Ajax form return variables to php

Question

I have another question:

Ajax Forms are working well. Most of them need to do mysql stuff and only return values if the entry could be written or not. I used just echo statements. For example echo "1"; if the values could be written and echo "2"; if the values could not be written.

Now I need to call back 3 variables. I know that I can write them in an array. My problem is just, that I can't return this variable into my visible site.

This is my JavaScript Code:

//Show statistic
$('.statistic_submit').click(function(){
    if ($('#month').val() == 'none' || $('#year').val() == 'none') {
        $("#dialog_empty").dialog( "open" );
        return false;
    }

    var form = $('#statistic_view');  
    var data = form.serialize(); 

    $.ajax({
        url: "include/scripts/user_statistic.php",
        type: "POST",
        data: data,
        success: function (reqCode) {
            if (reqCode == 1) {
                //Show generated table
                $('.done').fadeOut('slow'); 
                $('.done').fadeIn('slow');
            }
            if (reqCode == 2) {
                //No values found
                $('.done').fadeOut('slow');
                $("#dialog_error").dialog( "open" );
            }
        }
    });
    return false;
});

This is my html code:

<div>
    <form id="statistic_view" action="include/scripts/user_statistic.php" method="post">
        <select name="month" id="month">
            <option value="none" class="bold italic">Monat</option>
            <?php 
                for($i=1; $i<=12; $i++){
                    if($i == $month)
                        echo "<option value=\"".$i."\" selected>".$month_name[$i]."</option>\n";
                    else
                        echo "<option value=\"".$i."\">".$month_name[$i]."</option>\n";
                }
            ?>
        </select>
        <select name="year" id="year">
            <option value="none" class="bold italic">Jahr</option>
            <?php 
                for($i=2012; $i<=$year; $i++){
                    if($i == $year)
                         echo "<option value=\"".$i."\" selected>".$i."</option>\n";
                    else
                        echo "<option value=\"".$i."\">".$i."</option>\n";
                }
            ?>
        </select>
        <br/><br/>
        <div id="user_statistic">
            <input type="submit" id="small" class="statistic_submit" value="Daten anzeigen">
        </div>
    </form> 
    <br />
    <div class="done">
        <p class="bold center"><?php echo "Besucher ".$month_name[$month]." ".$year; ?></p>
        <canvas id="cvs" width="680" height="250">[No canvas support]</canvas>
        <script>
            chart = new RGraph.Line('cvs', <?php print($data_string) ?>);
            chart.Set('chart.tooltips', <?php print($labels_tooltip) ?>);
            chart.Set('chart.tooltips.effect', 'expand');
            chart.Set('chart.background.grid.autofit', true);
            chart.Set('chart.gutter.left', 35);
            chart.Set('chart.gutter.right', 5);
            chart.Set('chart.hmargin', 10);
            chart.Set('chart.tickmarks', 'circle');
            chart.Set('chart.labels', <?php print($labels_string) ?>);        
            chart.Draw();
        </script>
    </div>
</div>

And this my user_statistic.php:

... (mysql stuff)
    /******************************/
    /** Create diagram
    /******************************/
    $labels = array();
    $data   = array();

    for ($j=1; $j<=$days; $j++) {
        $labels[$j] =$j;
        $data[$j] = $day_value[$j];
    }

    // Aggregate all the data into one string
    $data_string = "[" . join(", ", $data) . "]";
    $labels_string = "['" . join("', '", $labels) . "']";
    $labels_tooltip =  "['" . join("', '", $data) . "']";

    //data written 
    echo "1";

So echo "1"; tells my script that everything is fine. But now I need $data_string, $labels_string and $labels_tooltip. So how can I return these values from user_statistic.php into my side?

Answer 1

The best approach here would be to return a json object. Create an array on server side -

$response['error_code'] = '1'; //everything ok. 0 if not ok
$response['data_string'] = 'this will have some data';
$response['labels_string'] = 'labels';
$response['labels_tooltip' = 'here goes the tooltips';
echo json_encode($response);

and in your javascript code, mention the return datatype as json -

$.ajax({
    url: "include/scripts/user_statistic.php",
    type: "POST",
    data: data,
    dataType: json,
    success: function (reqCode) {
        if (reqCode.error_code == 1) {
            alert('this is the data string '+resCode.data_string);
            //Show generated table
            $('.done').fadeOut('slow'); 
            $('.done').fadeIn('slow');
        }
        if (reqCode.error_code == 2) {
            //No values found
            $('.done').fadeOut('slow');
            $("#dialog_error").dialog( "open" );
        }
    }
});

Answer 2

Avoid converting arrays to strings on your own. If you need to pass a PHP array back to your jQuery, you should do so with the json_encode function:

echo json_encode( $array );

This will come through as a JSON object which you can then handle client-side. Your JSON string will be returned into the callback of your $.ajax method:

$.ajax({
  url: "include/scripts/user_statistic.php",
  type: "POST",
  data: data,
  dataType: 'json',
  success: function ( response ) {
     /* response is your array, in JSON form */
  }
});

For instance, if our PHP script did the following:

$response = array(
  'message' => 'Success',
  'allData' => array( 'Jonathan', 'Mariah', 'Samuel', 'Sally' )
);

echo json_encode( $response );

We could alert the message from our jQuery like this:

success: function ( response ) {
   alert( response.message );
}