show next value from mysql php in ajax

Question

I have the following Code which works good now,

<!DOCTYPE html>
<html>
<body> 
    <div id="one">
    echo "data 1";
   <input type="hidden" id="one" name ="one" value="'.$one.'">

  </div>

  <div id="two">
     echo "data 1";
    <input type="hidden" id="two" name="two" value="'.$one.'">

  </div>
</body> 
</html>

for each DIV I have data 1 2 3...more comes from MySQL php query. So, when the Hidden Id = one delay some second and show the next result data 2, until we get the input hidden to one again. that all come from database. automatically. I have all that working.

I am able to use ajax to show the data, but when i come to read the Input value and change accordingly. I have problem...

I have tried..

var value1  = $('one').attr( "value" ); //
 var value2  = $('two').attr( "value" ); //
 (function data() {
                $.ajax({
                    url: 'data.php',
                    if (value1 == 'one') { 
                    success: function(data) {
                    $('#one').html(data);
                    },
                    }
                    complete: function() {
                        // Schedule the next request when the current one's complete
                        setTimeout(data, 5000);
                    }
                });
            })();

But it seems not working....

The first problem I see is that you've got the div and the inputs as having the same id. These need to be unique. Perhaps label each as div-one and input-one, then div-two and input-two. — Paul
– Paul, Commented Oct 17, 2013 at 10:04
Also, when calling the ID's in jQuery don't forget the # sign. $('#one') not $('one') — Dan P.
– Dan P., Commented Oct 17, 2013 at 10:05

Paul · Accepted Answer · 2013-10-17 11:28:56Z

1

A couple of small modifications may help this work:

<!DOCTYPE html>
<html>
<body> 
    <div id="div-one">
    <?php echo "data 1"; ?>;
   <input type="hidden" id="one" name ="one" value="<?php $one; ?>;">

  </div>

  <div id="div-two">
     <?php echo "data 2"; ?>;
    <input type="hidden" id="two" name="two" value="<?php $two; ?>;">

  </div>
</body> 
</html>

Javascript:

var value1  = $('#one').val();
var value2  = $('#two').val();
 (function data() {
                $.ajax({
                    url: 'data.php',
                    success: function(data) {
                       if (value1 == 'one') { 
                          $('#div-one').html(data);
                       }
                    },
                    complete: function() {
                        // Schedule the next request when the current one's complete
                        setTimeout(data, 5000);
                    }
                });
            })();

edited Oct 17, 2013 at 11:28

answered Oct 17, 2013 at 10:11

Paul

8142 gold badges13 silver badges31 bronze badges

Sign up to request clarification or add additional context in comments.

5 Comments

MR.Internet Over a year ago

It seems doesn't load the next data, why?

Paul Over a year ago

Have you checked the result from the ajax request? Do a console.log(data); in the success to see what is being returned...

Paul Over a year ago

Ah, I just updated the javascript code to use .val() instead of attr. Give that a go.

MR.Internet Over a year ago

I am getting the following error, SyntaxError: missing : after property id if (value1 == 'one') {

Paul Over a year ago

Ok. I think I got this. I've now placed the "if (value1 == 'one') {" part inside the success function. This makes far more sense, because success is a property of the ajax: function api.jquery.com/jQuery.ajax

Collectives™ on Stack Overflow

show next value from mysql php in ajax

1 Answer 1

5 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

5 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related