Hello,
The code below is for a login that I'm using. It works fine, but when I first try logging
in after turning on
my computer, it only works the second time that I hit the "Login" button. Any idea how I can make it not require hitting the "Login" button twice
in this situation?
Thanks
in advance,
John
function isLoggedIn() {
if (session_is_registered('loginid') && session_is_registered('username')) {
return true; // the user is loged
in
} else {
return false; // not logged
in
}
return false;
}
if (!isLoggedIn())
{
if (isset($_POST['cmdlogin']))
{
if (checkLogin($_POST['username'], $_POST['password']))
{
show_userbox();
} else
{
echo "Incorrect Login information !";
show_loginform();
}
} else
{
show_loginform();
}
} else
{
show_userbox();
}
function show_loginform($disabled = false)
{
echo '<form name="login-form" id="login-form" method="post" action="./index.php">
<div class="usernameformtext"><label title="Username">Username: </label></div>
<div class="usernameformfield"><input tabindex="1" accesskey="u" name="username" type="text" maxlength="30" id="username" /></div>
<div class="passwordformtext"><label title="Password">Password: </label></div>
<div class="passwordformfield"><input tabindex="2" accesskey="p" name="password" type="password" maxlength="15" id="password" /></div>
<div class="registertext"><a href="http://www...com/sandbox/register.php" title="Register">Register</a></div>
<div class="lostpasswordtext"><a href="http://www...com/sandbox/lostpassword.php" title="
Lost Password">Lost password?</a></div>
<p class="loginbutton"><input tabindex="3" accesskey="l" type="submit" name="cmdlogin" value="Login" ';
if ($disabled == true)
{
echo 'disabled="disabled"';
}
echo ' /></p></form>';
}
EDIT: Here is another function that is used.
function isLoggedIn()
{
if (session_is_registered('loginid') && session_is_registered('username'))
{
return true; // the user is loged
in
} else
{
return false; // not logged
in
}
return false;
}
