assigning a php multidimensional array to a java script array [duplicate]

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Possible Duplicate:
assigning multidimensional php array to javascript array

I have a 2D php array as mentioned below and I'm trying to assign it to a java script array.

<?php 
$taskArray = array( 
           array( taskName => "t1", 
                      startDate => 11/01/2011,
                      duration => 10  ),
               array( taskName => "t2", 
                      startDate => 11/02/2011,
                      duration => 5  ),
               array( taskName => "t3", 
                      startDate => 11/05/2011,
                      duration => 8  ),
             );
?>



<script type="text/javascript"> 
function getArray(){

 var a=new Array;

 <?php

   $x = count($taskArray);

   for($i=1;$i<=count($taskArray); $i++){
     echo "a[$i][taskName]='".$taskArray[$i]["taskName"]."';\n";
     echo "a[$i][startDate]='".$taskArray[$i]["startDate"]."';\n";
     echo "a[$i][duration]='".$taskArray[$i]["duration"]."';\n";
   }

  ?>            
}
</script>

problem occurs when using the key strings in the above echo lines. what is the correct concatenation? Note:echo "a[$i]='".$taskArray[$i]."';\n"; works

If you have access to the JSON extension (usually enabled), then you can use json_encode with your array, which should get you an array that JavaScript can use. — jValdron
– jValdron, Commented Dec 2, 2011 at 19:18
Your on the correct path but the quotes in your echo statement are way off. see answer below. — vdbuilder
– vdbuilder, Commented Dec 2, 2011 at 19:57
Your keys aren't strings. Also, arrays in JS don't have keys, you're making an array of objects. As said above, just use JSON_ENCODE($taskArray) and assign it to some value. — Incognito
– Incognito, Commented Dec 2, 2011 at 20:44

Jeremy T · Accepted Answer · 2011-12-02 20:31:28Z

1

jValdron's comment would be the simplest way to go, but if you really want to do it yourself, I think the only think you're missing is some quotes in your generated javascript. a[$i][taskName] will treat taskName as a variable in the javascript. I believe you wanted

echo "a[$i]['taskName']=".$taskArray[$i]["taskName"]."';\n";
// etc...

Also, there are two other things I noticed. First, you're starting at index 1, which is actually the second thing in your array. This may be intentional, but most of the times that I've seen this it was done by accident. Second, why are you storing the count in $x and then recalculating it every time through the loop?

So, if I had to guess, I'd say the best way for the loop to be written would be

for($i=0; $i<$x; $i++) {
    //etc...
}

I'm not a huge fan of calling it $x (I'd prefer a more meaningful name like $len or $arrLen or something), but that's not a huge deal in this instance.

EDIT: I just noticed that your function has no return statement. So in all, the function should look like:

<?php 
$taskArray = array( 
           array( taskName => "t1", 
                      startDate => 11/01/2011,
                      duration => 10  ),
               array( taskName => "t2", 
                      startDate => 11/02/2011,
                      duration => 5  ),
               array( taskName => "t3", 
                      startDate => 11/05/2011,
                      duration => 8  ),
             );
?>



<script type="text/javascript"> 
function getArray(){

 var a=new Array;

 <?php

   $len = count($taskArray);

   for($i=1;$i<=$len; $i++){
     echo "a[$i]['taskName']='".$taskArray[$i]["taskName"]."';\n";
     echo "a[$i]['startDate']='".$taskArray[$i]["startDate"]."';\n";
     echo "a[$i]['duration']='".$taskArray[$i]["duration"]."';\n";
   }

  ?>
  return a;         
}
</script>

edited Dec 2, 2011 at 20:31

answered Dec 2, 2011 at 19:25

Jeremy T

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3 Comments

Jeremy T Over a year ago

@vdbuilder the only difference I see is that you added the parenthesis. It'll work fine without them, so why is your answer better?

sheng Over a year ago

@jearemy I start at index 1 purposely and there is no need of recounting the array size. But still the echo dosen't work with the above suggested line

Jeremy T Over a year ago

@sheng I just realized that there's no return statement in your javascript code. I edited my response to include that. Does it work now?

vdbuilder · Accepted Answer · 2011-12-02 21:59:18Z

1

Change your code to this:

<?php 
    $taskArray = array( 
       array( taskName => "t1", 
              startDate => 11/01/2011,
              duration => 10  ),
       array( taskName => "t2", 
              startDate => 11/02/2011,
              duration => 5  ),
       array( taskName => "t3", 
              startDate => 11/05/2011,
              duration => 8  ),
       );
?>


<script type="text/javascript"> 
function getArray(){

var a=new Array;

<?php
    $x = count($taskArray);

    for($i=1;$i<count($taskArray); $i++){
        echo ("a[$i]['taskName']='".$taskArray[$i]['taskName']."';\n");
        echo ("a[$i]['startDate']='".$taskArray[$i]['startDate']."';\n");
        echo ("a[$i]['duration']='".$taskArray[$i]['duration']."';\n");
    }
?>            
    return a;
}
</script>

edited Dec 2, 2011 at 21:59

answered Dec 2, 2011 at 19:53

vdbuilder

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1 Comment

sheng Over a year ago

I tried this but still it dosen't work

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