I have submitted some code to the redirected url and now trying to use this to echo some information out but can't figure out where I am going wrong.
I have the following code:
<?php $login_attempt = mysql_real_escape_string($_GET['login_attempt']);
if ($login_attempt) == '1'{
return 'failed';
}
?>
all I want to do is if the url has $login_attempt=1 I want to return the message 'failed' to the page.
There is no point of escaping anything if it doesn't enter anywhere important (like a database).
<?php
if ($_GET['login_attempt'] == '1') {
echo 'failed';
}
?>
Also, you have a problem in your if statement, that's corrected in the code above. Be sure to include all of the condition inside of parenthesis, and not just one side of the equality check.
Also, if you wish to display something on the screen, you should echo it, not return.
Try this one. Your error in $login_attempt == '1':
<?php $login_attempt = mysql_real_escape_string($_GET['login_attempt']);
if ($login_attempt == '1'){
echo 'failed';
return false;
}
?>
As others already mentioned you have several problems but the syntax error comes from this:
if ($login_attempt) == '1'{
it should be
if ($login_attempt == '1') {
Dont u think if ($login_attempt) == '1' should be something like this ($login_attempt == '1') Sorry...many others also suggested this :P
At the first, I must tell you that you have a mistake in your IF condition. You typed == outside of (). In addition, you have to be aware of status of setting your variable through your URL. Check the code below. In this code, I made a function to check the status. Default status is true, and we will check it just for a negative condition. I hope it could be useful for you:
<?php
function check() {
if (isset($_GET['login_attempt'])) {
$login_attempt = mysql_real_escape_string($_GET['login_attempt']);
if ($login_attempt == '1') {
return false;
} else {
return true;
}
} else {
return true;
}
}
if (!check()) echo('Error Message');
?>
