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Home/ Questions/Q 7630283
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Editorial Team
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Editorial Team
Asked: May 31, 20262026-05-31T06:03:47+00:00 2026-05-31T06:03:47+00:00

I want to pass array myWorkout to ‘play_workout.php’. I want ‘play_workout.php’ to open and

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I want to pass array myWorkout to ‘play_workout.php’.
I want ‘play_workout.php’ to open and display the contents of myWorkout (for this example).
(Once I see that this is working I will parse the data from myWorkout and write it back to a database).
I’m not getting any errors in firebug, but play_workout is not being opened nor is is capturing the array object myWorkout.

I would appreciate a second glance at this.
Thanks as always!

page workout_now.php

<div id="playworkout"><button onClick="playWorkout()">Play Workout</button></div>

JAVASCRIPT

function playWorkout(){

  var arr = $("#dropTargetframe > li").map(function(){
  return $(this).attr('data-id');}).get();

     var myRoutine = arr;
     var myWorkout = new Array();
        for (var i in myRoutine){
           if (myRoutine[i])
             myWorkout.push(myRoutine[i]);
            }
          //array appears like ["4", "5", "1", "4"]
          JSON.stringify(myWorkout);
          encodeURIComponent(myWorkout);

         var url = "http://localhost/RealCardio/play_workout.php";

         $.get(url, myWorkout); 

page play_workout.php

<?php
 ...
 $arrayWorkout = json_decode($_REQUEST['myWorkout']); 
 print_r($arrayWorkout);     
...
?>
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-31T06:03:49+00:00Added an answer on May 31, 2026 at 6:03 am

    Assuming, that as in your comment the array elements doesn’t contain any special chars, just numbers, simply do

    var myWorkoutPost=myWorkout.join('x');
    

    Transport this to the server the way you want (form hidden field, AJAX, ..) and in PHP do

    $myWorkout=explode('x',$_REQUEST['myWorkoutPost']);
    
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