trq

This topic has been moved to Application Frameworks. http://www.phpfreaks.com/forums/index.php?topic=317833.0

How exactly do you plan on editing it?

Your virtual host should be more like..... <VirtualHost *:80> ServerName helpcody.com ServerAlias www.helpcody.com DocumentRoot "/var/www/html/cody/" DirectoryIndex index.php </VirtualHost> You cannot use wildcards within the ServerName or ServerAlias directives.

You would need to store the name of the file (php) in a location that can be reached from the same scope as the page that requires it. One way to do this would be to store the image name in a database when it is uploaded, then retrieve that record on the page that needs the image name.

<?php $image = 'foo.jpg'; ?> <script type="javascript"> $arr = []; $arr[0] = '<?php echo $image; ?>'; </script>

We have a feelance board here: http://www.phpfreaks.com/forums/php-freelancing/ for such requests. You'll want to read the stickies within that board before posting.

And your problematic code is?

file_get_contents returns false on failure, not null.

You calling code is weird too, you might want to take another look at how inheritance actually works. include "config.php"; include "Library/database.php"; include "Library/module.php"; $db = new Module; $db->ConnectDB(MYSQL_HOST_NAME,MYSQL_USER_NAME,MYSQL_PASSWORD,MYSQL_DATABASE);//define in config.php $db->DoInsert("twitter_profiles", array(profile_user_name,profile_password,user_id), array($_POST['userName'],$_POST['password'],$userID) );//DoInsert(tableName,Fields,Values) echo sysMessage("NewProfile.txt");

This topic has been moved to Application Design. http://www.phpfreaks.com/forums/index.php?topic=317771.0

This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=317710.0

Change localhost to 127.0.0.1

I said mysqli_connect_error not mysqli_connect_errno.

What does mysqli_connect_error say?

Field names should not be surrounded in quotes.

This topic has been moved to Other Web Server Software. http://www.phpfreaks.com/forums/index.php?topic=317797.0

Post your code (in tags).

Why are you executing two queries on the same table in the first place?

Javascript executes on the client, long after your php is finished being processed on the server.

This topic has been moved to Miscellaneous. http://www.phpfreaks.com/forums/index.php?topic=317715.0

$somevar = 'foo'; Will not throw any notice. it's only if you try and concatenate to a variable that doesn't exist yet.

+ is Javascript's concatenation operator. window.location = "profile.php?user="+setHighlightedValue();

If $row is bigger than 0 then the user does already exist. There is nothing wrong with your code.

Again... what are you trying to do?