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Home/ Questions/Q 231733
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Editorial Team
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Editorial Team
Asked: May 11, 20262026-05-11T19:57:28+00:00 2026-05-11T19:57:28+00:00

I m new to CakePhp and JQuery. I am getting an error in using

  • 0

I m new to CakePhp and JQuery.
I am getting an error in using the cakephp code inside my JQuery.

My code

   <script type="text/javascript">
     $(document).ready(function(){
     var attributeid;var fieldname;
     $("#"+<?=$r['Attribute']['id'];?>).change(function () {

     fieldname=<?=$r['Attribute']['label'];?>; 
              alert(fieldname);//this show me that undefined 
             attributeid=<?=$r['Attribute']['id'];?>; 
             alert(attributeid);//But this works

    });//attribute change
});//ready function 

if I echoed ($r['Attribute']['label'];) this value is coming inside my <?php ?>.
But not inside my JQuery.

Note :

attributeid=<?=$r['Attribute']['id'];?>; 
alert(attributeid);//But this works  


Error: 
Name is not defined
fieldname=name; 
alert(fieldname);
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  1. Editorial Team
    Editorial Team
    2026-05-11T19:57:28+00:00Added an answer on May 11, 2026 at 7:57 pm

    You are not thinking about how this is translating over once the variables are echoed.

    If you have a variable $x with the contents “test”, doing this:

    var x = <?=$myvar?>;
    

    Will result in:

    var x = test;
    

    This is not valid (unless test is a variable) because you need quotations around it to make it a string:

    var x = "<?=$myvar?>";
    

    Which then results in the valid:

    var x = "test";
    

    The reason it works with the other variable is because you are echoing an ID, which is an integer:

    var x = <?=$myid?>;
    

    Would translate to:

    var x = 5;
    

    Which is perfectly valid.

    All this being said, you should put all the stuff you want to send over to Javascript in an array and call json_encode on it to easily and safely print the values over. Without it, you have to worry above about escaping quotes in the string and such.

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