shortysbest

And thankfully, you can avoid people who would hire them by having reasonable market rates. HTML/CSS, $20-30 is a fair rate. For PHP/MySQL, it can easily go up to 40-60. How long have you been doing it professionally? I started at 15, then each year went up by 5 or 10, until I found a good market to work in and jumped from 60 to 100 for freelance. (I also don't particularly want a lot of freelance work, only projects I really want to do, so $100/hr is a good way to weed out projects.) Well I have been doing this sort of work since I was about 16, between 16 and 17 I mostly did just basic HTML and CSS, then when I was about 17 I began PHP, MySQL, and a lot of javascript/jQuery. Since then I have done a lot of work with javascript/jQuery as I have been doing a great amount of work with ajax content loaded websites, the real source of my knowledge has been my social networking website that I started working on when I was about 17, which is the main source of my current skill level (which I believe to be quite high given my age and relatively short period of time doing this (I am 19, well just about 20 now)) I have done a few professional websites, and I have done a bit more freelance.

Hello, I am relatively new to doing professional website work, and I am trying to figure out how much I should generally charge for work? I feel like I am quite good at what I do, and could charge quite a lot, but how much is too much or too little? I am good in HTML, CSS, PHP, MySQL, JavaScript/jQuery, and node.js/socket.IO. I have generally charged about $20/hour for local work that I do for people that I know, however now I am getting into more business. On average what should I charge an hour to do this work? Also taking in consideration that I feel like I can get work done a lot quicker than most people (as far as I have seen), so perhaps I should take that into account? Would something like $40-$60/hour be a reasonable rate? I also have a question about what I should charge to do this work below: I do freelance work for a company, and this particular project they have for me is a very basic and simple one. A pre-designed 7 page html website that includes some javascript/jquery, but mostly just html/css. How much would you charge for something like this? And what about a basic html/css pre-designed E-mail job?

Thanks for you time and suggestion, but my client doesn't really want to go in this direction as far as image modification like so. I was more looking for design/layout help.

I am currently working on a website for a client. She is an artist and is going to be renting out studio spaces on the water, as well as an apartment for artists to come and visit and do their work. There will be workshops and other types of things going on as well. I have got a couple pages put together, but the homepage is what I really need to get underway. I am not the best with homepages no matter the project. Basically she would like to have an image of the ocean and land in front of the studio space to take up a good portion of the homepage, and she would also like to have a small description about what it is she is trying to do and everything. Any help would be greatly appreciated (perhaps you would sketch it up or something) as well as any feedback on the rest of the site if you wish. Link: http://www.kithell.com/test/sandy_v2

The way my friends table is it has one row for each friendship, there is no specific column for a user, it just depends on who adds who. So this just takes and whenever one row matches the users id, it sets the friend id as the other column.

I'm having trouble getting this to work: SELECT * FROM usersActivity (JOIN friends ON usersActivity.userID=friendID CASE WHEN friends.userID=$session THEN friends.userID2 ELSE friends.userID END AS friendID WHERE (friends.userID=friendID AND friends.userID2=$session) OR (friends.userID=$session AND friends.userID2=friendID) ) WHERE usersActivity.setActivity!=3 AND usersActivity.userID!=$session ORDER BY usersActivity.setActivity ASC I want to return who's online, based on who you're friends with. Currently I have to do this to get it to do what I want: $onlineUsers = mysql_query(" SELECT * FROM usersActivity WHERE setActivity!=3 AND userID!=$session ORDER BY setActivity ASC "); while($users = mysql_fetch_array($onlineUsers)) { $friendID = $users['userID']; $friend = mysql_fetch_assoc(mysql_query(" SELECT CASE WHEN userID=$session THEN userID2 ELSE userID END AS friendID FROM friends WHERE (userID=$friendID AND userID2=$session) OR (userID=$session AND userID2=$friendID)")); if($friendID==$friend['friendID']) { print $friendID; //userID that is online } }