LazyJones

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if ($cell == 3)

I think the reason is that $_REQUEST[category] is lost when you move to next page. Add that in query parameter list

simple answer: you cannot (you don't have to) use (declare) it in loop sequence. Just take it out of the while-loop

There's no such things as "php-page". Only html pages parsed with php But, I'm pretty sure this would work: index.php#text1?lessonNr=1&chapterNr=1

So many nifty solutions for this one, here's one http://www.tutorialized.com/tutorial/PHP-Random-String-Generator/13903 *couch*google*couch*

$q =$_POST['q']; $r =$_POST['r']; if(!empty($q)) { echo "http://www.mysite.com/images/$q/1"; } if(!empty($r)) { echo "http://www.mysite.com/images/$r/1"; }

min-function might work for you http://www.tizag.com/mysqlTutorial/mysqlmin.php that URL has a good example for your case

Hello Server upload limit (and the place to change it) depends on server. Nothing you can do in code

1. That code doesn't (even try to) print anything. 2. There's one mysql_query() before the actual sql expression is created. (might lead to unexpected errors)

strlen is a sufficient way of checking the length. To get rid of empty strings (like spaces or tabs), use trim

how about this: ... for(var i=1; i < trs.length; i++) ...

maybe src='' would work better

variable in a database? you mean column name? maybe you should check the return value of mysql_query to see if it really fails, or just seems to fail and if it fails, check mysql_error()

another reason I hate JavaScript, strange naming conventions also found a link that may help you in the future exploration http://www.comptechdoc.org/independent/web/cgi/javamanual/javastyle.html

sure $wagebill is not 0 ? btw. all those parenthesis in your calculations are meaningless, considering basic arithmetics

it echoes 'Array' because it is an array... if you want to echo all the values in the array, use something like foreach($value in $student_03) { echo $value; }

as warewolfe already said, $query2 doesn't even slightly resemble an SQL query

you can catch the array just like the other posted variables: e.g $student_03 = $_POST['student_03']; $student_03 is an array holding all the checked values

whole query is pretty much f*cked up, the amount of single quotes is hurting my eyes I think you are looking something like this (I persume $drugfact is variable, what else could it possibly be) "UPDATE players SET drugs = drugs + $drugfact*300, dpayout = $drugfact*300" You don't need single quotes unless you are working with varchars and alike

oh, sorry I missed that you could try .style.background-color Not on expert on JavaScript, but something I would try

Using single quotes with variables inside double quotes will print the variable name (not the value) you'll need to escape those single quotes (\)

document.getElementById('link').border could work document.getElementById('link').style.border should work

Your statements (opening and closing curly brackets) are a bit messed up: that code will never enter to checking for 'steakhouse' if the $step == 'add' btw. using e.g. switch-case statement would be cleaner and you would better avoid errors like such

pretty sure it should be end($numbers).... no more drinks for you!