[SOLVED] PHP arrays into arrays need help

[ashton321]

Hello

I am trying to highlight the days on my calendar based on the dates that i have in my database.  Currently I can only get it to display the last element in the database which leads me to believe that i have an error in my arrays

 

	
<?php
             ## My database
             $query = mysql_query("SELECT * FROM speaking_engagements");
while($result = mysql_fetch_array($query)){
$dates = $result['date'];
$dates = explode('-',$dates);
$dates = $dates[2];
}
#$dates = explode('-',$dates);
#$dates = $dates[2];
#$numdays = count($dates);
$time = time(); 
$today = date('j',$time); 
$days = array(
	$today=>array(NULL,NULL,'<span style="color: red; font-weight: bold; font-size: larger;">'.$today.'</span>'),
	$dates=>array('#'), 
        8=>array('#'),  ## a hard coded one that works
    );
$today = getdate();
$year = $today['year'];
$month = $today['mon'];
echo '<div id="calendar">';
    echo generate_calendar($year, $month, $days, 3, NULL, '/weblog/archive/2004/Jan'); 
echo '</div>';
?>

[abazoskib]

	
$dates = array();
$query = mysql_query("SELECT date_column FROM speaking_engagements");
while($result = mysql_fetch_array($query)){
   	$date = $result['date'];
   	$date_parts = explode('-',$dates);
   	$dates[] = $date_parts[2];
}

 

You should never use SELECT * in an application. It's trouble, besides, it looks like you only need one column anyway. Why waste the resources? Replace "date_column" with the name of your date column.

[ashton321]

Yeah, there is no reason to SELECT * I did that when I was frustrated and couldn't get it to work.  Now I have tried what you suggested and now I am not getting any results to show, not even the last.  I tried to put it into the calendar array and there was no error, but i tried to echo $dates and it output Array and when I tried to output $dates[0] there was nothing.  Here is the updated code i tried.

 

<?php 
#database connection here
$dates = array();
$query = mysql_query("SELECT date FROM speaking_engagements");
while($result = mysql_fetch_array($query)){
	$date = $result['date'];
	$date_parts = explode('-',$dates);
	$dates[] = $date_parts[2];
}
$time = time(); 
$today = date('j',$time); 
$days = array(
	$today=>array(NULL,NULL,'<span style="color: red; font-weight: bold; font-size: larger;">'.$today.'</span>'),
	$dates[0]=>array('#'),
	$dates[1]=>array('#'),
	$dates[2]=>array('#'),
	$dates[3]=>array('#'),		
        8=>array('#'),
    );
echo $dates[0];

$today = getdate();
$year = $today['year'];
$month = $today['mon'];
echo '<div id="calendar">';
    echo generate_calendar($year, $month, $days, 3, NULL, '#'); 
echo '</div>';
?> 

[scripterdx]

it was the same proble that i had

[scripterdx]

the array when is filled?

[knsito]

try

$query = mysql_query("SELECT `date` FROM speaking_engagements");

 

[added accent]

Are you sure you are getting dates in format you are expecting?

[DavidAM]

You could just use DayOfMonth(`date`) in the SQL to get the day number instead of picking it apart in PHP:

SELECT DayOfMonth(`date`) AS CalDay FROM speaking_engagements

 

Note: Are the back-ticks still necessary?  I think so, but I'm not sure.  You should always avoid using reserved words for column names, table names, database names, etc.  It just makes things harder on you when accessing the table.

[abazoskib]

It could be that "date" is a mysql keyword? Did you try the backticks like kinsito said? Try echoing mysql_error()?

[ashton321]

Found the problem now!

<?php	$dates = array();
$query = mysql_query("SELECT date FROM speaking_engagements");
while($result = mysql_fetch_array($query)){
	$date = $result['date'];
	$date_parts = explode('-',$dates);
	$dates[] = $date_parts[2];
}
?>

 

is now

<?php	$dates = array();
$query = mysql_query("SELECT eventdate FROM speaking_engagements") or die(mysql_error());

while($result = mysql_fetch_array($query)){
	$date = $result['eventdate'];
	$date_parts = explode('-',$date);
	$dates[] = $date_parts[2];
}?>

 

I changed the column name in the database from the keyword date and that was not the issue.  Then I look at what I was exploding, and saw that i was exploding dates with an "s" when it needed to be without the "s".  Thanks for all the help with the arrays

[abazoskib]

my mistake!! it must have had you going nuts. glad you got it sorted.

[ashton321]

It has been driving me nuts for a few days now!  But, now that I got my data out, I need to make 01 for a day turn into 1 but I donk know how to run like and if statement or something on the entire array because I will not know its position in the array?

[abazoskib]

It has been driving me nuts for a few days now!  But, now that I got my data out, I need to make 01 for a day turn into 1 but I donk know how to run like and if statement or something on the entire array because I will not know its position in the array?

 

You can either use ltrim: $number = ltrim($number, '0');

 

Or you can just cast your $number variable as (int) and the zero will be removed. Your choice.

[ashton321]

I am going to try the (int) cast because I am not sure if this is true but the other may remove the 0 from a day like 20 and make it 2?

[mikesta707]

with an ltrim it won't take any zeros on the right of the string, so they would both work the same

[ashton321]

Sorry to be a pain here but im not that great with arrays.  i tried doing the int cast and it wasnt working so I tried the ltrim and now it is echoing A r r a as my array rather than the numbers.

<?php $dates = array();
$query = mysql_query("SELECT eventdate FROM speaking_engagements") or die(mysql_error());

while($result = mysql_fetch_array($query)){
	$date = $result['eventdate'];
	$date_parts = explode('-',$date);
	$dates[] = $date_parts[2];
}
$dates = ltrim($dates, '0');
$time = time(); 
$today = date('j',$time); 
$days = array(
	$today=>array(NULL,NULL,'<span style="color: red; font-weight: bold; font-size: larger;">'.$today.'</span>'),
	$dates[0]=>array('#'),
	$dates[1]=>array('#'),
	$dates[2]=>array('#'),
	$dates[3]=>array('#'),		
        );
echo "$dates[0]  ";
echo "$dates[1]  ";
echo "$dates[2]  ";
echo "$dates[3]  ";

[abazoskib]

you cant apply it to an array like that. you can do a few things.

 

while($result = mysql_fetch_array($query)){
      $date = $result['eventdate'];
      $date_parts = explode('-',$date);
      $dates[] = ltrim($date_parts[2],'0');
   }

 

Or you can setup a foreach loop by reference afterwards to do it. Or you could even use array_map. Best case is above though, its the fastest.

[ashton321]

ok thank you it works!

 

one last question is it possible to use some time of loop in

<?php $days = array(
	$today=>array(NULL,NULL,'<span style="color: red; font-weight: bold; font-size: larger;">'.$today.'</span>'),
	$dates[0]=>array('#'),
	$dates[1]=>array('#'),
	$dates[2]=>array('#'),
	$dates[3]=>array('#'),		
        );?>

for when I add elemts to the database they all get displayed and i dont have to change the script?

[abazoskib]

can you explain youre question a little bit better?

[ashton321]

I'll try.  This is a list of days that are going to become links on a calendar once i get it to work.  The calendar makes the link based on the

<?php $days = array(
	$today=>array(NULL,NULL,'<span style="color: red; font-weight: bold; font-size: larger;">'.$today.'</span>'),
	$dates[0]=>array('#'),
	$dates[1]=>array('#'),
	$dates[2]=>array('#'),
	$dates[3]=>array('#'),		
        );?>

array.  i would like to know if I am allowed to put like a for or while type loop inside the $days array to make every element of the $dates array show up.  right now i had to add

$dates[0]=>array('#'),
	$dates[1]=>array('#'),
	$dates[2]=>array('#'),
	$dates[3]=>array('#'),

for each element.  i would like it to be that if i add a fifth element to the database it also becomes a link on the calendar without me going in and adding

$dates[4]=>array(#),

 

Hope that is more clear

[abazoskib]

<?php
$days = array(
      $today=>array(NULL,NULL,'<span style="color: red; font-weight: bold; font-size: larger;">'.$today.'</span>');

foreach($dates as $key => $value) {
     $days[$key] = array('#');
}

?>

 

Not tested.

[ashton321]

I dont have much experience with foreach loops and i got

Parse error: syntax error, unexpected ';', expecting ')' in /homepages/44/d263144020/htdocs/webdev/inc/sidebar.php on line 15

 

when I tested

$days = array(
	$today=>array(NULL,NULL,'<span style="color: red; font-weight: bold; font-size: larger;">'.$today.'</span>');
	foreach($dates as $key => $value) {
	$days[$key] = array('#');
	});

[abazoskib]

<?php
$days = array(
      $today=>array(NULL,NULL,'<span style="color: red; font-weight: bold; font-size: larger;">'.$today.'</span>'));

foreach($dates as $key => $value) {
     $days[$key] = array('#');
}

?>

that should work

[ashton321]

It works without an error now, but it is only highlighting the first 3 days rather than the values

[abazoskib]

ok one more time

 

<?php
$days = array(
      $today=>array(NULL,NULL,'<span style="color: red; font-weight: bold; font-size: larger;">'.$today.'</span>'));

foreach($dates as $key => $value) {
     $days[$value] = array('#');
}

?>

[ashton321]

Thank you so much it works!