Hi all,
I am trying to set up a PHP dynamic pop, e.g.
http://www.domain.com/?message=logged-in
would produce a javascript pop up to say "You are now logged in" and
http://www.domain.com/?message=logged-out
would produce a javascript pop up to say "You are now logged out"
If this was being done in ASP the code would look something like
<% If Request("message") = "logged-in" Then %>
{do login popup}
<% End If %>
What code would I use for PHP to achieve the same effect?
Thanks everyone,
James
<?php if (isset($_GET['message'])) { ?>
{do you popup here, using $_GET['message'] for the popup text,
that way you don't need separate conditionals for each message}
<?php } ?>Thanks Joe,
I have put this together:
<?php if (isset($_GET['message'])) { ?>
<script language="JavaScript" type="text/javascript">
alert("<?php $_GET['message'] ?>")
</script>
<?php } ?>
and have tried it with /?message=logged-out.
The popup appeared, but there was no text. Where am I going wrong?
Sorry - I'm not too competent with PHP...
Thanks.
<?php echo $_GET['message'] ?>
This tell php to print out the variable to the HTML, where JS will pick it up
Thanks, I was missing "echo"!
That is very insecure. You should validate the text because someone could do http://www.domain.com/?message="); //any javascript OR PHP can go here insert bad serving trojan horses to all your vistors it is bad. use urlencode to avoid this http://us2.php.net/urlencode.
<?php if (isset($_GET['message'])) { ?>
<script language="JavaScript" type="text/javascript">
alert("<?php echo urlencode($_GET['message']) ?>")
</script>
<?php } ?>Thanks Alec!
I am wondering whether I would be better going back to plan 1 to avoid SQL injection and having tried Joe's method my links are looking something like:
http://www.domain.com/?message=You+have+now+been+logged+out
which I don't like, I admit
http://www.domain.com/?message=logged-out
isn't too pretty but the spiders should never find that link.
But if someone ever sees that link and they are well versed in XSS (cross site scripting, the example above) they will try it and may or may not report it. Going back to plan one:
<?php
if($_GET['message'] != "") {
switch ($_GET['message']) {
case "logged-out":
$message = "You have now been logged out";
break;
case "logged-in":
$message = "You have now been logged in";
break;
default :
$message = "Whoa! That message doesn't exist!";
break;
}
echo '<script type="text/javascript">
alert("'.$message.'");
</script>';
}
?>Thanks Alec you're a star.
Except....
Parse error: syntax error, unexpected T_VARIABLE, expecting '(' in \wp-content\themes\freshnews\includes\popup.php on line 3
By the looks of it, PHP wants brackets somewhere...
Edit: Fixed, I worked out where the brackets needed to be...
oops! I forgot the ('s around $_GET['message']
revised code:
<?php
if($_GET['message'] != "") {
switch ($_GET['message']) {
case "logged-out":
$message = "You have now been logged out";
break;
case "logged-in":
$message = "You have now been logged in";
break;
default :
$message = "Whoa! That message doesn't exist!";
break;
}
echo '<script type="text/javascript">
alert("'.$message.'");
</script>';
}
?>I am having a strange isssue with this code. It works when on the homepage but not on any other page.
I tested that the file is actually being loaded and it definately is but can't understand why it won't display the message on other pages.
Does anyone have any idea what could cause this?
An example would be useful!
I've moved the site onto an Apache Server and the problem has gone!
Thanks anyway.
How do I change this code so that if a user already has "?message={whatever message}" in the current URL it removes "?message={whatever message}"?
<?php echo wp_logout_url(); ?>&redirect_to=http://<?php echo $_SERVER["HTTP_HOST"] . $_SERVER["REQUEST_URI"] ?>?message=logged-out
Basically the return URL is keeping the current "message" and adding "?message=logged-out" and is not showing the correct message e.g.
http://www.domain.com/?message=logged-in?message=logged-out
I have had a look on Google and found lots of pages showing ways to grab the current page URL but none on how to remove queries.
Thanks!
$_SERVER["SERVER_NAME"].$_SERVER["REQUEST_URI"]
That is the current URL minus the query string.
Thanks for the quick reply but it's still not working properly.
The url is http://techtrends.co.uk
If you register then try to login you can see the problem.
$url = $_SERVER["SERVER_NAME"].$_SERVER["REQUEST_URI"];
if(substr_count($url, '?') > 0) {
$url = strstr($visual[0], '?', true);
}
<?php echo wp_logout_url(); ?>&redirect_to=http://<?php echo $url ?>?message=logged-outThat should work. It finds the first occurrence of ? and returns everything in front. Should work for this application.
Ok thanks, it works on pages without a query string but on pages with a query string the URL $url produces looks like http://?message=logged-out
Would this be easy to fix?
yes:
$url = $_SERVER['SERVER_NAME'].$_SERVER['REQUEST_URI'];
if(substr_count($url, '?') > 0) {
$url = explode('?', $url, 1);
$url = $url[0];
}
<?php echo wp_logout_url(); ?>&redirect_to=http://<?php echo $url ?>?message=logged-outI'm sorry to be a pain...
The original problem has come back with this code:
http://techtrends.co.uk/?message=logged-out?message=logged-in
I've put a link to Lemur CMS in the Useful Link section. If you want this updated (e.g. if you want it to your homepage) let me know.
hmm, I tested it myself and it did work.
I'll try some more.
Thanks Alec
Sorry for the delay, try this:
$url = $_SERVER['SERVER_NAME'].$_SERVER['REQUEST_URI'];
if(substr_count($url, '?') > 0) {
$url = explode('?', $url, 2);
$url = $url[0];
}
<?php echo wp_logout_url(); ?>&redirect_to=http://<?php echo $url ?>?message=logged-out
Thanks, its working properly now.
great!