Php and $_GET array?

There is no point of escaping anything if it doesn't enter anywhere important (like a database).

There is no point of escaping anything if it doesn't enter anywhere important (like a database). 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.

Thanks worked with the echo instead of return. – user968808 7 hours ago echoing instead of returning isn't the only thing I changed in the code. Read the answer thoroughly to understand what I've done.

– Truth 7 hours ago.

How about: if ($login_attempt == '1'){ echo 'failed'; }.

Try this one. Your error in $login_attempt == '1.

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') {.

Don't you 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.

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