Using ajax to display what is being typed not working, PHP, AJAX, JAVASCRIPT

Question

Using ajax, I'm trying to display what is being typed in the text box, but it's not displaying anything at all for some reason. I know the ajax function itself got called, by using alert inside the function, and I think the real problem is actually in test2.php, but I'm not sure what I did wrong. Please take a look:

test1.php

<?php

include('ajax.php');

echo "<input type = 'text' name = 'select' onkeyup = 'ajax(\"test2.php\",\"select\",\"output\")'>";
echo "<div id = 'output'/>";

?>

test2

<?php

$select = $_POST['select'];
echo $select;

?>

ajax.php

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>

<script type = "text/javascript"> 

function ajax(url,select,id) {

      $.ajax({
           type: "POST",
           url: url,
           data: { select: $('select[name="select"]').val()},
           error: function(xhr,status,error){alert(error);},
           success:function(data) {
             document.getElementById( id ).innerHTML = data;
           }

      });

}

</script>

select[name="select"] is invalid. Shouldn't it be "input[name='select']" — ksealey
– ksealey, Commented Jun 16, 2015 at 18:38

ksealey · Accepted Answer · 2015-06-16 18:49:25Z

3

function ajax(url,unused,id) {
    $.post(url,{select: $('input[name="select"]').val()})
    .error(function(e){
        alert(e);
    })
    .success(function(d){
        $("#"+id).text(d);
    });
}

edited Jun 16, 2015 at 18:49

answered Jun 16, 2015 at 18:45

ksealey

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3 Comments

frosty Over a year ago

@ksealey Anyway for me to pass in the name in the parameter?

ksealey Over a year ago

@frosty The name of the input field?

frosty Over a year ago

@ksealey Actually that's resolved. But I have a recent question open, if you want? stackoverflow.com/questions/30905603/…

CodeGodie · Accepted Answer · 2015-06-16 18:47:39Z

0

The problem is here:

 data: {select: $('select[name="select"]').val()},

There is no select element. And if you meant to get the id named element, then you need to change it to:

data: {select: $('#select[name="select"]').val()},

or in your case:

data: {select: $('input[name="select"]').val()},

answered Jun 16, 2015 at 18:47

CodeGodie

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