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Last Active:: Oct 01 2012 08:10 PM
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Posts I've Made

In Topic: Find the weekend dates in Calendar

Posted 1 Oct 2012
CTphpnwb, on 01 October 2012 - 08:40 PM, said:
Because 6 is true.
Try:
```
    if ($day_of_week == 0 || $day_of_week == 6)
```
See...I just learned something. Kudos to you. I mean it, you deserve it,
In Topic: Find the weekend dates in Calendar

Posted 1 Oct 2012
Well I thought I had this logic for a second but I was sadly mistaken.
```
for($day = 1; $day < 32; $day++) {
	$date_time = strtotime($year.'/'.$month.'/'.$day);
	$day_of_week = date("w",$date_time);
	if ($day_of_week == 0 || 6)
	echo $day_of_week." = ".$day_name[$day_of_week]."<br>";
}
```
This still printed out all the dates? Why did this not work.

I have soooo much to learn!
In Topic: Find the weekend dates in Calendar

Posted 1 Oct 2012

CTphpnwb, on 01 October 2012 - 07:27 PM, said:
Date() can return the day number, 0 - 6, starting with Sunday.

Yes, I've been reading about this function, but I'm not sure how to implement it!
In Topic: Login Form not authenticating!?

Posted 25 Sep 2012
JackOfAllTrades, on 25 September 2012 - 05:47 AM, said:
First, learn to use the Javascript error console.
Second, learn to use Firefox and Firebug (or another browser and JS debugger, like Google Chrome and its Developer Tools) to debug your own errors.
Third, and most importantly, learn how everything in web development fits together. It really looks to me like you're just copying and pasting code from random sources without really understanding the big picture.

So, what's the value of the variable url in your AJAX call?

I am, nor do I want to be a web designer. I have been given this task from my boss to figure out. Our company is trying to design a intranet with login. We have bought a "Admin Template" which is what you see below the PHP. It came with the html and javascript already there. I just added the PHP to try to get it to work. I have no preconceived notions that I know what I'm doing which is why I came here for help. I'm sorry if my ingnorance upsets you, but I have no problem helping ohters out when I have the expertise, or asking for help when I don't.

as far as the url variable, I have changed it to
```
					 $.ajax(this.action, {
							data: {
								username:	username,
								pass:	pass
							},
							success: function(data)
							{
								if (data.logged)
								{
									document.location.href = 'dashboard.php';
								}
								else
								{
									formWrapper.clearMessages();
									displayError(data.error || 'Invalid user/password, please try again');
								}
							},
							error: function()
							{
								formWrapper.clearMessages();
								displayError('Error while contacting server, please try again');
							}
					  });
				}
			});
```
Now it always says "Invalid user/password, please try again" so something is not doing the proper call.

Once again, I am a novice at this and I'm looking to everyone here for guidance.

Thanks for your time and patience.
In Topic: Remember User From Session

Posted 24 Sep 2012
Thanks guys!
```
$username = $_SESSION['user']['username'];
```
This did the trick.