Maq

You can do it all in MySQL: SELECT * FROM applicant WHERE AppliedDate >= DATE_SUB(NOW(), INTERVAL 30 day) ORDER BY AppliedDate DESC;

So what's the problem? What do you want the end result to look like?

Yes, please do. The issue is most likely output to the browser, this includes whitespace, before your header call. This sticky (http://www.phpfreaks.com/forums/index.php/topic,37442.0.html) should explain everything.

What version of FF are you guys using? Looks fine for me in FF3.0.

The only other suggestions is to try doing it through the command line.

Hmm, still don't see anything peculiar. Are you inputting your query exactly like this? SELECT first_name FROM users WHERE last_name ='Simpsons'; The error doesn't seem to relate to your query. Maybe I'm overlooking something, but everything looks valid to me.

Apparently you haven't tried to Google this question. There are plenty of relevant hits. "cakePHP application multilanguage"

Can you show me your table structure? describe users;

Looks correct to me. Are you querying directly in phpMyAdmin?

Why don't people read the stickied threads? Ninja EDIT: http://www.phpfreaks.com/forums/index.php/topic,58799.0.html Which is clearly a stickied thread in this subforum. I moved it from another section. Still doesn't excuse the fact that they haven't taken the time to post in the proper section or search for similar threads.

Have you read this sticky yet? http://www.phpfreaks.com/forums/index.php/topic,37442.0.html

What does Send() return? How does it handle errors within the method?

Yes, you will probably get more responses that way, rather than people seeing 10+ posts and disregarding your thread.

I don't really see anything wrong with it. Try putting DISTINCT first: $get_attacks = "select DISTINCT attack, id, identity, level from attacks where identity = '$identity'"; If that doesn't work then you can try a GROUP BY: $get_attacks = "select * from attacks where identity = '$identity' GROUP BY attack"; Other than that I'm not sure.

Change this line: $get_attacks = "select * from attacks where identity = '$identity'"; to: $get_attacks = "select id, identity, level, DISTINCT attack from attacks where identity = '$identity'";

Looks like this line: echo $attack; should be: $display_block .= "$attack"; Or, if you wanted to display it directly after the if is executed: echo "$attack"; If you turned error reporting on you would have seen this. ini_set ("display_errors", "1"); error_reporting(E_ALL);

Missing a single quote after this value: $data['description'] A couple of side notes. - You only need single quotes if the type is not numeric. - You can use curly braces to escape associative arrays rather than concatenating all the variables.

Show us your query. Looks like you're mixing up some quotes.

$result is a resource id. Not what the OP is looking for.

The '@' symbol suppresses the error messages for that line. What is the point of that? They are equivalent. Are you sure it's not returning anything? Your code seems fine to me. Add an else statement with output and see if it displays.

Not sure exactly what you're asking but: (.25 * $multiplier);

You can use mysql_insert_id. All the information you need is in the manual link.

Your "if" statements are missing the second argument after the greater than sign.

date

You couldn't find any examples in the manual? mail.