floridaflatlander

Is the image floated to the right? Is the image inside div or fieldset that you have in the picture?

I think you just change $image in imagejpeg($newImage,$image,90); and for the thumb one too. I use time and for thumbs I use thbtime.

What image, what text, all I really see for text is "There was a problem, please click Here to return to the main page and try again"

Well not to be a smart a but something has changed. Get and use fire fox's firebug(I think chrome does the same thing), open fire bug click the div tags until you get to the tags in question. Click them, firebug well tell you where the style is coming from if any, what the css file name is, what line the style is coming from, the inheritance if any and the name of the style.

A browser will load and read the file even if there is no styles on it and it loads images even if they aren't displayed on your page. Go here http://www.websiteoptimization.com/services/analyze/ and enter and test your url and you'll see that the browser loads files and images even if they are not used.

I would think your file would need some type of Get statement for search and if it was true echo the new header Also where is your doc types, in the file above and in the headers? You're starting your header then calling headers.

Also besides the docs there are videos on youtube that can help walk you through installing xamp and others on your computer. When you make a password for you mysql db make it an easy one to remember and never change it(this is for your local computer). If you don't know what your doing it can be a bear to fix if you change your pw at the control panel later on. Take your time, you can do it.

Yeah that would be nice if you need it.

This is how I do it UPDATE table_name set column_name = replace (column_name, 'this_content', 'with_this_content'); remember to back it up first.

Why don't you just enter your info in a db table. Assign a $last_id. Use a redirect to www.mysite.com/show.php?item=$last_id

I use one parameter for mysqli_insert_id $lastid = mysqli_insert_id($dbc); place it directly below $r = mysqli_query($dbc, $q);

Okay I was reading about third party cookies, that's not what I want, I think I'll go the db route. Thanks again premiso

Thanks, I just read that somewhere else too. I may use a db table. As an example, how does one site on the internet know I was looking at motorcycles on another web site?(this has happened to me)

I know this has been asked before and I have searched but I can't find an answer. I have a cookie that is set on a customers web site(yes I've tested it and it is defined) and I was going to use it to display certain info on my web site. Sooo ... How do I get a cookie to work across multiple (2) web sites? Thanks

That'll do it, thanks

I use a simple machines forum and want to combine a table I have and a smf table and I need the dates in a numberic format(like smf uses). Sooo ... how do I get the time in a format like this 1336392447 Thanks

I mean like test I allways put echo "Whatever: $whatever"; so at least something prints and lets me know the echo is working.

Then go over your code(not just the echo statement) very carefully after you move it, something is missing.

Place an echo for $totalshares outside your loop and see what or if $totalshares has value.

Look at the source code and see what it has "or doesn't have" as a link for the image.

Try echo "<a href=".$row['url']."><img class="image1" src=".$row['image_path']." alt=".$row['summary']."></a><br />";

Were is you destination file, it's not here <a href=\'\'>. This should be something like <a href="http:\\www.webpage.com" >. have you named columns correctly? Put the echo statement you're using now on here.

So why are there single quotes and no double quotes, why are the slashes in your code? Play with it.

You add double quotes for the html not php and you have to escape them for php, your going to have to play with it but basiclly so it would be something like href=\"{$row['url']}\" and the same for src and alt I use single quotes so echo '<a href='.$row['url'].'></a>'; Like I said, play with it.

I don't see any added and escaped quotation marks echo "<a href='{$row['url']}'><img class='image1' src='{$row['image_path']}' alt='{$row['summary']}'> </a><br />";