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Can you modify the sitebuilder at all to trigger these events?

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Wouldn't have anything to do with php. incrond is what your looking for though, assuming your talking about a Linux OS.

You shouldn't have any trouble running a dev server on the system you have. I just wouldn't install any desktop without a bit more ram, but you shouldn't really be running a desktop on a server anyway.

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You need to change the links yourself. mod_rewrite is only there to handle the new links, it doesn't convert anything.

Sounds to me like your probably trying to learn too much. In order to build web applications yourself you will need to have a good understanding of client side languages such as html & css first. Javascript will also come in handy on the client side. From there, to learn to actually program, you are much better off getting a good book and reading it from start to finish (probably two or three times). Most books are written in such a way that they take you on a learning journey, with each chapter building on the previous. Jumping from site to site, tutorial to tutorial won't really get you anywhere but frustrated (or here asking stupid questions like allot of other lost souls), you need structure. I understand that the idea of reading through a 700 page book a few times cover to cover can seem daunting, but seriously, you need a good grasp of the basics before you can move on. Without them, you'll just keep hitting dead ends and asking questions on forums over and over and over and over and...... Your right, your not going to learn to be an expert programmer overnight. With this in mind, be patient. It's not difficult, you don't need to be any kind of genius. You just need some patience.

Have another a read of the 'Return Values' section of the manual. here.

Right clicking and viewing source does show the actual code. Your css is NOT in an external file. Either you are lying or have no idea what you are talking about. Again: You should use id's to identify unique elements. Do you know what unique means?

mysql -uroot will get you to a prompt. (mysql>) if your already at a prompt it will cause errors because it is not part of a valid command / query.

The errors are pretty self explanatory.

This topic has been moved to Miscellaneous. http://www.phpfreaks.com/forums/index.php?topic=317142.0

Either view or $pageInfo[0] are not objects.

Sounds pretty inflexible to me. You really should take a look at some MVC implementations.

That will not give you a total count of the word 'YES'. It will only give you a count of how many rows contain the word 'YES'. Like I said, You'll need to read some tutorials on database normalization (or even take a look at sweeb's reply), and fix your db design before you can move forward.

It is indeed missing the table name.

Look at your code. You have 5 set string positions %s followed by 6 arguments containing mysql_real_escape_string. Your also missing the closing ) on your query.

if is a control structure. Its part of the language itself. Besides, you still need to rearrange your code.

Not without recompiling php. You'll need to execute this as root. mkdir -p /myapp/conf/ && cd $! && ln -s /opt/myapp/conf/php.ini php.ini

They are refers to what exactly? Id's are for identifying unique elements. You work it out.

There is no such thing as an 'if' function.

To do that your going to have to do it the hard way. however, before going down that path, I would put it to you that your database is poorly designed. Why would you have 5 fields all containing the same or similar data? I should have mentioned this in my previous post. Any time you see field names like..... row1, row2, row3, row4 in a database says its a poor design. Look up some tutorials on database normalization and come back when your data is stored properly. Actually, you probably won't need to because a simple query will get you the answers you need. That's one of the benefits of good DB design.

mysql_fetch_assoc() returns an array just like mysql_fetch_row() does. *assoc returns an associative array though while *row returns a numerically indexed array so your syntax would fail because the cnt index does not exist. And in both cases there is no need for any while loop. Maybe you where thinking of.... $sql = "SELECT COUNT(*) AS cnt WHERE 'YES' IN(rowBa, rowBb, rowBc, rowBd, rowBe)"; if ($result = mysql_query($sql)) { echo mysql_result($result, 0); } which is shorter, but IMO less clear.

<?php $sql = "SELECT COUNT(*) AS cnt WHERE 'YES' IN(rowBa, rowBb, rowBc, rowBd, rowBe)"; if ($result = mysql_query($sql)) { $row = mysql_fetch_assoc($result)) { echo $row['cnt']; } }