How to use the data passed with ajax in the php script?

Question

I am not sure how I should use the data passed along with an ajax call in the PHP script. Specifically, here is the code for the ajax call:

if (node_selected!=0 & node_selected!=null){
                $.ajax({
                    type: "POST",
                    url: "php/fetch_sensors.php",
                    data: node_selected,
                    dataType:'json',
                    success: function( options ){
                        ...
                    }
                });
            }

'node_selected' is an integer variable.

And here is the PHP script (the fetch_sensors.php):

$query = "SELECT SensorID,Variable FROM sensors WHERE SensorID IN (SELECT SensorID FROM nodesensors WHERE NodeID=node_selected)";
$result = mysql_query($query, $con) or die('query not made');

while ($row = mysql_fetch_assoc($result)) {
    $sensors[] = $row;
}

echo json_encode($sensors);

If I replace 'NodeID=node_selected' with 'NodeID=2' (2 is just an example) everything works fine. So, I figure that I don't use correctly the 'node_selected'. Any ideas?

Thanks a lot!

add in top of your php an echo '<pre>';print_r($_POST);echo '</pre>'; to see what we have . — Mimouni
– Mimouni, Commented Oct 22, 2014 at 14:17

Hackerman · Accepted Answer · 2014-10-22 14:27:27Z

3

Jquery:

 if (node_selected!=0 && node_selected!=null){
            $.ajax({
                type: "POST",
                url: "php/fetch_sensors.php",
                data: {myNode: node_selected}, //like this
                dataType:'json',
                success: function( options ){
                    ...
                }
            });
  }

PHP:

$myNode = mysql_escape_string($_POST["myNode"]);
$query = "SELECT SensorID,Variable FROM sensors WHERE SensorID IN ";
$query.= "(SELECT SensorID FROM nodesensors WHERE NodeID= '".$myNode."')";
$result = mysql_query($query, $con) or die('query not made');

while ($row = mysql_fetch_assoc($result)) {
  $sensors[] = $row;
}

echo json_encode($sensors);

edited Oct 22, 2014 at 14:27

answered Oct 22, 2014 at 14:21

Hackerman

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4 Comments

jeroen Over a year ago

+1 Although you should either cast to int if you don't quote the value in the query or use a string escape function and quote it. Now your query can break.

Hackerman Over a year ago

You are right @jeroen, i post my answer under the assumtion that SensorID is an int field, but it could be alphanumeric...editing..

DimitrisChousiadas Over a year ago

Yep, it works, thanks a lot. But why do I have to pass the data like that?? Why "data: node_selected" isn't enough??

Hackerman Over a year ago

According to jquery documentation:

data Type: PlainObject or String or Array Data to be sent to the server. It is converted to a query string, if not already a string. It's appended to the url for GET-requests. See processData option to prevent this automatic processing. Object must be Key/Value pairs. If value is an Array, jQuery serializes multiple values with same key based on the value of the traditional setting (described below).

...source: api.jquery.com/jQuery.ajax

charlietfl · Accepted Answer · 2014-10-22 14:21:36Z

3

You have to send a key/value pair. You are only sending a value.

If you want to receive $_POST['NodeID'] in php then the easiest way in jQuery AJAX is to create an object using the same key

$.ajax({
    type: "POST",
    url: "php/fetch_sensors.php",
    data: {
        NodeID: node_selected
    },
    dataType: 'json',
    success: function (options) {...
    }
});

The easiest way to think of it is the object keys are the same as using name in a form control. The processing of most ajax at server is identical to processing of forms

answered Oct 22, 2014 at 14:21

charlietfl

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Comments

Steve · Accepted Answer · 2014-10-22 14:20:46Z

1

Give your data a key, so you can fetch is in php via $_POST global:

//js
data: {node_selected: node_selected},

//php
$query = "SELECT SensorID,Variable FROM sensors WHERE SensorID IN (SELECT SensorID FROM nodesensors WHERE NodeID=" . $_POST['node_selected'] . ")";

answered Oct 22, 2014 at 14:20

Steve

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