Array inside an array PHP SQL

Question

I have tried everything I can think of for the last two days. I'm truly lost. I need to get this line of code to change based on my database. I have tried everything I could find on php.net and many forums.

$Myday=>array('/index.php?day=$Myday&year=$MYyear','linked-day'),

Any ideas or help would be much appericated


$SQL = "SELECT * FROM DATE_BOOK'"; 
$result = mysql_query( $SQL ); 
while( $row = mysql_fetch_array( $result ) ) { 
    $agenda_id = $row["agenda_id"]; 
    $MYyear = $row["agenda_year"]; 
    $Myday = $row["agenda_day"];  
    $agenda_month = $row["agenda_month"]; }

    $days = array( 
        $Myday=>array('/index.php?day=$Myday&year=$MYyear','linked-day'),
        $Myday=>array('/index.php?day=$Myday&year=$MYyear','linked-day')); 

    echo generate_calendar(2009, 1, $days, 3, '/weblog/archive/2004/Jan');

Tim · Accepted Answer · 2009-10-20 00:33:30Z

2

You want this out side your loop:

$days = array();

And this inside your while loop:

$days[$Myday] = array('/index.php?day=$Myday&year=$MYyear','linked-day');

Your loop isn't actually adding to your array its just assigning a bunch of variables that get rewritten anyways.

answered Oct 20, 2009 at 0:33

Tim

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1 Comment

Michael Howey Over a year ago

Funny, I over looked the simple option. I thought I needed to use foreach to do this. Thank you so much.

takrl · Accepted Answer · 2011-07-15 14:40:30Z

0

For anyone else trying to get http://keithdevens.com/software/php_calendar to work here is the simple code. When trying to figure it out I saw a lot of people who were looking for an answer.

$xgenda_real_date = date("m/d/Y");
$xgenda_month = date("m");
$xgenda_day = date("d");
$xgenda_year = date("Y");

$days = array();

$query = "SELECT * FROM DATE_BOOK";
$result = mysql_query( $query );
while( $row = mysql_fetch_array($result ) ) {
    $Myday = $row["agenda_day"];
    $agenda_id = $row["agenda_id"];
    $agenda_year = $row["agenda_year"];
    $agenda_month = $row["agenda_month"];

    $days[$Myday] = array("/index.php?day=$Myday&year=$MYyear",'linked-day');
}

echo generate_calendar($xgenda_year, $xgenda_month, $days, 3,'index.php?page_id=$page_id', $pn);

echo "<p>$xgenda_real_date";

edited Jul 15, 2011 at 14:40

takrl

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answered Oct 20, 2009 at 1:30

Michael Howey

1

1 Comment

random Over a year ago

Isn't this exactly what Tim answered with?

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