如何通过AJAX更改变量？

Question

I need advice with AJAX.

I have page where i use to show info in block like:

<div1 container>
   <div2>
      some data
   </div2>
   <div3>
    <?php echo $variable;?>
   </div3>
</div>

and have another page that produce data:

function example( $array )
{
   ...some code...
return $answer
}

$variable = example($array);

i am sending AJAX query with some POST data from form and i need after operations get new variable and put it in same new params.

Here is my AJAX function:

function sendDataChild(btn) {
var form = $(btn).closest('FORM[name=answer-form]');
var data = form.serialize();
$.ajax({
    type: "POST",
    url: "get.php",
    dataType: "post",
    data: data,
    cache: false,
    success: function (data) {
        form[0].reset();
        //document.location.reload();
    },
    error: function (xhr, str) {
        alert('Возникла ошибка: ' + xhr.responseCode);
    }
});
return false;};

when i clear // and just reload page it works, but it is wrong way cause my page is quite big. Can some body advise how to make it?

Answer 1

I think you want the success function to output the data into the div where you want the variable to show:

success: function (data) {
    form[0].reset();
    $("#divId").html(data);
}

Answer 2

Give a id to the div where you want to show the result

<div id="div3">

</div>

And you need to change your ajax

 function sendDataChild(btn) {
var form = $(btn).closest('FORM[name=answer-form]');
var data = form.serialize();
$.ajax({
    type: "POST",
    url: "get.php",
    dataType: "post",
    data: data,
    cache: false,
    success: function (data) {
        $('div3').html(data);
    },
    error: function (xhr, str) {
        alert('Возникла ошибка: ' + xhr.responseCode);
    }
});
}

Answer 3

HTML page

<div1 container>
   <div2>
      <!--some data-->
   </div2>
   <div3 id="div3">

   </div3>
</div>

PHP page

function example( $array )
{
   //...some code...
   echo $answer;
   exit;
}

And your JavaScript

function sendDataChild(btn) {
var form = $(btn).closest('FORM[name=answer-form]');
var data = form.serialize();
$.ajax({
    type: "POST",
    url: "get.php",
    dataType: "post",
    data: data,
    cache: false,
    success: function (data) {
        $('div3').html(data);
    }
});
}