CODEWITHSUNDEEP
phpjqueryvariables
Robert Mrsic
Robert Mrsic

Reputation: 53

Php to Jquery Variable

I have this in my config.php file

     $DOMAIN = 'http://domain.com';
 $DOMAIN_SSL = 'http://domain.com';
$DOMAIN_ROOT = 'var/domain/root/html';

config.php is included in each page on first line

Now I have this jQuery code

var country = '<?php echo $_GET['country'];?>'; 
var category = 'Cat' ;
var category = '<?php echo $DOMAIN.'/link'; ?>';

How can I get variables from php with jQuery code to make jQuery code like this

var country = '$country'; 
var category = 'Cat' ;
var url= '$domain','/link';

ORIGINAL CODE I NEED TO EXLUDE PHP

var country = '<?php echo $_GET['country'];?>'; 
                var category = 'CAT1' ;               
                function loadData(page){
                    loading_show();                    
                    $.ajax
                    ({
                        type: "POST",
                        url: "<?php echo $DOMAIN ?>/post.php",
                        data: "page="+page +"&category="+category+"&country="+country,
                        success: function(msg)
                        {
                        loading_hide();
                        $("#content-large").html(msg);
                        }
                    });
                }
                loadData(1);  // For first time page load default results
                $('#content-large .pagination li.active').live('click',function(){
                    var page = $(this).attr('p');
                    loadData(page);

                });           
                $('#go_btn').live('click',function(){
                    var page = parseInt($('.goto').val());
                    var no_of_pages = parseInt($('.total').attr('a'));
                    if(page != 0 && page <= no_of_pages){
                        loadData(page);
                    }else{
                        alert('Enter a PAGE between 1 and '+no_of_pages);
                        $('.goto').val("").focus();
                        return false;
                    }

                });
            });

Upvotes: 1

Views: 3870

Answers (3)

Mizuho
Mizuho

Reputation: 90

Assuming this is on the page and not in some separate JS file, you could just write the entire statement in PHP.

<?php echo "var country = " . json_encode($_GET['country']) . ";"; ?>

Upvotes: 0

Armatus
Armatus

Reputation: 2191

Frankly, I think you don't need to overcomplicate things here so much. This may not be the fanciest solution or anything but it should work.

The idea is, on the actual page (say, home.php) you would echo an HTML element containing the values you need, such as:

//Fill this array with all the values that you want available in JS
$JSONarray = array('all'=>'the','values'=>'you','want'=>'in','JS');
//Later print it on the page
echo "<span class='hidden' id='transferred_values'>"
    .json_encode($JSONarray)."</span>";

(Side note: you would set the class "hidden" to be invisible to the user.)

Now your external JS (say, scripts.js) can retrieve the values using JS only which will allow you to use the compressor:

$(function(){
    var transfer = $.parseJSON( $('#transferred_values').text() );
});

Let me know if this violates good practice or any rules I'm not aware of :)

Upvotes: 0

Sampson
Sampson

Reputation: 268462

Updated Answer

If you're attempting to get values from the URL, much like you would using $_GET in PHP, I would encourage you to use something like the jQuery URL Parser:

var country = $.url().param("country"); // Gets the 'country' parameter

Download jQuery URL Parser: https://github.com/allmarkedup/jQuery-URL-Parser

Original Answer

I would encourage you to keep it simple; just use formatted strings and json_encode. For example:

$country  = "United States";
$domain   = "http://stackoverflow.com";
$category = 5;

printf( "var country  = %s;", json_encode( $country ) );
printf( "var category = %s;", json_encode( $category ) );
printf( "var domain   = %s;", json_encode( $domain . "/link" ) );

Which outputs:

var country  = "United States";
var category = 5;
var domain   = "http:\/\/stackoverflow.com\/link";

Upvotes: 4

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