Jquery .load function. Passing a php variable to .load()

Question

Im trying to pass a PHP variable, ($name) through the .load function. Every time i try to call similar.php, instead of search for whatever is in $name, its searching for

Heres my java code;

    <script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js">
</script>
<script>
$(document).ready(function(){
  $("button").click(function(){
    $("#div1").load("similar.php?search="+"&name");
  });
});
</script>

Ive also tried, orginally:

<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js">
    </script>
    <script>
    $(document).ready(function(){
      $("button").click(function(){
        $("#div1").load("similar.php?search=<?php $name ?>);
      });
    });
    </script>

Any tips?

I really don't understand what you want to. But you make weird things in your code like load('string'+'string') or syntax error load("string). And Java != JavaScript — Novelist
– Novelist, Commented May 11, 2013 at 19:24
sorry, the post about filtered out php tags. similar.php takes an argument, ie, similar.php?search=snoop%dog snop%dog is stored in $name im calling similar.php from video.php to display similar artists. But when i attempt to call the above code(which you mentioned are both strings) which then returns a php page as a result, which would make sense. So to review, what i need to do, is call the string, similar.php?search= followed by the variable $name Can anyone, or you help me out please? — Alex Emile
– Alex Emile, Commented May 11, 2013 at 19:29
So, that should work $("#div1").load("similar.php?search=<?php $name ?>"); Maybe it's all about that syntax error (missed "). — Novelist
– Novelist, Commented May 11, 2013 at 19:33

adeneo · Accepted Answer · 2013-05-11 19:53:59Z

1

You forgot the last quote and to echo the $name, change this :

$("#div1").load("similar.php?search=<?php $name ?>);

to:

$("#div1").load("similar.php?search=<?php echo urlencode($name) ?>");

And use the DOM inspector to view the code and see that the PHP is outputted properly.

edited May 11, 2013 at 19:53

answered May 11, 2013 at 19:33

adeneo

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10 Comments

Alex Emile Over a year ago

Thanks, but now it doesnt load anything, as can be seen here, soundwavez.net/video/sVaRHJn_dgM Any other suggestions?

adeneo Over a year ago

That's because your URL is not valid, you need to URL encode the strings when there are spaces and special characters.

Alex Emile Over a year ago

Well, going by what you said, ive removed the spaces, $("#div1").load("similar.php?search=<?$name?>"); and now its back to square one, returning a php related page. see the same link

adeneo Over a year ago

The spaces are in your strings, you can't remove the echo and the spaces inside the php code, as that has nothing to do with it ?

Alex Emile Over a year ago

oh im following you now! So i'd have to call $name before the function, switch the space with % assign it to a new variable, then call the new variable? You sir, are a genious, Thanks

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