SESSION['SESS_MEMBER_ID']) || (trim($_SESSION['SESS_MEMBER_ID']) == '')) {
extract($_POST);
$username = "";
$password = "";
$userrole = "";
$userid ="";
$login_query = "SELECT user_id, user_role, user_username FROM users WHERE user_username = '$_POST[logInUsername]' AND user_password = '$_POST[logInPassword]'";
if(!($database = mysql_connect("localhost","root","")))
die("<p>Could not connect to database</p></div></div>
</body>
</html>");
if(!mysql_select_db("mydatabase", $database))
die("<p>Could not open my databases database</p></div>
</div>
</body>
</html>");
if(!($result = mysql_query($login_query, $database)))
{
print("Could not execute query!<br/>");
die(mysql_error()."</div>
</div>
</body>
</html>");
}
if (mysql_num_rows($result) == 0) {
print("Please verify your login information<br/>");
}
while ($row = mysql_fetch_assoc($result)) {
$username = $row["user_username"];
$userrole = $row["user_role"];
$userid = $row["user_id"];
}
echo "Hello - '$username'";
mysql_close($database);
session_regenerate_id();
$_SESSION['SESS_MEMBER_ID'] = $userid;
$_SESSION['SESS_NAME'] = $username;
//Write session to disc
session_write_close();
echo '<div id="login" class="login">
<label for="login">User Name</label>
<input type="text" name="logInUsername" />
<label for="Password">Password</label>
<input type="password" name="logInPassword" />
<input type="submit" value="Submit" class="button" />
</div>';
}
else
{
$sessionName = $_SESSION['SESS_NAME'];
echo '<div id="login" class="login">
<label for="welcome">Welcome '. $sessionName.'!</label>
</div>';
}
?>
答案 0 :(得分：1)
所以，不要处理此处的任何安全或样式问题......
现在您正在查看会话是否已设置。如果不是，则处理登录。处理完登录后，您将显示表单字段。
你应该检查3个州......
有人已登录吗？ 你需要处理登录吗？ 如果这两者都不显示正常形式......
您可以使用会话字段的现有isset来执行此操作。
然后，如果未设置，请检查是否设置了帖子字段...如果已设置，则处理登录。
否则，请显示基本登录表单。
编辑：
完整的代码示例（抱歉可怕的格式化，主要是剪切和粘贴......：
<?php
echo "<form method=\"post\" action=\"index.php?form_type=$page_vals\">";
echo "<body>";
//Start session
session_start();
//Check whether the session variable SESS_MEMBER_ID is present or not
if(isset($_SESSION['SESS_MEMBER_ID']) || (trim($_SESSION['SESS_MEMBER_ID']) != '')) {
$sessionName = $_SESSION['SESS_NAME'];
echo '<div id="login" class="login">
<label for="welcome">Welcome '. $sessionName.'!</label>
</div>';
}
else if ($_POST[logInPassword] != null && $_POST[logInUsername] != null)
{
extract($_POST);
$username = "";
$password = "";
$userrole = "";
$userid ="";
$login_query = "SELECT user_id, user_role, user_username FROM users WHERE user_username = '$_POST[logInUsername]' AND user_password = '$_POST[logInPassword]'";
if(!($database = mysql_connect("localhost","root","")))
die("<p>Could not connect to database</p></div></div>
</body>
</html>");
if(!mysql_select_db("mydatabase", $database))
die("<p>Could not open my databases database</p></div>
</div>
</body>
</html>");
if(!($result = mysql_query($login_query, $database)))
{
print("Could not execute query!<br/>");
die(mysql_error()."</div>
</div>
</body>
</html>");
}
if (mysql_num_rows($result) == 0) {
print("Please verify your login information<br/>");
}
while ($row = mysql_fetch_assoc($result)) {
$username = $row["user_username"];
$userrole = $row["user_role"];
$userid = $row["user_id"];
}
echo "Hello - '$username'";
mysql_close($database);
session_regenerate_id();
$_SESSION['SESS_MEMBER_ID'] = $userid;
$_SESSION['SESS_NAME'] = $username;
//Write session to disc
session_write_close();
$sessionName = $_SESSION['SESS_NAME'];
echo '<div id="login" class="login">
<label for="welcome">Welcome '. $sessionName.'!</label>
</div>';
}
else
{
echo '<div id="login" class="login">
<label for="login">User Name</label>
<input type="text" name="logInUsername" />
<label for="Password">Password</label>
<input type="password" name="logInPassword" />
<input type="submit" value="Submit" class="button" />
</div>';
}
?>
答案 1 :(得分：1)
这里的问题只是你的代码没有按顺序排列。我已经纠正了现在试试。
<?php
session_start();
echo "<body>";
//Start session
//print_r($_SESSION);exit;
//Check whether the session variable SESS_MEMBER_ID is present or not
extract($_POST);
$username = "";
$password = "";
$userrole = "";
$userid ="";
if(isset($_POST))
{
$login_query = "SELECT reg_id, role_id, f_name FROM registration WHERE f_name = '$_POST[logInUsername]' AND password = '$_POST[logInPassword]'";
if(!($database = mysql_connect("sunlinux","pukhraj","pukhraj123")))
die("<p>Could not connect to database</p></div></div>
</body>
</html>");
if(!mysql_select_db("testbaj", $database))
die("<p>Could not open my databases database</p></div>
</div>
</body>
</html>");
if(!($result = mysql_query($login_query, $database)))
{
print("Could not execute query!<br/>");
die(mysql_error()."</div>
</div>
</body>
</html>");
}
if (mysql_num_rows($result) == 0) {
print("Please verify your login information<br/>");
}
while ($row = mysql_fetch_assoc($result)) {
$username = $row["f_name"];
$userrole = $row["role"];
$userid = $row["reg_id"];
}
$_SESSION['SESS_MEMBER_ID'] = $userid;
$_SESSION['SESS_NAME'] = $username;
}
if(!isset($_SESSION['SESS_MEMBER_ID']) || (trim($_SESSION['SESS_MEMBER_ID']) == '')) {
echo "Hello - '$username'";
mysql_close($database);
session_regenerate_id();
//Write session to disc
session_write_close();
echo "<form method=\"post\" ><div id=\"login\" class=\"login\">
<label for=\"login\">User Name</label>
<input type=\"text\" name=\"logInUsername\" />
<label for=\"Password\">Password</label>
<input type=\"password\" name=\"logInPassword\" />
<input type=\"submit\" value=\"Submit\" class=\"button\" />
</div>";
}
else
{
$sessionName = $_SESSION['SESS_NAME'];
echo "<div id=\"login\" class=\"login\">
<label for=\"welcome\">Welcome '$sessionName' !</label>
</div>";
}
?>
小变化：
：
进行以下更改：
提供名称以提交按钮<input type=\"submit\" name=\"submit\" value=\"Submit\" class=\"button\" />
首先使用if(isset($_POST['submit']))
答案 2 :(得分：0)
你的逻辑需要重新考虑。这样的事怎么样？ （伪代码）
if( user is NOT logged in) // Check via session
{
$errors = array();
if( user submitted the form and is trying to log in) // Can be checked with a POST'd variable
{
// Set the session correctly here, query DB, etc.
// If there are any errors, add them to the $error array
}
if( !empty( $errors) || form was not submitted)
{
// Print the form and any errors (like invalid username / password combo)
}
exit; // Stop here
}
// Print welcome message here (since we know if we get here, the user is logged in)