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I have a slight problem. I have several arrays in php with different team names. Each array contains teams of a certain league. When I click an add button I get the option to add a new entry to the calendar. I want the drop down to have only the teams for that league. onclick of the add button I call a javascript function that knows what division was clicked. However in order to give the javascript the information for which teams to display I have to pass it one of the php arrays. The problem I am having is telling php which array to pass to javascript depending on which league javascript is on. I don't want to specify the array myself because there is an option to add a league and this would mean having to code in more code each time a league is added. The point of the site is being dynamic.

here is some code.

for ($i = 0;$i<$sizeof($leaguesarray);$i++){
    $htmlimploded[$i] = implode($html[$i]);
}

here I have used emplode to make all of my php arrays readable into javascript.

for (var h = 0; h<size; h++){ // goes through every league 
    if(h == leaguenum){ // finds the league for the clicked add button
        // this is the line that I have trouble with I can't think of 
        //anyway of telling it which array to use since it is serverside code.
        var myarray = ["<? echo $htmlimploded[]?>"]; 
    }
}

Javascript code above.

share|improve this question
    
Do you not have unique identifiers assigned to each league that would also be the key in one of your arrays? –  prodigitalson Apr 12 '12 at 4:05
    
I don't believe so. Either way I would still have to specify it. I'm thinking I'm just going to have to change my approach and just pass the data to a form and refresh the page with the new parameters and get rid of javascript. –  user541597 Apr 12 '12 at 4:10
    
Well you can do that but you should have a way to reference each league uniquely. Why are you not using a database? –  prodigitalson Apr 12 '12 at 4:11
    
the leagues are in a database. I pull each leagues teams from a database and put it into a php array. which all works fine. However the javascript is what detects what league I want to add to not php. –  user541597 Apr 12 '12 at 4:16

3 Answers 3

up vote 0 down vote accepted

Imploding works but why not json_encode($array)? It's a simpler, built in way to turn php arrays into javascript objects or arrays. If you have something like:

$league1 = array('team1', 'team2');
$league2 = array('team3, 'team4') ;

Then make a multidimensional associative array of these:

$all_teams = array('league1'=>$league1, 'league2'=>$league2);

encode it into a Javascript object and print it into your JS:

$encoded = json_encode($all_teams);
print 'var teamObject = '.$encoded.';';

If you were to console.log(teamObject) you'd see something like this:

{"league1": ["team1", "team2"], "league2": ["team3", "team4"]}

Looks complicated, but now you can pull out the array you desire very easily. The league 1 array is teamObject.league1 and the league2 array is teamObject.league2, and so on.

share|improve this answer
up vote 0 down vote

i think you missed something in the following code:

var myarray = ["<? echo $htmlimploded[]?>"];

By right, it should be:

var myarray = ["<?php echo $htmlimploded[]?>"];
share|improve this answer
up vote 0 down vote

Assuming that PHP knows the names of the leagues and the teams and that JavaScript knows the league name that is clicked, You can wrap the arrays of the team names inside an object with the league as the name of the property.

<?php 
$arr = array("League1" => array("Team 1", "Team 2"), 
             "League2" => array("Team 3", "Team 4")
       );
?>
var obj = {};
<?php foreach ($arr as $k => $v): ?>
obj.<?php echo $k; ?> = ["<?php echo implode('","', $v); ?>"];
<?php endforeach; ?>

Then when a user selects a league, you can loop through the array of the property (which is the league name) of the object.

clickedLeague = "League1";
for (var i = 0; i < obj[clickedLeague].length; i++)
{
    console.log(obj[clickedLeague][i]); // Logs the team name to console
}
share|improve this answer

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